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GCSE Maths · EdexcelChapter 14 · Quadratics and simultaneous equations
Chapter 14 · Foundation tier

GCSE Maths Quadratic and Simultaneous Equations (Foundation): worked examples and questions with answers

By the end of this chapter you will be able to

  • Factorise quadratic expressions, including the difference of two squares (and ax2 + bx + c at Higher)
  • Solve quadratic equations by factorising, using the null factor law, after rearranging if necessary
  • Form and solve quadratic equations from areas, consecutive numbers and other problems
  • Solve simultaneous linear equations by elimination and by substitution
  • Solve simultaneous equations from graphs, and form them from words

You should already know

  • Expanding single and double brackets (Chapter 7)
  • Solving linear equations (Chapter 9)
  • Plotting straight-line and quadratic graphs (Chapters 12 and 13)
  • Square roots and surds (Chapter 6)

Where this leads

  • Quadratic inequalities (Chapter 10)
  • Equations of circles and tangents (Chapter 25)
  • Algebraic proof and rates of change in higher mathematics
  • Pythagoras and area problems (Chapters 18 and 20)

Edexcel GCSE (9–1) Mathematics 1MA1, Foundation tier. Content references: A4, A18, A19, A21.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

Exercise P  Do you remember?

Non-calculator
1
FExpand and simplify (x + 3)(x + 5).
2
FSolve 3x − 7 = 11.
3
FFactorise 6x + 15.
4
FWork out the value of x2 − 3x when x = −2.
5
FIs the point (2, 5) on the line y = 2x + 1? Show how you decide.
14.1

Factorising quadratic expressions

Edexcel 1MA1: A4
Grade 3-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

In Chapter 7 you expanded brackets: (x + 2)(x + 3) = x2 + 5x + 6. Factorising is the same journey backwards. You start with x2 + 5x + 6 and look for the brackets that multiply to make it. You need this skill to solve quadratic equations, so it is worth becoming really fluent at it before you move on.

A quadratic expression is an expression whose highest power is 2, such as x2 + 5x + 6. A factor is something that is multiplied: (x + 2) and (x + 3) are factors of x2 + 5x + 6.

x²3x2x6x3x2Area = x² + 5x + 6
A rectangle with sides (x + 2) and (x + 3) is cut into four pieces. The two middle pieces, 3x and 2x, join to make 5x.

Look at the picture. The whole rectangle has area x2 + 3x + 2x + 6 = x2 + 5x + 6, and its sides are (x + 2) and (x + 3). Now look at the numbers: 2 × 3 = 6, which is the last number, and 2 + 3 = 5, which is the middle number (the number in front of x). This happens every time, and it gives us a method.

To factorise x2 + bx + c: find two numbers that multiply to make c and add to make b. Call them p and q. Then the answer is (x + p)(x + q). The signs follow a pattern. If c is positive, both numbers have the same sign as b. If c is negative, one number is positive and one is negative, and the bigger one has the sign of b.

Always look for a common factor first. In 3x2 − 12x every term divides by 3x, so it factorises as 3x(x − 4). Check by expanding.

Difference of two squares. An expression with two terms, both perfect squares, with a minus between them, factorises as a2 − b2 = (a − b)(a + b). Why? Expand (a − b)(a + b) and the two middle terms, +ab and −ab, cancel out. So x2 − 49 = (x − 7)(x + 7). There is no middle term because there is no x term.

Key ideas

  • Factorising is the reverse of expanding. Always check by expanding your answer.
  • x2 + bx + c = (x + p)(x + q) where p × q = c and p + q = b.
  • Take out a common factor first: 3x2 − 12x = 3x(x − 4).
  • Difference of two squares: a2 − b2 = (a − b)(a + b).
  • Higher: for ax2 + bx + c, find two numbers that multiply to ac and add to b, split the middle term, and factorise in pairs.
Worked example 14.1.1
Factorise x2 + 7x + 12.
▶ Watch this solution as a video

Key words. To factorise is the opposite of expanding: we turn a sum like x² + 5x + 6 into a product like (x + 2)(x + 3).

Method. Factorising means writing an expression as a product (brackets multiplied together).

  1. Look at the expression and decide which method to use. Check in this order: (1) is there a common factor? (2) is it a difference of two squares (two square terms with a minus between)? (3) is it a quadratic with three terms?
    x2 + 7x + 12
  2. Factorise. Follow the method for the pattern you found.
    Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 7 and c = 12.
    We need two numbers that multiply to 12 (the last number) and add to 7 (the middle number). List the pairs that multiply to 12:
    1 × 12 = 12,   add: 1 + 12 = 13 ✗
    -1 × -12 = 12,   add: -1 + -12 = -13 ✗
    2 × 6 = 12,   add: 2 + 6 = 8 ✗
    -2 × -6 = 12,   add: -2 + -6 = -8 ✗
    3 × 4 = 12,   add: 3 + 4 = 7 ✓ this pair works
    Put the pair into two brackets: (x + 3)(x + 4).
    Check by expanding: x² + 4x + 3x + 12 = x2 + 7x + 12. ✓
    x2 + 7x + 12 = (x + 3)(x + 4)
  3. Check by expanding. Factorising is the reverse of expanding, so multiplying the brackets out must give the original expression back.
    First: x × x = x2
    Outer: x × 4 = 4x
    Inner: 3 × x = 3x
    Last: 3 × 4 = 12
    (x + 3)(x + 4) = x2 + 7x + 12  ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong.

Worked example 14.1.2
Factorise x2 − 2x − 15.
▶ Watch this solution as a video

Key words. To factorise is the opposite of expanding: we turn a sum like x² + 5x + 6 into a product like (x + 2)(x + 3).

Method. Factorising means writing an expression as a product (brackets multiplied together).

  1. Look at the expression and decide which method to use. Check in this order: (1) is there a common factor? (2) is it a difference of two squares (two square terms with a minus between)? (3) is it a quadratic with three terms?
    x2 − 2x − 15
  2. Factorise. Follow the method for the pattern you found.
    Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -2 and c = -15.
    We need two numbers that multiply to -15 (the last number) and add to -2 (the middle number). List the pairs that multiply to -15:
    1 × -15 = -15,   add: 1 + -15 = -14 ✗
    -1 × 15 = -15,   add: -1 + 15 = 14 ✗
    3 × -5 = -15,   add: 3 + -5 = -2 ✓ this pair works
    Put the pair into two brackets: (x + 3)(x − 5).
    Check by expanding: x² − 5x + 3x − 15 = x2 − 2x − 15. ✓
    x2 − 2x − 15 = (x + 3)(x − 5)
  3. Check by expanding. Factorising is the reverse of expanding, so multiplying the brackets out must give the original expression back.
    First: x × x = x2
    Outer: x × 3 = 3x
    Inner: (−5) × x = −5x
    Last: (−5) × 3 = −15
    (x + 3)(x − 5) = x2 − 2x − 15  ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong.

Worked example 14.1.3
Factorise x2 − 49.
▶ Watch this solution as a video

Key words. To factorise is the opposite of expanding: we turn a sum like x² + 5x + 6 into a product like (x + 2)(x + 3).

Method. Factorising means writing an expression as a product (brackets multiplied together).

  1. Look at the expression and decide which method to use. Check in this order: (1) is there a common factor? (2) is it a difference of two squares (two square terms with a minus between)? (3) is it a quadratic with three terms?
    x2 − 49
  2. Factorise. Follow the method for the pattern you found.
    Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
    The rule is: a² − b² = (a − b)(a + b).
    Find a and b. x2 = (x)², so a = x. And 49 = 7², so b = 7.
    Put them into the rule: (x − 7)(x + 7).
    Check by expanding: (x − 7)(x + 7) = x2 + 7x − 7x − 49 = x2 − 49. ✓ (the middle terms cancel)
    x2 − 49 = (x − 7)(x + 7)
  3. Check by expanding. Factorising is the reverse of expanding, so multiplying the brackets out must give the original expression back.
    First: x × x = x2
    Outer: x × 7 = 7x
    Inner: (−7) × x = −7x
    Last: (−7) × 7 = −49
    (x − 7)(x + 7) = x2 − 49  ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong.

Worked example 14.1.4
Factorise fully 5x2 − 20.
▶ Watch this solution as a video

Key words. To factorise is the opposite of expanding: we turn a sum like x² + 5x + 6 into a product like (x + 2)(x + 3).

Method. Factorising means writing an expression as a product (brackets multiplied together).

  1. Look at the expression and decide which method to use. Check in this order: (1) is there a common factor? (2) is it a difference of two squares (two square terms with a minus between)? (3) is it a quadratic with three terms?
    5x2 − 20
  2. Factorise. Follow the method for the pattern you found.
    Look for a common factor. A common factor is something that divides into every term.
    The terms are 5x2, −20. The biggest number and the highest power of x that divide into all of them is 5.
    Take it outside a bracket, and inside the bracket put each term divided by it:
    5x2 ÷ 5 = x2
    −20 ÷ 5 = −4
    So 5x2 − 20 = 5(x2 − 4).
    Check by expanding: 5 × (x2 − 4) = 5 × x2, 5 × (−4) → 5x2 − 20. ✓
    Can the bracket be factorised again? Look at (x2 − 4).
    Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
    The rule is: a² − b² = (a − b)(a + b).
    Find a and b. x2 = (x)², so a = x. And 4 = 2², so b = 2.
    Put them into the rule: (x − 2)(x + 2).
    Check by expanding: (x − 2)(x + 2) = x2 + 2x − 2x − 4 = x2 − 4. ✓ (the middle terms cancel)
    5x2 − 20 = 5(x − 2)(x + 2)
  3. Check by expanding. Factorising is the reverse of expanding, so multiplying the brackets out must give the original expression back.
    First: 5x × x = 5x2
    Outer: 5x × 2 = 10x
    Inner: (−10) × x = −10x
    Last: (−10) × 2 = −20
    5(x − 2)(x + 2) = 5x2 − 20  ✓

Watch out. Always check by expanding. If you do not get the original expression back, the factorising is wrong.

Watch out

The expression x2 + 9 cannot be factorised, because there is a plus between the squares. Only a minus gives a difference of two squares. Also, do not forget to take out a common factor first: 2x2 − 50 is 2(x − 5)(x + 5), not just 2(x2 − 25).

Exercise 14A  Factorising quadratic expressions

Non-calculator
1
FFactorise by taking out the highest common factor.
(a)6x + 9
(b)10x − 15
(c)x2 + 5x
(d)4x2 − 12x
(e)9x2 + 6x
(f)8x2 − 8x
2
FFactorise.
(a)x2 + 5x + 6
(b)x2 + 7x + 10
(c)x2 + 8x + 15
(d)x2 + 9x + 20
(e)x2 + 6x + 8
(f)x2 + 10x + 24
3
FWrite the missing number in each bracket.
(a)(x + 3)(x +  ? ) = x2 + 7x + 12
(b)(x − 2)(x −  ? ) = x2 − 7x + 10
(c)(x + 6)(x −  ? ) = x2 + 3x − 18
(d)(x − 4)(x +  ? ) = x2 + x − 20
4
FFactorise.
(a)x2 − 6x + 8
(b)x2 − 7x + 10
(c)x2 − 8x + 15
(d)x2 − 9x + 14
(e)x2 − 10x + 21
(f)x2 − 5x + 4
5
FFactorise.
(a)x2 + 2x − 8
(b)x2 + x − 12
(c)x2 − 3x − 10
(d)x2 − 2x − 15
(e)x2 + 4x − 21
(f)x2 − x − 20
6
FFactorise these differences of two squares.
(a)x2 − 9
(b)x2 − 16
(c)x2 − 100
(d)x2 − 1
(e)4x2 − 25
(f)9x2 − 4
7
RZara says that x2 + 8x + 12 = (x + 3)(x + 4). Explain why she is wrong and write the correct factorisation.
8
PSThree of these four expressions can be factorised into two brackets. Which one cannot?
x2 + 6x + 9,   x2 − 25,   x2 + 4x + 5,   x2 − 5x + 6
9
PSThe area of this rectangle is (x2 + 9x + 20) cm2. The width is (x + 4) cm. Find an expression for the length.
length = ?(x + 4)Area = x² + 9x + 20
10
FFactorise.
(a)x2 + 12x + 35
(b)x2 − 4x
(c)7x − 21
(d)x2 − 144
(e)x2 − 11x + 28
(f)x2 + 3x − 28
11
RUse the difference of two squares to work out, without a calculator.
(a)522 − 482
(b)1012 − 992
12
RThe expression x2 + bx + 12 can be factorised into two brackets with whole numbers in them for certain values of b. How many different values of b (positive or negative) are possible? List them.
14.2

Solving quadratic equations by factorising

Edexcel 1MA1: A18
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

A quadratic equation is an equation that contains an x2 term (and no higher power), for example x2 + 5x + 6 = 0. To solve it means to find the values of x that make it true. A linear equation such as 2x + 1 = 9 has one solution. A quadratic equation usually has two solutions. You can see why from its graph (Chapter 13): the curve crosses the x-axis in two places, and each crossing is a solution of the equation “expression = 0”.

xy-4-3-2-11234567-14-12-10-8-6-4-2246810
The curve y = x2 − 3x − 10 crosses the x-axis where x = −2 and x = 5. These are the two solutions of x2 − 3x − 10 = 0.

So how do we find them without drawing a graph? The key is an idea called the null factor law. Think about multiplying two numbers together. If the answer is 0, one of the numbers must have been 0, because 7 × 3 = 21, 7 × 1 = 7, and only 7 × 0 = 0. This is also true when the two things are brackets. If (x − 5)(x + 2) = 0 then either (x − 5) = 0 or (x + 2) = 0. So x = 5 or x = −2.

Most quadratics are not written as brackets, so first we have to factorise (section 14.1). The method is:

  1. Make sure the equation has 0 on one side (the null factor law only works for a product that equals 0).
  2. Factorise the other side.
  3. Set each bracket equal to 0.
  4. Solve each of the two little equations.
  5. Check both answers in the original equation.

Some special cases are worth knowing. If there is no number term, such as x2 − 5x = 0, take out the common factor x: x(x − 5) = 0, so x = 0 or x = 5. If it is a difference of two squares, such as x2 − 49 = 0, then (x − 7)(x + 7) = 0 gives x = 7 or x = −7, usually written x = ±7. And if both brackets are identical, such as (x − 3)2 = 0, there is only one solution, x = 3. This is called a repeated root; the curve just touches the x-axis.

Key ideas

  • A quadratic equation has an x2 term. It usually has two solutions (also called roots).
  • Null factor law: if A × B = 0 then A = 0 or B = 0.
  • Steps: (1) 0 on one side, (2) factorise, (3) each bracket = 0, (4) solve, (5) check.
  • Never divide by x. For x2 − 5x = 0 factorise to x(x − 5) = 0 so that you do not lose the answer x = 0.
  • The solutions of “expression = 0” are where the graph crosses the x-axis.
Worked example 14.2.1
Solve (x + 3)(x − 5) = 0.
▶ Watch this solution as a video

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

  1. Write down the equation. The quadratic is already written as brackets multiplied together and equal to 0.
    (x + 3)(x − 5) = 0
  2. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    (x + 3)(x − 5) = 0
    So (x − 5) = 0 or (x + 3) = 0.
  3. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    x − 5 = 0
    Add 5 to both sides: x = 5

    x + 3 = 0
    Subtract 3 from both sides: x = −3
  4. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = −3 or x = 5

Check. When x = −3: ((−3) − 5)((−3) + 3) = (−8) × 0 = 0. ✓
When x = 5: (5 − 5)(5 + 3) = 0 × 8 = 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.2.2
Solve x2 + 7x + 12 = 0.
▶ Watch this solution as a video

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

  1. Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
    x2 + 7x + 12 = 0
  2. Check that one side is 0. We can only use the null factor law when one side of the equation is exactly 0.
    The right-hand side is already 0, so there is nothing to move.
  3. Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
    Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 7 and c = 12.
    We need two numbers that multiply to 12 (the last number) and add to 7 (the middle number). List the pairs that multiply to 12:
    1 × 12 = 12,   add: 1 + 12 = 13 ✗
    -1 × -12 = 12,   add: -1 + -12 = -13 ✗
    2 × 6 = 12,   add: 2 + 6 = 8 ✗
    -2 × -6 = 12,   add: -2 + -6 = -8 ✗
    3 × 4 = 12,   add: 3 + 4 = 7 ✓ this pair works
    Put the pair into two brackets: (x + 3)(x + 4).
    Check by expanding: x² + 4x + 3x + 12 = x2 + 7x + 12. ✓
    x2 + 7x + 12 = (x + 3)(x + 4)
  4. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    (x + 3)(x + 4) = 0
    So (x + 3) = 0 or (x + 4) = 0.
  5. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    x + 3 = 0
    Subtract 3 from both sides: x = −3

    x + 4 = 0
    Subtract 4 from both sides: x = −4
  6. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = −4 or x = −3

Check. When x = −4: left side = (−4)2 + 7 × (−4) + 12 = 16 − 28 + 12 = 0. It equals 0. ✓
When x = −3: left side = (−3)2 + 7 × (−3) + 12 = 9 − 21 + 12 = 0. It equals 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.2.3
Solve x2 − 5x = 0.
▶ Watch this solution as a video

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

  1. Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
    x2 − 5x = 0
  2. Check that one side is 0. We can only use the null factor law when one side of the equation is exactly 0.
    The right-hand side is already 0, so there is nothing to move.
  3. Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
    Look for a common factor. A common factor is something that divides into every term.
    The terms are x2, −5x. The biggest number and the highest power of x that divide into all of them is x.
    Take it outside a bracket, and inside the bracket put each term divided by it:
    x2 ÷ x = x
    −5x ÷ x = −5
    So x2 − 5x = x(x − 5).
    Check by expanding: x × (x − 5) = x × x, x × (−5) → x2 − 5x. ✓
    x2 − 5x = x(x − 5)
  4. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    x(x − 5) = 0
    So (x − 5) = 0 or x = 0.
  5. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    x − 5 = 0
    Add 5 to both sides: x = 5

    x = 0
    x = 0 is a solution.
  6. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = 0 or x = 5

Check. When x = 0: left side = 02 − 5 × 0 = 0 + 0 = 0. It equals 0. ✓
When x = 5: left side = 52 − 5 × 5 = 25 − 25 = 0. It equals 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.2.4
Solve x2 − 81 = 0.
▶ Watch this solution as a video

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

  1. Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
    x2 − 81 = 0
  2. Check that one side is 0. We can only use the null factor law when one side of the equation is exactly 0.
    The right-hand side is already 0, so there is nothing to move.
  3. Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
    Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
    The rule is: a² − b² = (a − b)(a + b).
    Find a and b. x2 = (x)², so a = x. And 81 = 9², so b = 9.
    Put them into the rule: (x − 9)(x + 9).
    Check by expanding: (x − 9)(x + 9) = x2 + 9x − 9x − 81 = x2 − 81. ✓ (the middle terms cancel)
    x2 − 81 = (x − 9)(x + 9)
  4. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    (x − 9)(x + 9) = 0
    So (x − 9) = 0 or (x + 9) = 0.
  5. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    x − 9 = 0
    Add 9 to both sides: x = 9

    x + 9 = 0
    Subtract 9 from both sides: x = −9
  6. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = −9 or x = 9

Check. When x = −9: left side = (−9)2 − 81 = 81 − 81 = 0. It equals 0. ✓
When x = 9: left side = 92 − 81 = 81 − 81 = 0. It equals 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.2.5
Solve x2 − 6x + 9 = 0.
▶ Watch this solution as a video

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

  1. Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
    x2 − 6x + 9 = 0
  2. Check that one side is 0. We can only use the null factor law when one side of the equation is exactly 0.
    The right-hand side is already 0, so there is nothing to move.
  3. Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
    Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -6 and c = 9.
    We need two numbers that multiply to 9 (the last number) and add to -6 (the middle number). List the pairs that multiply to 9:
    1 × 9 = 9,   add: 1 + 9 = 10 ✗
    -1 × -9 = 9,   add: -1 + -9 = -10 ✗
    3 × 3 = 9,   add: 3 + 3 = 6 ✗
    -3 × -3 = 9,   add: -3 + -3 = -6 ✓ this pair works
    Put the pair into two brackets: (x − 3)(x − 3).
    Check by expanding: x² − 3x − 3x + 9 = x2 − 6x + 9. ✓
    x2 − 6x + 9 = (x − 3)2
  4. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    (x − 3)2 = 0
    So (x − 3) = 0.
    A bracket that appears twice gives the same answer twice, so there is only one solution from it (a repeated root).
  5. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    x − 3 = 0
    Add 3 to both sides: x = 3
  6. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = 3

Check. When x = 3: left side = 32 − 6 × 3 + 9 = 9 − 18 + 9 = 0. It equals 0. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Watch out

When you solve (x − 5) = 0 the answer is x = +5; for (x + 2) = 0 it is x = −2. The sign in the solution is the opposite of the sign in the bracket. And never set a bracket equal to a number other than 0: if (x − 1)(x + 4) = 10, you cannot say x − 1 = 10. You must rearrange to equal 0 first (see the next section).

Exercise 14B  Solving quadratic equations by factorising

Non-calculator
1
FSolve. These equations are already factorised.
(a)(x + 2)(x + 5) = 0
(b)(x − 3)(x + 7) = 0
(c)(x − 4)(x − 1) = 0
(d)x(x + 6) = 0
(e)(x + 8)(x − 8) = 0
(f)(x − 5)2 = 0
2
FSolve by factorising.
(a)x2 + 6x + 5 = 0
(b)x2 + 7x + 12 = 0
(c)x2 + 8x + 15 = 0
(d)x2 + 9x + 14 = 0
(e)x2 + 11x + 10 = 0
(f)x2 + 10x + 21 = 0
3
FSolve by factorising.
(a)x2 − 7x + 10 = 0
(b)x2 − 6x + 8 = 0
(c)x2 − 8x + 12 = 0
(d)x2 − 9x + 18 = 0
(e)x2 − 10x + 16 = 0
(f)x2 − 11x + 24 = 0
4
FSolve by factorising.
(a)x2 + 2x − 8 = 0
(b)x2 − x − 6 = 0
(c)x2 + 3x − 10 = 0
(d)x2 − 4x − 12 = 0
(e)x2 + x − 20 = 0
(f)x2 − 2x − 24 = 0
5
FSolve. Take out the common factor first.
(a)x2 + 3x = 0
(b)x2 − 7x = 0
(c)x2 + 10x = 0
(d)2x2 − 8x = 0
(e)3x2 + 9x = 0
(f)x2 − x = 0
6
FSolve these differences of two squares.
(a)x2 − 4 = 0
(b)x2 − 25 = 0
(c)x2 − 64 = 0
(d)x2 − 121 = 0
(e)x2 − 1 = 0
(f)x2 − 400 = 0
7
RJo solves (x − 3)(x + 2) = 0 and writes “x = 3 or x = 2”. Explain her mistake and write down the correct solutions.
8
RHow many solutions does each equation have? You do not need to find them.
(a)(x − 1)(x + 4) = 0
(b)(x − 6)2 = 0
(c)x(x − 2) = 0
(d)x2 + 4 = 0
9
PSThe graph of y = x2 − 2x − 3 is shown. Use the graph to write down the solutions of x2 − 2x − 3 = 0. Then check your answer by factorising.
xy-3-2-112345-5-4-3-2-1123456
10
PSFind the x-coordinates of the points where the curve y = x2 + 4x − 5 crosses the x-axis.
11
FSolve.
(a)x2 + 9x + 18 = 0
(b)x2 − 3x = 0
(c)x2 − 49 = 0
(d)x2 − 8x + 7 = 0
(e)x2 + 5x − 14 = 0
(f)x2 − 6x = 0
12
RWrite down a quadratic equation, in the form x2 + bx + c = 0, whose solutions are x = 3 and x = −7.
13
RThe equation x2 + kx + 16 = 0 has exactly one solution. Find the two possible values of k.
14.3

Rearranging first, and harder quadratics

Edexcel 1MA1: A18
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Rearranging first, and harder quadraticsComing soon

The null factor law only works when one side is exactly 0. Many quadratic equations do not start in that form, for example x2 + 3x = 10. If you just factorised the left side you would get x(x + 3) = 10, and there is nothing special about a product equal to 10. So the first job is to rearrange the equation to make one side 0.

Use the same balance method as for linear equations (Chapter 9): do the same thing to both sides. To get x2 + 3x = 10 into the right form, subtract 10 from both sides: x2 + 3x − 10 = 0. Now factorise and carry on as before. A good habit is to move everything to the side where the x2 term is positive, so that the number in front of x2 is positive and the factorising is easier.

If the equation has brackets, expand them first, then collect like terms, then rearrange. For example (x + 2)(x + 5) = 28 becomes x2 + 7x + 10 = 28, and then x2 + 7x − 18 = 0.

Why is it wrong to say “x + 2 = 28 or x + 5 = 28”? Try it: if x + 2 = 28 then x = 26, and then x + 5 = 31, so the product is 28 × 31, which is nowhere near 28. Only a product equal to zero forces one of the factors to be zero.

Key ideas

  • The null factor law needs 0 on one side. Rearrange first: do the same to both sides.
  • Expand brackets and collect like terms before you rearrange.
  • Move everything to the side where x2 is positive, then factorise, set each bracket to 0 and solve.
  • Never set the brackets equal to a number other than 0.
  • Higher: for ax2 + bx + c = 0, split the middle term. The solutions can be fractions.
Worked example 14.3.1
Solve x2 + 3x = 10.
▶ Watch this solution as a video

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

  1. Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
    x2 + 3x = 10
  2. Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
    Subtract 10 from both sides: x2 + 3x − 10 = 0
  3. Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
    Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 3 and c = -10.
    We need two numbers that multiply to -10 (the last number) and add to 3 (the middle number). List the pairs that multiply to -10:
    1 × -10 = -10,   add: 1 + -10 = -9 ✗
    -1 × 10 = -10,   add: -1 + 10 = 9 ✗
    2 × -5 = -10,   add: 2 + -5 = -3 ✗
    -2 × 5 = -10,   add: -2 + 5 = 3 ✓ this pair works
    Put the pair into two brackets: (x − 2)(x + 5).
    Check by expanding: x² + 5x − 2x − 10 = x2 + 3x − 10. ✓
    x2 + 3x − 10 = (x − 2)(x + 5)
  4. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    (x − 2)(x + 5) = 0
    So (x − 2) = 0 or (x + 5) = 0.
  5. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    x − 2 = 0
    Add 2 to both sides: x = 2

    x + 5 = 0
    Subtract 5 from both sides: x = −5
  6. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = −5 or x = 2

Check. When x = −5: left side = (−5)2 + 3 × (−5) = 25 − 15 = 10; right side = 10 = 10. Both sides equal. ✓
When x = 2: left side = 22 + 3 × 2 = 4 + 6 = 10; right side = 10 = 10. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.3.2
Solve x2 = 2x + 15.

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

  1. Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
    x2 = 2x + 15
  2. Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
    Subtract 2x from both sides: x2 − 2x = 15
    Subtract 15 from both sides: x2 − 2x − 15 = 0
  3. Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
    Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -2 and c = -15.
    We need two numbers that multiply to -15 (the last number) and add to -2 (the middle number). List the pairs that multiply to -15:
    1 × -15 = -15,   add: 1 + -15 = -14 ✗
    -1 × 15 = -15,   add: -1 + 15 = 14 ✗
    3 × -5 = -15,   add: 3 + -5 = -2 ✓ this pair works
    Put the pair into two brackets: (x + 3)(x − 5).
    Check by expanding: x² − 5x + 3x − 15 = x2 − 2x − 15. ✓
    x2 − 2x − 15 = (x + 3)(x − 5)
  4. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    (x + 3)(x − 5) = 0
    So (x − 5) = 0 or (x + 3) = 0.
  5. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    x − 5 = 0
    Add 5 to both sides: x = 5

    x + 3 = 0
    Subtract 3 from both sides: x = −3
  6. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = −3 or x = 5

Check. When x = −3: left side = (−3)2 = 9 = 9; right side = 2 × (−3) + 15 = 9. Both sides equal. ✓
When x = 5: left side = 52 = 25 = 25; right side = 2 × 5 + 15 = 25. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.3.3
Solve (x + 2)(x + 5) = 28.

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

  1. Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
    (x + 2)(x + 5) = 28
  2. Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
    Expand (x + 2)(x + 5):
    First: x × x = x2
    Outer: x × 5 = 5x
    Inner: 2 × x = 2x
    Last: 2 × 5 = 10
    so (x + 2)(x + 5) = x2 + 7x + 10
    The equation is now: x2 + 7x + 10 = 28
  3. Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
    Subtract 28 from both sides: x2 + 7x + 10 − 28 = 0
    Collect like terms: x2 + 7x − 18 = 0
  4. Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
    Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 7 and c = -18.
    We need two numbers that multiply to -18 (the last number) and add to 7 (the middle number). List the pairs that multiply to -18:
    1 × -18 = -18,   add: 1 + -18 = -17 ✗
    -1 × 18 = -18,   add: -1 + 18 = 17 ✗
    2 × -9 = -18,   add: 2 + -9 = -7 ✗
    -2 × 9 = -18,   add: -2 + 9 = 7 ✓ this pair works
    Put the pair into two brackets: (x − 2)(x + 9).
    Check by expanding: x² + 9x − 2x − 18 = x2 + 7x − 18. ✓
    x2 + 7x − 18 = (x − 2)(x + 9)
  5. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    (x − 2)(x + 9) = 0
    So (x − 2) = 0 or (x + 9) = 0.
  6. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    x − 2 = 0
    Add 2 to both sides: x = 2

    x + 9 = 0
    Subtract 9 from both sides: x = −9
  7. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = −9 or x = 2

Check. When x = −9: left side = (−9)2 + 7 × (−9) + 10 = 81 − 63 + 10 = 28; right side = 28 = 28. Both sides equal. ✓
When x = 2: left side = 22 + 7 × 2 + 10 = 4 + 14 + 10 = 28; right side = 28 = 28. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Worked example 14.3.4
Solve x(x + 4) = 21.

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

  1. Write down the equation. A quadratic equation contains a squared letter. We want every value of the letter that makes the two sides equal.
    x(x + 4) = 21
  2. Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
    Expand x(x + 4):
    x × x = x2
    x × 4 = 4x
    so x(x + 4) = x2 + 4x
    The equation is now: x2 + 4x = 21
  3. Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
    Subtract 21 from both sides: x2 + 4x − 21 = 0
  4. Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
    Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 4 and c = -21.
    We need two numbers that multiply to -21 (the last number) and add to 4 (the middle number). List the pairs that multiply to -21:
    1 × -21 = -21,   add: 1 + -21 = -20 ✗
    -1 × 21 = -21,   add: -1 + 21 = 20 ✗
    3 × -7 = -21,   add: 3 + -7 = -4 ✗
    -3 × 7 = -21,   add: -3 + 7 = 4 ✓ this pair works
    Put the pair into two brackets: (x − 3)(x + 7).
    Check by expanding: x² + 7x − 3x − 21 = x2 + 4x − 21. ✓
    x2 + 4x − 21 = (x − 3)(x + 7)
  5. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    (x − 3)(x + 7) = 0
    So (x − 3) = 0 or (x + 7) = 0.
  6. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    x − 3 = 0
    Add 3 to both sides: x = 3

    x + 7 = 0
    Subtract 7 from both sides: x = −7
  7. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = −7 or x = 3

Check. When x = −7: left side = (−7)2 + 4 × (−7) = 49 − 28 = 21; right side = 21 = 21. Both sides equal. ✓
When x = 3: left side = 32 + 4 × 3 = 9 + 12 = 21; right side = 21 = 21. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Watch out

Do not divide both sides by x to “simplify”. If x2 = 9x you must not divide by x, because that loses the solution x = 0. Move everything to one side: x2 − 9x = 0, then factorise x(x − 9) = 0, which gives x = 0 or x = 9. Also take care with signs when you move terms across the equals sign.

Exercise 14C  Rearranging first, and harder quadratics

Non-calculator
1
FRearrange each equation so that it equals 0. Do not solve it.
(a)x2 + 5x = 6
(b)x2 = 4x + 12
(c)x2 − 3x = 2x − 4
(d)x2 + 7 = 8x
(e)3x = −x2 + 10
2
FSolve by rearranging and factorising.
(a)x2 + 2x = 15
(b)x2 + 5x = 14
(c)x2 − x = 12
(d)x2 = 3x + 10
(e)x2 = 7x − 12
(f)x2 + 6x = −8
3
FSolve. The letters appear on both sides.
(a)x2 + 3x = 2x + 6
(b)x2 − 5 = 4x
(c)x2 + 8 = 6x
(d)x2 − 2x = 3x − 6
(e)x2 + 4x = 2x + 3
(f)x2 − 3 = 2x
4
FExpand the brackets, then solve.
(a)x(x + 3) = 10
(b)x(x − 2) = 24
(c)(x + 1)(x + 2) = 6
(d)(x − 1)(x + 4) = 6
(e)(x + 3)(x + 4) = 30
(f)(x − 2)(x + 1) = 4
5
RAli solves (x + 1)(x + 3) = 15 like this: “x + 1 = 15, so x = 14; or x + 3 = 15, so x = 12.” Explain why this method is wrong and then solve the equation correctly.
6
PSShow that x(x + 5) = 14 can be written as x2 + 5x − 14 = 0, and then solve it.
7
RAmir solves x2 = 9x by dividing both sides by x, and gets “x = 9”. Find all the solutions of the equation and say what Amir missed.
8
PSA number squared, plus six times the number, makes 7. Find the number. (There are two possible answers.)
9
RSam solves (x + 2)(x + 3) = (x + 2)(2x − 1) by dividing both sides by (x + 2). He finds only x = 4. Solve the equation properly and explain what Sam missed.
14.4

Problems leading to quadratic equations

Edexcel 1MA1: A18, A21
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Problems leading to quadratic equationsComing soon

Quadratic equations appear whenever two lengths are multiplied together, for example in areas, or whenever two numbers are multiplied together, for example “consecutive numbers”. Here is a strategy that works for every problem in this section.

  1. Choose a letter for the unknown and write down what it stands for, with units.
  2. Write every other quantity in terms of that letter (a rectangle that is 3 cm longer than it is wide has length x + 3 if the width is x).
  3. Form an equation from the information in the question (for instance “area = 54”).
  4. Expand, rearrange to equal 0, factorise and solve exactly as in the last two sections.
  5. Decide which answers make sense. A length cannot be negative, so a negative solution is rejected. Then answer the question that was asked.
  6. Check in the words of the problem, not just in the equation.

Consecutive numbers follow each other with no gaps: 8, 9, 10. If the first is n the next is n + 1. Consecutive odd numbers (or consecutive even numbers) are 2 apart: n and n + 2. The product of two consecutive numbers is n(n + 1).

(x + 3) cmx cmArea = 54 cm²
A rectangle that is 3 cm longer than it is wide. If the area is 54 cm2, then x(x + 3) = 54.

Key ideas

  • Choose a letter, write the other quantities using it, form an equation, solve, then interpret.
  • Reject answers that cannot happen (a negative length or a negative number of items).
  • Consecutive whole numbers: n, n + 1. Consecutive odd (or even) numbers: n, n + 2.
  • Area of a rectangle = length × width; area of a triangle = ½ × base × height.
  • Check your answer against the words of the question.
Worked example 14.4.1
A rectangle has width x cm and length (x + 3) cm. Its area is 54 cm2. Find the width and the length.

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

  1. Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
    Let the width be x cm. Then the length is (x + 3) cm.
  2. Form an equation from the information. Area of a rectangle = length × width.
    x(x + 3) = 54
  3. Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
    Expand x(x + 3):
    x × x = x2
    x × 3 = 3x
    so x(x + 3) = x2 + 3x
    The equation is now: x2 + 3x = 54
  4. Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
    Subtract 54 from both sides: x2 + 3x − 54 = 0
  5. Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
    Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 3 and c = -54.
    We need two numbers that multiply to -54 (the last number) and add to 3 (the middle number). List the pairs that multiply to -54:
    1 × -54 = -54,   add: 1 + -54 = -53 ✗
    -1 × 54 = -54,   add: -1 + 54 = 53 ✗
    2 × -27 = -54,   add: 2 + -27 = -25 ✗
    -2 × 27 = -54,   add: -2 + 27 = 25 ✗
    3 × -18 = -54,   add: 3 + -18 = -15 ✗
    -3 × 18 = -54,   add: -3 + 18 = 15 ✗
    6 × -9 = -54,   add: 6 + -9 = -3 ✗
    -6 × 9 = -54,   add: -6 + 9 = 3 ✓ this pair works
    Put the pair into two brackets: (x − 6)(x + 9).
    Check by expanding: x² + 9x − 6x − 54 = x2 + 3x − 54. ✓
    x2 + 3x − 54 = (x − 6)(x + 9)
  6. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    (x − 6)(x + 9) = 0
    So (x − 6) = 0 or (x + 9) = 0.
  7. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    x − 6 = 0
    Add 6 to both sides: x = 6

    x + 9 = 0
    Subtract 9 from both sides: x = −9
  8. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = −9 or x = 6
  9. Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −9.
    x = 6 cm
  10. Find the length and answer the question. The length is x + 3.
    Length = 6 + 3 = 9 cm
    Width = 6 cm, length = 9 cm

Check. Check in the words: width 6 cm and length 9 cm differ by 3 cm and the area is 6 × 9 = 54 cm2. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Worked example 14.4.2
The product of two consecutive positive whole numbers is 156. Find the numbers.

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

  1. Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
    Let the smaller number be n. The next whole number is n + 1.
  2. Form an equation from the information. The product of the two numbers means they are multiplied.
    n(n + 1) = 156
  3. Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
    Expand n(n + 1):
    n × n = n2
    n × 1 = n
    so n(n + 1) = n2 + n
    The equation is now: n2 + n = 156
  4. Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
    Subtract 156 from both sides: n2 + n − 156 = 0
  5. Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
    Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 1 and c = -156.
    We need two numbers that multiply to -156 (the last number) and add to 1 (the middle number). List the pairs that multiply to -156:
    1 × -156 = -156,   add: 1 + -156 = -155 ✗
    -1 × 156 = -156,   add: -1 + 156 = 155 ✗
    2 × -78 = -156,   add: 2 + -78 = -76 ✗
    -2 × 78 = -156,   add: -2 + 78 = 76 ✗
    3 × -52 = -156,   add: 3 + -52 = -49 ✗
    -3 × 52 = -156,   add: -3 + 52 = 49 ✗
    4 × -39 = -156,   add: 4 + -39 = -35 ✗
    -4 × 39 = -156,   add: -4 + 39 = 35 ✗
    6 × -26 = -156,   add: 6 + -26 = -20 ✗
    -6 × 26 = -156,   add: -6 + 26 = 20 ✗
    12 × -13 = -156,   add: 12 + -13 = -1 ✗
    -12 × 13 = -156,   add: -12 + 13 = 1 ✓ this pair works
    Put the pair into two brackets: (n − 12)(n + 13).
    Check by expanding: n² + 13n − 12n − 156 = n2 + n − 156. ✓
    n2 + n − 156 = (n − 12)(n + 13)
  6. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    (n − 12)(n + 13) = 0
    So (n − 12) = 0 or (n + 13) = 0.
  7. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    n − 12 = 0
    Add 12 to both sides: n = 12

    n + 13 = 0
    Subtract 13 from both sides: n = −13
  8. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    n = −13 or n = 12
  9. Decide which solution makes sense. The numbers are positive, so reject n = −13.
    n = 12
  10. Find both numbers. The numbers are n and n + 1.
    12 and 12 + 1 = 13
    The numbers are 12 and 13.

Check. Check: 12 × 13 = 156. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Worked example 14.4.3
A triangle has base (x + 4) cm and height x cm. Its area is 30 cm2. Find the height and the base.

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x2 term), like x2 + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

  1. Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
    Let the height of the triangle be x cm. Then the base is (x + 4) cm.
  2. Form an equation from the information. The area of a triangle is ½ × base × height. Here ½ × (x + 4) × x = 30. Multiply both sides by 2 to remove the half, so (x + 4) × x = 60.
    x(x + 4) = 60
  3. Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
    Expand x(x + 4):
    x × x = x2
    x × 4 = 4x
    so x(x + 4) = x2 + 4x
    The equation is now: x2 + 4x = 60
  4. Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
    Subtract 60 from both sides: x2 + 4x − 60 = 0
  5. Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
    Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 4 and c = -60.
    We need two numbers that multiply to -60 (the last number) and add to 4 (the middle number). List the pairs that multiply to -60:
    1 × -60 = -60,   add: 1 + -60 = -59 ✗
    -1 × 60 = -60,   add: -1 + 60 = 59 ✗
    2 × -30 = -60,   add: 2 + -30 = -28 ✗
    -2 × 30 = -60,   add: -2 + 30 = 28 ✗
    3 × -20 = -60,   add: 3 + -20 = -17 ✗
    -3 × 20 = -60,   add: -3 + 20 = 17 ✗
    4 × -15 = -60,   add: 4 + -15 = -11 ✗
    -4 × 15 = -60,   add: -4 + 15 = 11 ✗
    5 × -12 = -60,   add: 5 + -12 = -7 ✗
    -5 × 12 = -60,   add: -5 + 12 = 7 ✗
    6 × -10 = -60,   add: 6 + -10 = -4 ✗
    -6 × 10 = -60,   add: -6 + 10 = 4 ✓ this pair works
    Put the pair into two brackets: (x − 6)(x + 10).
    Check by expanding: x² + 10x − 6x − 60 = x2 + 4x − 60. ✓
    x2 + 4x − 60 = (x − 6)(x + 10)
  6. Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
    (x − 6)(x + 10) = 0
    So (x − 6) = 0 or (x + 10) = 0.
  7. Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
    x − 6 = 0
    Add 6 to both sides: x = 6

    x + 10 = 0
    Subtract 10 from both sides: x = −10
  8. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
    x = −10 or x = 6
  9. Decide which solution makes sense. A height cannot be negative, so reject x = −10.
    x = 6 cm
  10. Find the base and answer. The base is x + 4.
    Base = 10 cm
    Height = 6 cm, base = 10 cm

Check. Area = ½ × 10 × 6 = 30 cm2. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Watch out

Always reject impossible answers, and always answer the question that was asked. If x is the width, the question may want the length or the area, so do not stop at x. Also remember (x + 1)2 is not x2 + 1: it is x2 + 2x + 1.

Exercise 14D  Problems leading to quadratic equations

Non-calculator
1
PSSolve these number problems. In each case form an equation first.
(a)The product of two consecutive positive whole numbers is 72. Find the numbers.
(b)The product of two consecutive positive whole numbers is 110. Find the numbers.
(c)The product of two consecutive positive whole numbers is 210. Find the numbers.
2
PSSolve these rectangle problems.
(a)A rectangle has width x cm and length (x + 2) cm. Its area is 24 cm2. Find the width and the length.
(b)A rectangle has width x cm and length (x + 5) cm. Its area is 36 cm2. Find the width and the length.
(c)A rectangle has width x cm and length (x + 3) cm. Its area is 40 cm2. Find the width and the length.
(d)A rectangle has width x cm and length (x + 1) cm. Its area is 56 cm2. Find the width and the length.
3
PSThe square of a number is 5 more than four times the number. Find the possible values of the number.
4
PSA number added to its square gives 42. Find the two possible numbers.
5
PSA rectangle has width w cm and length (w + 5) cm. Its area is 14 cm2. Find the width and the length.
6
PSA rectangular garden has a perimeter of 34 m and an area of 60 m2. Find the length and the width of the garden.
7
PSSam is x years old. His sister is 3 years younger. The product of their ages is 130. Find their ages.
8
PSA triangle has base (x + 4) cm and height x cm. Its area is 30 cm2. Find the base and the height.
(x + 4) cmx cm
9
PSA photograph measures 12 cm by 8 cm. It is put in a frame with a border of width x cm all round. The area of the whole frame, including the photograph, is 140 cm2. Find x.
photo 12 cm by 8 cmxwhole frame: (12 + 2x) cm by (8 + 2x) cm
10
RExplain why two consecutive whole numbers cannot have a product of 50.
14.5

Simultaneous equations: elimination

Edexcel 1MA1: A19
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Simultaneous equations: eliminationComing soon

One equation with two letters, such as x + y = 7, has lots of solutions: (1, 6), (2, 5), (3, 4), and so on. But if a second equation must be true at the same time, such as x − y = 1, there is usually only one pair of values that works for both: x = 4 and y = 3. Two equations that must both be true at the same time are called simultaneous equations. (“Simultaneous” means “at the same time”.)

2 pens + 1 book£51 pen + 1 book£3
Simultaneous equations from prices: 2p + b = 5 and p + b = 3. The first basket has one more pen, and costs £2 more, so a pen costs £2.

The picture shows the idea of elimination. The two baskets are the same except that the first has one extra pen, and it costs £2 more. So one pen costs £2, and then the book costs £1. We got rid of the book by comparing the baskets: that is elimination. In algebra we do this by adding or subtracting the two equations. This is allowed because if you add equal amounts to equal amounts, the results are equal.

The method.

  1. Number the equations (1) and (2). Make sure each has the form (number)x + (number)y = (number).
  2. Choose a letter to eliminate. Make the numbers in front of it the same size in both equations. If they are not, multiply one or both equations (multiply every term, including the right-hand side).
  3. If the matching terms have the same sign, subtract the equations. If they have different signs, add them. One letter disappears.
  4. Solve the equation that is left.
  5. Substitute the value into one of the original equations to find the other letter.
  6. Check in the other original equation.

A memory aid: Same Signs Subtract. If the equations are not lined up, rearrange them first: 2x = 13 − y becomes 2x + y = 13.

Key ideas

  • Simultaneous equations: two equations, two letters, both true at once. The solution is a pair of values.
  • Eliminate a letter by adding or subtracting the equations. Same signs: subtract. Different signs: add.
  • If no letter has matching numbers, multiply one or both equations first. Multiply every term.
  • Find one letter, substitute to find the other, and always check in the other equation.
  • Line up the equations first: letters on the left, numbers on the right.
Worked example 14.5.1
Solve the simultaneous equations 3x + 2y = 12 and 3x − y = 3.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Simultaneous equations: make the number in front of one letter the same size in both equations, then add or subtract to eliminate (remove) that letter. Solve for the other letter, substitute back, and check.

  1. Number the equations. We will keep referring to them, so call them (1) and (2).
    (1)   3x + 2y = 12
    (2)   3x − y = 3
  2. Choose a letter to eliminate: x. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of x are 3 and 3: they are already the same size, so no multiplying is needed.
  3. Subtract (2) from (1). The x terms are 3x and 3x. They have the same sign, so subtract the equations: 3x − 3x = 0, so x disappears. Do the same thing to the left sides and to the right sides.
    (1)   3x + 2y = 12
    (2)   3x − y = 3
    x terms: 3x − 3x = 0 (they cancel)
    y terms: 2y − (−y) = 3y
    Numbers: 12 − 3 = 9
    3y = 9
  4. Solve for y. 3y means 3 multiplied by y, so divide both sides by 3.
    Divide both sides by 3: y = 9 ÷ 3 = 3
  5. Substitute y = 3 into equation (1) to find x. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
    Equation (1): 3x + 2y = 12
    3x + 2 × 3 = 12
    3x + 6 = 12
    Subtract 6 from both sides: 3x = 6
    Divide both sides by 3: x = 6 ÷ 3 = 2
  6. Write the solution. Give both values. They belong together as a pair.
    x = 2, y = 3

Check. Put x = 2, y = 3 into the other equation (equation (2), which we did not use to find x): 3 × 2 − 3 = 6 − 3 = 3. This equals 3, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Worked example 14.5.2
Solve 2x + 3y = 12 and 4x − 3y = 6.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Simultaneous equations: make the number in front of one letter the same size in both equations, then add or subtract to eliminate (remove) that letter. Solve for the other letter, substitute back, and check.

  1. Number the equations. We will keep referring to them, so call them (1) and (2).
    (1)   2x + 3y = 12
    (2)   4x − 3y = 6
  2. Choose a letter to eliminate: y. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of y are 3 and −3: they are already the same size, so no multiplying is needed.
  3. Add (1) and (2). The y terms are 3y and −3y. They have different signs, so add the equations: 3y + (−3y) = 0, so y disappears. Do the same thing to the left sides and to the right sides.
    (1)   2x + 3y = 12
    (2)   4x − 3y = 6
    y terms: 3y + (−3y) = 0 (they cancel)
    x terms: 2x + 4x = 6x
    Numbers: 12 + 6 = 18
    6x = 18
  4. Solve for x. 6x means 6 multiplied by x, so divide both sides by 6.
    Divide both sides by 6: x = 18 ÷ 6 = 3
  5. Substitute x = 3 into equation (1) to find y. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
    Equation (1): 2x + 3y = 12
    2 × 3 + 3y = 12
    3y + 6 = 12
    Subtract 6 from both sides: 3y = 6
    Divide both sides by 3: y = 6 ÷ 3 = 2
  6. Write the solution. Give both values. They belong together as a pair.
    x = 3, y = 2

Check. Put x = 3, y = 2 into the other equation (equation (2), which we did not use to find y): 4 × 3 − 3 × 2 = 12 − 6 = 6. This equals 6, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Worked example 14.5.3
Solve 2x + y = 7 and 3x − 2y = 7. (Multiply one equation first.)

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Simultaneous equations: make the number in front of one letter the same size in both equations, then add or subtract to eliminate (remove) that letter. Solve for the other letter, substitute back, and check.

  1. Number the equations. We will keep referring to them, so call them (1) and (2).
    (1)   2x + y = 7
    (2)   3x − 2y = 7
  2. Choose a letter to eliminate: y, and make the numbers match. The numbers in front of y are 1 and 2. The lowest common multiple of 1 and 2 is 2, so we will make both equal to 2. So multiply equation (1) by 2. Multiply every term, including the number on the right-hand side.
  3. Multiply to make the numbers in front of y match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
    Equation (1) × 2: 2 × 2x, 2 × y on the left and 2 × 7 on the right.
    (3)   4x + 2y = 14
  4. Add (3) and (2). The y terms are 2y and −2y. They have different signs, so add the equations: 2y + (−2y) = 0, so y disappears. Do the same thing to the left sides and to the right sides.
    (3)   4x + 2y = 14
    (2)   3x − 2y = 7
    y terms: 2y + (−2y) = 0 (they cancel)
    x terms: 4x + 3x = 7x
    Numbers: 14 + 7 = 21
    7x = 21
  5. Solve for x. 7x means 7 multiplied by x, so divide both sides by 7.
    Divide both sides by 7: x = 21 ÷ 7 = 3
  6. Substitute x = 3 into equation (1) to find y. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
    Equation (1): 2x + y = 7
    2 × 3 + y = 7
    y + 6 = 7
    Subtract 6 from both sides: y = 1
  7. Write the solution. Give both values. They belong together as a pair.
    x = 3, y = 1

Check. Put x = 3, y = 1 into the other equation (equation (2), which we did not use to find y): 3 × 3 − 2 × 1 = 9 − 2 = 7. This equals 7, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Worked example 14.5.4
Solve 3x + 2y = 16 and 5x − 3y = 14. (Multiply both equations.)

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Simultaneous equations: make the number in front of one letter the same size in both equations, then add or subtract to eliminate (remove) that letter. Solve for the other letter, substitute back, and check.

  1. Number the equations. We will keep referring to them, so call them (1) and (2).
    (1)   3x + 2y = 16
    (2)   5x − 3y = 14
  2. Choose a letter to eliminate: y, and make the numbers match. The numbers in front of y are 2 and 3. The lowest common multiple of 2 and 3 is 6, so we will make both equal to 6. So multiply equation (1) by 3 and multiply equation (2) by 2. Multiply every term, including the number on the right-hand side.
  3. Multiply to make the numbers in front of y match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
    Equation (1) × 3: 3 × 3x, 3 × 2y on the left and 3 × 16 on the right.
    (3)   9x + 6y = 48

    Equation (2) × 2: 2 × 5x, 2 × (−3y) on the left and 2 × 14 on the right.
    (4)   10x − 6y = 28
  4. Add (3) and (4). The y terms are 6y and −6y. They have different signs, so add the equations: 6y + (−6y) = 0, so y disappears. Do the same thing to the left sides and to the right sides.
    (3)   9x + 6y = 48
    (4)   10x − 6y = 28
    y terms: 6y + (−6y) = 0 (they cancel)
    x terms: 9x + 10x = 19x
    Numbers: 48 + 28 = 76
    19x = 76
  5. Solve for x. 19x means 19 multiplied by x, so divide both sides by 19.
    Divide both sides by 19: x = 76 ÷ 19 = 4
  6. Substitute x = 4 into equation (1) to find y. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
    Equation (1): 3x + 2y = 16
    3 × 4 + 2y = 16
    2y + 12 = 16
    Subtract 12 from both sides: 2y = 4
    Divide both sides by 2: y = 4 ÷ 2 = 2
  7. Write the solution. Give both values. They belong together as a pair.
    x = 4, y = 2

Check. Put x = 4, y = 2 into the other equation (equation (2), which we did not use to find y): 5 × 4 − 3 × 2 = 20 − 6 = 14. This equals 14, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Worked example 14.5.5
Solve 2x = 13 − y and x + 3y = 14. (Rearrange first.)

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Simultaneous equations: make the number in front of one letter the same size in both equations, then add or subtract to eliminate (remove) that letter. Solve for the other letter, substitute back, and check.

  1. Rearrange each equation so the letters are on the left and the plain number is on the right. Elimination needs the equations lined up in the same pattern: (number)x + (number)y = (number). Use inverse operations on both sides.
    2x = 13 − y  →  2x + y = 13
    x + 3y = 14  →  x + 3y = 14
  2. Number the equations. Call the tidied equations (1) and (2).
    (1)   2x + y = 13
    (2)   x + 3y = 14
  3. Choose a letter to eliminate: x, and make the numbers match. The numbers in front of x are 2 and 1. The lowest common multiple of 2 and 1 is 2, so we will make both equal to 2. So multiply equation (2) by 2. Multiply every term, including the number on the right-hand side.
  4. Multiply to make the numbers in front of x match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
    Equation (2) × 2: 2 × x, 2 × 3y on the left and 2 × 14 on the right.
    (3)   2x + 6y = 28
  5. Subtract (1) from (3). The x terms are 2x and 2x. They have the same sign, so subtract the equations: 2x − 2x = 0, so x disappears. Do the same thing to the left sides and to the right sides.
    (3)   2x + 6y = 28
    (1)   2x + y = 13
    x terms: 2x − 2x = 0 (they cancel)
    y terms: 6y − y = 5y
    Numbers: 28 − 13 = 15
    5y = 15
  6. Solve for y. 5y means 5 multiplied by y, so divide both sides by 5.
    Divide both sides by 5: y = 15 ÷ 5 = 3
  7. Substitute y = 3 into equation (2) to find x. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
    Equation (2): x + 3y = 14
    x + 3 × 3 = 14
    x + 9 = 14
    Subtract 9 from both sides: x = 5
  8. Write the solution. Give both values. They belong together as a pair.
    x = 5, y = 3

Check. Put x = 5, y = 3 into the other equation (equation (1), which we did not use to find x): 2 × 5 + 3 = 10 + 3 = 13. This equals 13, as it should. ✓

Watch out. Same signs subtract, different signs add. Multiply every term of the equation, including the number on the right, and take care with negative numbers when you subtract.

Watch out

When you subtract, you subtract every term, and subtracting a negative number is the same as adding. For example (2y) − (−y) = 3y, not y. A very common mistake is to subtract only the first term, or to forget to multiply the number on the right-hand side when you multiply an equation.

Exercise 14E  Simultaneous equations: elimination

Non-calculator
1
FSolve each pair of simultaneous equations. Just add or subtract: no multiplying is needed.
(a)x + y = 9  and  x − y = 3
(b)2x + y = 11  and  x + y = 7
(c)3x + y = 14  and  x + y = 8
(d)x + 2y = 9  and  x + y = 6
(e)4x + 3y = 27  and  4x − y = 7
(f)5x + 2y = 24  and  3x − 2y = 8
2
FSolve the simultaneous equations by adding the two equations.
x + y = 5
x − y = 1
3
FSolve the simultaneous equations by subtracting.
3x + y = 10
x + y = 4
4
FSolve the simultaneous equations
2x + 3y = 12
2x + y = 8
5
FSolve the simultaneous equations
5x + 2y = 19
3x − 2y = 5
6
FSolve each pair. Multiply one equation first.
(a)2x + y = 8  and  3x − 2y = 5
(b)x + 3y = 11  and  2x − y = 1
(c)3x + 2y = 14  and  x − y = 3
(d)5x − y = 13  and  2x + 3y = 12
(e)x + 2y = 8  and  3x + 4y = 20
(f)4x + y = 14  and  3x − 2y = 5
7
FSolve each pair. Multiply both equations.
(a)2x + 3y = 13  and  3x + 2y = 12
(b)3x + 4y = 24  and  2x + 3y = 17
(c)5x + 2y = 19  and  3x + 5y = 19
(d)4x − 3y = 6  and  3x + 2y = 13
(e)7x + 2y = 20  and  3x − 5y = −9
(f)2x − 5y = −4  and  3x + 2y = 13
8
FSolve the simultaneous equations.
x + y = 10
3x + y = 22
9
FSolve the simultaneous equations
4x + 3y = 18
x + 3y = 9
10
FRearrange each pair into the form ax + by = c, then solve.
(a)2x = 13 − y  and  x + 3y = 14
(b)y = 8 − x  and  3x − y = 4
(c)2x + 3y = 19  and  x = y + 2
(d)3x = 11 − y  and  x + 2y = 7
11
RMaya has the equations 4x + 3y = 18 and x + 3y = 9. She adds them but both letters are still there. Explain why adding does not help, and solve the equations correctly.
12
RExplain why the simultaneous equations x + y = 5 and x + y = 7 have no solution.
13
PSSolve the simultaneous equations and then write down the value of x + y.
3x + 5y = 21
2x + 5y = 19
14
Rx + y = 10 and x − y = 2. Without finding x or y first, show that x2 − y2 = 20. Then check by solving the equations.
14.6

Simultaneous equations: substitution

Edexcel 1MA1: A19
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Simultaneous equations: substitutionComing soon

Elimination is not the only way to solve simultaneous equations. Sometimes one equation already tells you what a letter equals, for example y = 2x + 1. Then it is quick to substitute, which means “replace”: wherever you see y in the other equation, write 2x + 1 instead. Because y and 2x + 1 are equal, this does not change the equation, but it removes y, leaving an ordinary equation with one letter.

The method.

  1. Number the equations. Choose one that has a letter on its own (y = … or x = …). If there is not one, rearrange one equation to make a letter the subject. The easiest choice is a letter with no number in front of it.
  2. Substitute this expression into the other equation, in brackets.
  3. Expand the brackets and collect like terms.
  4. Solve the one-letter equation.
  5. Substitute the answer back into the rearranged equation to find the other letter.
  6. Check in the other equation.

Brackets are very important. If y = 2x + 1 and the other equation is 4x − 3y = 7, then you must write 4x − 3(2x + 1) = 7. Without the brackets, the −3 would only multiply the 2x and you would get the wrong answer.

Which method should I use? If one equation has a letter on its own (or with a number 1 in front of it), substitution is usually quickest. If both equations are in the form ax + by = c and the numbers are awkward, elimination is usually quicker. Either method always gives the same answer, so you can use one to check the other.

Key ideas

  • Substitution: replace a letter by an expression that equals it.
  • Make a letter the subject of one equation (choose a letter with 1 in front of it).
  • Put the expression in brackets when you substitute, then expand and collect like terms.
  • Solve for one letter, then find the other, then check in the other equation.
  • Use whichever method (substitution or elimination) is easier. They give the same answer.
Worked example 14.6.1
Solve y = 2x + 1 and 3x + y = 11.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true. To substitute means to replace a letter with something equal to it.

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

  1. Write the equations. We have two equations with two unknown letters. We will call them (1) and (2).
    (1)   y = 2x + 1
    (2)   3x + y = 11
  2. Equation (1) already tells us what y equals. y is on its own on the left, so we can use it as it is.
    y = 2x + 1
  3. Substitute into equation (2). Equation (2) contains y. Wherever y appears, write what it equals (inside a bracket). Now there is only one letter, x, so we can solve.
    Equation (2): 3x + y = 11
    3x + (2x + 1) = 11
  4. Expand the bracket and collect like terms. Multiply the bracket out, then add the x terms together and the plain numbers together.
    There is no number in front of the bracket, so the brackets can just be removed: (2x + 1) = 2x + 1
    The equation becomes: 2x + 1 + 3x = 11
    Collect like terms: 5x + 1 = 11
  5. Solve the equation for x. It is now a one-letter equation. Use inverse operations on both sides.
    Subtract 1 from both sides: 5x = 10
    Divide both sides by 5: x = 10 ÷ 5 = 2
  6. Substitute x = 2 into the rearranged equation to find y. The rearranged equation y = 2x + 1 is the easiest to use.
    y = 2x + 1
    y = 2 × 2 + 1
    y = 5
  7. Write the solution. Give both values as a pair.
    x = 2, y = 5

Check. Put x = 2, y = 5 into equation (1): −2 × 2 + 5 = −4 + 5 = 1. It equals 1, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Worked example 14.6.2
Solve x = 3y − 2 and 2x + y = 10.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true. To substitute means to replace a letter with something equal to it.

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

  1. Write the equations. We have two equations with two unknown letters. We will call them (1) and (2).
    (1)   x = 3y − 2
    (2)   2x + y = 10
  2. Rearrange equation (2) to make y the subject. The subject is the letter on its own on one side. We choose the equation and letter that is easiest to rearrange (a letter with no number in front of it, so there is no dividing).
    Equation (2): 2x + y = 10
    Subtract 2x from both sides: y = 10 − 2x
  3. Substitute into equation (1). Equation (1) contains y. Wherever y appears, write what it equals (inside a bracket). Now there is only one letter, x, so we can solve.
    Equation (1): x − 3y = −2
    x − 3(10 − 2x) = −2
  4. Expand the bracket and collect like terms. Multiply the bracket out, then add the x terms together and the plain numbers together.
    Expand −3(10 − 2x):
    (−3) × (−2x) = 6x
    (−3) × 10 = −30
    so −3(10 − 2x) = 6x − 30
    The equation becomes: 6x − 30 + x = −2
    Collect like terms: 7x − 30 = −2
  5. Solve the equation for x. It is now a one-letter equation. Use inverse operations on both sides.
    Add 30 to both sides: 7x = 28
    Divide both sides by 7: x = 28 ÷ 7 = 4
  6. Substitute x = 4 into the rearranged equation to find y. The rearranged equation y = 10 − 2x is the easiest to use.
    y = 10 − 2x
    y = 10 − 2 × 4
    y = 2
  7. Write the solution. Give both values as a pair.
    x = 4, y = 2

Check. Put x = 4, y = 2 into equation (2): 2 × 4 + 2 = 8 + 2 = 10. It equals 10, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Worked example 14.6.3
Solve x + 2y = 8 and 3x − y = 3. (Rearrange one equation first.)

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true. To substitute means to replace a letter with something equal to it.

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

  1. Write the equations and number them. We have two unknown letters, so we need two equations.
    (1)   x + 2y = 8
    (2)   3x − y = 3
  2. Rearrange equation (1) to make x the subject. The subject is the letter on its own on one side. We choose the equation and letter that is easiest to rearrange (a letter with no number in front of it, so there is no dividing).
    Equation (1): x + 2y = 8
    Subtract 2y from both sides: x = 8 − 2y
  3. Substitute into equation (2). Equation (2) contains x. Wherever x appears, write what it equals (inside a bracket). Now there is only one letter, y, so we can solve.
    Equation (2): 3x − y = 3
    3(8 − 2y) − y = 3
  4. Expand the bracket and collect like terms. Multiply the bracket out, then add the y terms together and the plain numbers together.
    Expand 3(8 − 2y):
    3 × (−2y) = −6y
    3 × 8 = 24
    so 3(8 − 2y) = −6y + 24
    The equation becomes: −6y + 24 − y = 3
    Collect like terms: −7y + 24 = 3
  5. Solve the equation for y. It is now a one-letter equation. Use inverse operations on both sides.
    Subtract 24 from both sides: −7y = −21
    Divide both sides by −7: y = −21 ÷ (−7) = 3
  6. Substitute y = 3 into the rearranged equation to find x. The rearranged equation x = 8 − 2y is the easiest to use.
    x = 8 − 2y
    x = 8 − 2 × 3
    x = 2
  7. Write the solution. Give both values as a pair.
    x = 2, y = 3

Check. Put x = 2, y = 3 into equation (1): 2 + 2 × 3 = 2 + 6 = 8. It equals 8, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Worked example 14.6.4
Solve y = x − 5 and 3x + 2y = 5. (The answer has a negative number.)

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true. To substitute means to replace a letter with something equal to it.

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

  1. Write the equations. We have two equations with two unknown letters. We will call them (1) and (2).
    (1)   y = x − 5
    (2)   3x + 2y = 5
  2. Equation (1) already tells us what y equals. y is on its own on the left, so we can use it as it is.
    y = x − 5
  3. Substitute into equation (2). Equation (2) contains y. Wherever y appears, write what it equals (inside a bracket). Now there is only one letter, x, so we can solve.
    Equation (2): 3x + 2y = 5
    3x + 2(x − 5) = 5
  4. Expand the bracket and collect like terms. Multiply the bracket out, then add the x terms together and the plain numbers together.
    Expand 2(x − 5):
    2 × x = 2x
    2 × (−5) = −10
    so 2(x − 5) = 2x − 10
    The equation becomes: 2x − 10 + 3x = 5
    Collect like terms: 5x − 10 = 5
  5. Solve the equation for x. It is now a one-letter equation. Use inverse operations on both sides.
    Add 10 to both sides: 5x = 15
    Divide both sides by 5: x = 15 ÷ 5 = 3
  6. Substitute x = 3 into the rearranged equation to find y. The rearranged equation y = x − 5 is the easiest to use.
    y = x − 5
    y = 3 − 5
    y = −2
  7. Write the solution. Give both values as a pair.
    x = 3, y = −2

Check. Put x = 3, y = −2 into equation (1): −3 + (−2) = −3 − 2 = −5. It equals −5, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Watch out

Do not forget the brackets, and do not stop when you find the first letter. A solution has two values. Also, when you finish, put the values back into the other equation to check, not only into the one you rearranged.

Exercise 14F  Simultaneous equations: substitution

Non-calculator
1
FSolve by substitution. One letter is already the subject.
(a)y = x + 3  and  x + y = 11
(b)y = 2x  and  3x + y = 15
(c)y = x − 2  and  2x + y = 10
(d)y = 3x − 1  and  x + y = 7
(e)x = y + 4  and  x + 2y = 10
(f)x = 2y  and  x + y = 9
2
FSolve the simultaneous equations
y = x + 5
x + y = 9
3
FSolve the simultaneous equations
x = y + 2
2x + y = 13
4
FSolve by substitution. Remember the brackets.
(a)y = 2x + 1  and  3x + 2y = 16
(b)y = 3x − 2  and  2x + y = 13
(c)x = y − 1  and  3x + 2y = 7
(d)y = 5 − x  and  2x + 3y = 12
(e)y = 3x − 5  and  2x + y = 5
(f)x = 7 − 2y  and  3x − y = 7
5
FRearrange one of the equations, then solve by substitution.
(a)x + y = 8  and  2x − y = 4
(b)x + 3y = 11  and  2x − y = 1
(c)3x + y = 10  and  x − 2y = −6
(d)x − y = 3  and  2x + 3y = 16
6
FSolve the simultaneous equations
x + y = 12
3x + y = 20
7
FSolve the simultaneous equations
2x − y = 1
x + y = 8
8
RExplain which method (substitution or elimination) looks easier for y = x + 7 and 2x + y = 13. Then solve the equations.
9
RTom is solving y = 2x − 1 and 4x − 3y = 7. He writes “4x − 3 × 2x − 1 = 7”. Explain his mistake, then solve the equations correctly.
10
PSSolve by substitution, and then check your answer using elimination.
3x + 2y = 12
x + y = 5
11
RThe lines y = 2x + 1 and y = ax − 5 cross at the point where x = 3. Find the value of a.
14.7

Solving equations with graphs

Edexcel 1MA1: A18, A19
Grade 3-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Solving equations with graphsComing soon

In Chapter 12 you learned that the equation of a straight line, such as y = 2x + 1, is a rule that every point on the line obeys. Now suppose we draw two lines on the same grid. A point that is on both lines obeys both equations at once. So the point where the lines cross is the solution of the simultaneous equations. This is the graphical method.

xy-1123456-1123456789y = x + 1y = -x + 7(3, 4)
The lines y = x + 1 and y = 7 − x cross at (3, 4). So the solution is x = 3, y = 4.

To draw a straight line make a table of values: choose three values of x, work out y for each, plot the points and join them with a ruler. If the equation is like x + y = 8, first rearrange it to y = 8 − x. Another quick way is to find where the line crosses the axes: put x = 0 to find where it crosses the y-axis, and y = 0 to find where it crosses the x-axis.

Graphs give approximate answers, because you can only read a graph to a certain accuracy. When the crossing point is at a grid corner, you can read it exactly, and you should always check by substituting into both equations. If two lines are parallel (same gradient) they never cross, so the simultaneous equations have no solution.

Solving quadratic equations with a graph. The solutions of x2 − 4x + 1 = 0 are the values of x where the curve y = x2 − 4x + 1 meets the line y = 0 (the x-axis). Read the x-values. These are usually not whole numbers, so give them to 1 decimal place. To solve x2 − 2x − 3 = 2, draw the line y = 2 on the same graph and read the x-values where it meets the curve. In general, to solve an equation with a graph, plot each side as a graph and find where they cross.

xy-112345-4-3-2-11234567
The curve y = x2 − 4x + 1 crosses the x-axis at about x = 0.3 and x = 3.7. These are approximate solutions of x2 − 4x + 1 = 0.

Key ideas

  • The solution of simultaneous equations is the point where their graphs cross: (x, y).
  • Draw a line using a table of values (or the axis crossings). Rearrange to y = … first if you need to.
  • Parallel lines never cross, so there is no solution.
  • To solve f(x) = 0 with a graph, read the x-values where the curve meets the x-axis (y = 0).
  • To solve f(x) = k, draw the line y = k and read the x-values where it meets the curve.
  • Graph answers are approximate: give them to 1 decimal place when they are not whole numbers, and check by substituting.
Worked example 14.7.1
The graph shows the lines y = x + 1 and y = 7 − x. Use it to solve the simultaneous equations y = x + 1 and y = 7 − x.

Key words. The solution of two simultaneous equations is the pair (x, y) that makes both true. On a graph it is where the two lines intersect (cross).

Method. The solution of two simultaneous equations is the point where their graphs cross.

  1. Find the point where the two lines cross. Every point on a line fits that line's equation. The one point that is on both lines fits both equations, so it is the solution of the simultaneous equations.
    Look for the crossing point of the two lines.
  2. Read the coordinates of the crossing point. The coordinates are written (x, y): read along the x-axis first, then up the y-axis.
    The lines cross at (3, 4).
  3. Write the solution. The x-coordinate is the value of x and the y-coordinate is the value of y.
    x = 3, y = 4

Check. Put x = 3 into both equations. y = x + 1: y = 3 + 1 = 4. y = 7 − x: y = −3 + 7 = 4. Both give y = 4. ✓

Watch out. Give both coordinates and say which is x and which is y. Graph answers are only as accurate as your reading, so check by substituting.

Worked example 14.7.2
By drawing graphs, solve y = 2x − 1 and x + y = 8.

Method. Draw both straight lines on the same grid. The solution is the point where they cross.

  1. Rearrange to the form y = mx + c (where necessary). It is much easier to find points when y is on its own.
    x + y = 8  →  subtract x from both sides  →  y = 8 − x
  2. Make a table of values for line 1: y = 2x − 1. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.
    When x = 0: y = 2 × 0 − 1 = 0 − 1 = −1
    When x = 2: y = 2 × 2 − 1 = 4 − 1 = 3
    When x = 4: y = 2 × 4 − 1 = 8 − 1 = 7
    When x = 6: y = 2 × 6 − 1 = 12 − 1 = 11
  3. Make a table of values for line 2: x + y = 8. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.
    When x = 0: y = −0 + 8 = 0 + 8 = 8
    When x = 2: y = −2 + 8 = 6
    When x = 4: y = −4 + 8 = 4
    When x = 6: y = −6 + 8 = 2
  4. Plot the points and draw both lines. Plot each pair (x, y), then join the points of each line with a ruler. Extend the lines so that they cross.
    xy-1123456-2-112345678910y = -x + 8(3, 5)
  5. Read off the crossing point. The point where the lines cross is on both lines, so it is the solution.
    The lines cross at (3, 5).
    x = 3, y = 5

Check. Check by substituting x = 3: line 1 gives y = 2 × 3 − 1 = 5 and line 2 gives y = −3 + 8 = 5. They agree. ✓

Watch out. Use a sharp pencil and a ruler, and extend the lines far enough to see where they cross.

Worked example 14.7.3
Explain why the simultaneous equations y = 2x + 1 and y = 2x + 4 have no solution.

Method. Compare the gradients. If two straight lines have the same gradient they are parallel.

  1. Write the gradient of each line. In y = mx + c, the number m in front of x is the gradient.
    y = 2x + 1 has gradient 2
    y = 2x + 4 has gradient 2
  2. Compare. The gradients are equal, so the lines slope the same way and stay the same distance apart. They never cross.
    The lines are parallel.
  3. Conclude. A solution would be a point on both lines, and there is no such point.
    There is no solution.

Check. Algebra agrees: if 2x + 1 = 2x + 4 then 1 = 4, which is false. ✓

Watch out. Parallel lines have the same m but different c.

Worked example 14.7.4
The graph of y = x2 − 4x + 1 is shown above. Use it to find estimates of the solutions of x2 − 4x + 1 = 0.

Method. The solutions of x2 − 4x + 1 = 0 are the x-values where the curve y = x2 − 4x + 1 crosses the x-axis.

  1. Find where the curve meets y = 0. The x-axis is the line y = 0, and an equation “expression = 0” asks where the expression is 0.
    The curve crosses the x-axis twice: just right of 0, and just left of 4.
  2. Read the x-values as accurately as you can. Each small square on the grid is 1 unit, so estimate the position between the grid lines. The first crossing is about a quarter of the way from 0 to 1. The second is about three-quarters of the way from 3 to 4.
    x ≈ 0.3 and x ≈ 3.7
  3. Write the answers to 1 decimal place. These are approximate answers because a graph cannot be read exactly.
    x = 0.3 or x = 3.7 (to 1 d.p.)

Check. Substitute x = 0.3: 0.32 − 4 × 0.3 + 1 = 0.09 − 1.2 + 1 = −0.11, close to 0. Substitute x = 3.7: 3.72 − 4 × 3.7 + 1 = 13.69 − 14.8 + 1 = −0.11, close to 0. ✓

Watch out. Graph readings are only approximate, so any answer within about 0.1 of the true value is acceptable.

Watch out

Always read both coordinates and write the answer as x = …, y = … (not just a point). And remember a graph gives an approximate answer unless the crossing is exactly at a grid corner: check by substituting your values into both equations.

Exercise 14G  Solving equations with graphs

Non-calculator
1
FUse the graph to solve the simultaneous equations
y = x + 1
y = 7 − x
xy-1123456-1123456789y = x + 1y = -x + 7
2
FUse the graph to solve the simultaneous equations
y = 2x
y = x + 3
xy-1123456-112345678y = x + 3
3
FUse the graph to solve the simultaneous equations
y = x − 1
y = 5 − x
xy-1123456-2-112345678y = x - 1y = -x + 5
4
FUse the graph to solve the simultaneous equations
y = 2x + 1
y = 11 − 3x
xy-1123456-112345678910111213y = 2x + 1
5
FDraw the graphs of y = 2x − 1 and x + y = 8 for values of x from 0 to 6. Use your graphs to solve the simultaneous equations.
6
FOn the same grid draw the graphs of y = x + 2 and y = 8 − x for x from 0 to 6. Write down the solution of the simultaneous equations.
7
FDraw the graphs of y = 3x and x + y = 8 for x from 0 to 4. Use your graph to solve the simultaneous equations.
8
FIs x = 2, y = 5 the solution of y = x + 3 and y = 2x + 1? Show how you decide.
9
RWithout drawing, explain how you know that y = 3x + 2 and y = 3x − 5 have no solution.
10
PSPlan A for a mobile phone costs a fixed £5 plus 10p per minute. Plan B costs a fixed £1 plus 20p per minute. The graph shows the cost of each plan for up to 60 minutes. For how many minutes do the two plans cost the same, and what is that cost?
minutes£1020304050602468101214Plan APlan B
11
PSThe graph of y = x2 − 4x + 1 is drawn. Use it to estimate the solutions of x2 − 4x + 1 = 0 to 1 decimal place.
xy-112345-4-3-2-11234567
12
PSThe graphs of y = x2 and y = 2x + 3 are drawn. Use them to solve x2 = 2x + 3.
xy-3-2-11234-2-11234567891011
13
PSThe graph of y = x2 − 2x − 3 is drawn. Use it to estimate the solutions of x2 − 2x − 3 = 2.
xy-3-2-112345-5-4-3-2-112345678
14
RSally draws the graphs of y = x + 1 and y = 2x − 3 and says they cross at (4, 5). Use algebra to show that she is correct.
14.8

Forming simultaneous equations from words

Edexcel 1MA1: A19, A21
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Forming simultaneous equations from wordsComing soon

Simultaneous equations are used whenever there are two unknown quantities and two pieces of information about them: two shopping baskets, two journeys, two ages, two shapes. The skill is translating the words into algebra. Then you solve with elimination (section 14.5) or substitution (14.6).

Strategy.

  1. Choose letters for the two unknowns and write down what each stands for, with units (for example “Let a be the cost of one adult ticket in pounds”).
  2. Write equation (1) from the first piece of information, then equation (2) from the second. Check that both sides use the same units.
  3. Solve the equations.
  4. Answer the question in words, with units. The question may ask for something different from the letters, such as the total cost of a different basket.
  5. Check your values in the original words, not just in your equations.

Useful translations: “the sum of” means add, “the difference between” means subtract, “a is 3 more than b” means a = b + 3 (or a − b = 3), “twice” means ×2, and the perimeter of a rectangle is 2l + 2w. For ages, remember that after n years each person is n years older. For a fixed charge plus a rate, the total is (fixed charge) + (rate × amount).

Key ideas

  • Two unknowns need two equations. Choose a letter for each unknown and say what it stands for, with units.
  • Turn each sentence of information into one equation. Keep the units the same.
  • Solve with elimination or substitution, then give the answer in words with units.
  • The question may not ask for the letters themselves. Read the last sentence again.
  • Check in the original sentences.
Worked example 14.8.1
2 adult tickets and 3 child tickets cost £31. 3 adult tickets and 2 child tickets cost £34. Work out the cost of one adult ticket and the cost of one child ticket.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

  1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
    Let one adult ticket cost £a and one child ticket cost £c.
  2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
    2 adult tickets and 3 child tickets cost £31.
    Total cost in pounds = (2 × a) + (3 × c).
    2a + 3c = 31
  3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
    3 adult tickets and 2 child tickets cost £34.
    Total cost in pounds = (3 × a) + (2 × c).
    3a + 2c = 34
  4. Number the equations. We will keep referring to them, so call them (1) and (2).
    (1)   2a + 3c = 31
    (2)   3a + 2c = 34
  5. Choose a letter to eliminate: a, and make the numbers match. The numbers in front of a are 2 and 3. The lowest common multiple of 2 and 3 is 6, so we will make both equal to 6. So multiply equation (1) by 3 and multiply equation (2) by 2. Multiply every term, including the number on the right-hand side.
  6. Multiply to make the numbers in front of a match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
    Equation (1) × 3: 3 × 2a, 3 × 3c on the left and 3 × 31 on the right.
    (3)   6a + 9c = 93

    Equation (2) × 2: 2 × 3a, 2 × 2c on the left and 2 × 34 on the right.
    (4)   6a + 4c = 68
  7. Subtract (4) from (3). The a terms are 6a and 6a. They have the same sign, so subtract the equations: 6a − 6a = 0, so a disappears. Do the same thing to the left sides and to the right sides.
    (3)   6a + 9c = 93
    (4)   6a + 4c = 68
    a terms: 6a − 6a = 0 (they cancel)
    c terms: 9c − 4c = 5c
    Numbers: 93 − 68 = 25
    5c = 25
  8. Solve for c. 5c means 5 multiplied by c, so divide both sides by 5.
    Divide both sides by 5: c = 25 ÷ 5 = 5
  9. Substitute c = 5 into equation (1) to find a. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
    Equation (1): 2a + 3c = 31
    2a + 3 × 5 = 31
    2a + 15 = 31
    Subtract 15 from both sides: 2a = 16
    Divide both sides by 2: a = 16 ÷ 2 = 8
  10. Write the solution. Give both values. They belong together as a pair.
    a = 8, c = 5
  11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Adult ticket: a = 8, so one adult ticket costs £8.
    Child ticket: c = 5, so one child ticket costs £5.
    One adult ticket costs £8 and one child ticket costs £5.

Check. Put a = 8, c = 5 into the other equation (equation (2), which we did not use to find a): 3 × 8 + 2 × 5 = 24 + 10 = 34. This equals 34, as it should. ✓

Watch out. Write what each letter stands for, including the units (pounds here). Check the answer in both sentences of the question.

Worked example 14.8.2
The sum of Mia’s age and her brother’s age is 31. Mia is 4 years older than twice her brother’s age. Find both ages.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

  1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
    Let Mia be m years old and her brother be b years old.
  2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
    Their ages add up to 31.
    Mia’s age is 4 more than twice her brother’s age: m = 2b + 4, which rearranges to m − 2b = 4.
    m + b = 31
  3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
    Mia is 4 years older than twice her brother’s age.
    m = 2b + 4, so m − 2b = 4.
    m − 2b = 4
  4. Number the equations. We will keep referring to them, so call them (1) and (2).
    (1)   m + b = 31
    (2)   m − 2b = 4
  5. Choose a letter to eliminate: m. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of m are 1 and 1: they are already the same size, so no multiplying is needed.
  6. Subtract (2) from (1). The m terms are m and m. They have the same sign, so subtract the equations: m − m = 0, so m disappears. Do the same thing to the left sides and to the right sides.
    (1)   m + b = 31
    (2)   m − 2b = 4
    m terms: m − m = 0 (they cancel)
    b terms: b − (−2b) = 3b
    Numbers: 31 − 4 = 27
    3b = 27
  7. Solve for b. 3b means 3 multiplied by b, so divide both sides by 3.
    Divide both sides by 3: b = 27 ÷ 3 = 9
  8. Substitute b = 9 into equation (1) to find m. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
    Equation (1): m + b = 31
    m + 9 = 31
    m + 9 = 31
    Subtract 9 from both sides: m = 22
  9. Write the solution. Give both values. They belong together as a pair.
    m = 22, b = 9
  10. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Mia: m = 22 years.
    Her brother: b = 9 years.
    Mia is 22 and her brother is 9.

Check. Put m = 22, b = 9 into the other equation (equation (2), which we did not use to find m): 22 − 2 × 9 = 22 − 18 = 4. This equals 4, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Worked example 14.8.3
The perimeter of a rectangle is 38 cm. The length is 5 cm more than the width. Find the length and the width.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

  1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
    Let the length be l cm and the width be w cm.
  2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
    The perimeter is two lengths and two widths: 2l + 2w = 38.
    2l + 2w = 38
  3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
    The length is 5 more than the width: l = w + 5, so l − w = 5.
    l − w = 5
  4. Number the equations. We will keep referring to them, so call them (1) and (2).
    (1)   2l + 2w = 38
    (2)   l − w = 5
  5. Choose a letter to eliminate: l, and make the numbers match. The numbers in front of l are 2 and 1. The lowest common multiple of 2 and 1 is 2, so we will make both equal to 2. So multiply equation (2) by 2. Multiply every term, including the number on the right-hand side.
  6. Multiply to make the numbers in front of l match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
    Equation (2) × 2: 2 × l, 2 × (−w) on the left and 2 × 5 on the right.
    (3)   2l − 2w = 10
  7. Subtract (3) from (1). The l terms are 2l and 2l. They have the same sign, so subtract the equations: 2l − 2l = 0, so l disappears. Do the same thing to the left sides and to the right sides.
    (1)   2l + 2w = 38
    (3)   2l − 2w = 10
    l terms: 2l − 2l = 0 (they cancel)
    w terms: 2w − (−2w) = 4w
    Numbers: 38 − 10 = 28
    4w = 28
  8. Solve for w. 4w means 4 multiplied by w, so divide both sides by 4.
    Divide both sides by 4: w = 28 ÷ 4 = 7
  9. Substitute w = 7 into equation (2) to find l. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
    Equation (2): l − w = 5
    l − 7 = 5
    l − 7 = 5
    Add 7 to both sides: l = 12
  10. Write the solution. Give both values. They belong together as a pair.
    l = 12, w = 7
  11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Length = 12 cm, width = 7 cm.
    The rectangle is 12 cm by 7 cm.

Check. Put l = 12, w = 7 into the other equation (equation (1), which we did not use to find l): 2 × 12 + 2 × 7 = 24 + 14 = 38. This equals 38, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Watch out

Check that you have used both pieces of information, and that your final answer is what the question asked for. If the question asks for “the total cost of 1 adult and 1 child ticket”, finding the two separate prices is only part of the job. Also keep units consistent: do not mix pounds and pence in the same equation.

Exercise 14H  Forming simultaneous equations from words

Non-calculator
1
PS2 pens and 1 book cost £7. 1 pen and 1 book cost £5. Work out the cost of one pen and the cost of one book.
2
PS2 adult tickets and 3 child tickets cost £45. 3 adult tickets and 1 child ticket cost £43. Work out the cost of one adult ticket and the cost of one child ticket.
3
PS2 coffees and 1 cake cost £8. 1 coffee and 3 cakes cost £9. Work out the cost of one coffee and the cost of one cake.
4
PS3 pens and 5 pencils cost £14. 2 pens and 4 pencils cost £10. Work out the total cost of 4 pens and 4 pencils.
5
PS3 burgers and 2 portions of chips cost £19. 2 burgers and 3 portions of chips cost £16. Work out the cost of one burger and the cost of one portion of chips.
6
PS3 notebooks and 2 rulers cost £18. 1 notebook and 4 rulers cost £16. Work out the total cost of 1 notebook and 1 ruler.
7
FThe sum of two numbers is 47 and their difference is 9. Find the numbers.
8
PSTwo numbers add to 52. Twice the larger number plus the smaller number is 80. Find the numbers.
9
PSSam is 28 years older than his son. In 6 years’ time Sam will be three times as old as his son. Find their ages now.
10
PSThe perimeter of a rectangular field is 140 m. The length is 10 m more than the width. Find the area of the field.
11
PSA jar contains 24 coins. They are all 10p and 20p coins and are worth £3.40 altogether (340p). How many coins of each kind are there?
12
PSA taxi company charges a fixed fee plus a rate per kilometre. A 5 km journey costs £11 and an 8 km journey costs £17. Find the fixed fee and the charge per kilometre.
13
PSMia thinks of two numbers. Three times the first number plus the second number is 17. The first number plus twice the second is 9. What are the numbers?
14
RThe digits of a two-digit number add up to 11. When the digits are reversed, the number increases by 27. Find the original number.

Chapter summary

SkillMethod
Factorise x2 + bx + cFind two numbers that multiply to c and add to b: (x + p)(x + q). Take out a common factor first.
Difference of two squaresa2 − b2 = (a − b)(a + b)
Factorise ax2 + bx + c (Higher)Multiply a × c, find two numbers with that product that add to b, split the middle term and factorise in pairs.
Solve a quadratic by factorisingRearrange to 0 on one side, factorise, set each bracket to 0 (null factor law), solve, check.
Problems leading to quadraticsChoose a letter, form an equation, solve, reject impossible answers, answer in words.
Simultaneous equations: eliminationMake one letter match in size, subtract if signs are the same and add if they are different, solve, substitute back, check.
Simultaneous equations: substitutionMake a letter the subject of one equation, substitute it (in brackets) into the other, solve, find the other letter, check.
Graphical solutionsThe solution is where the graphs cross. Roots of a quadratic are where the curve meets y = 0.
Equations from wordsChoose letters, write one equation per sentence, solve, answer in words with units.
Quadratic formula (Higher)x = (−b ± √(b2 − 4ac)) ÷ 2a; the discriminant decides how many solutions there are.
Completing the square (Higher)x2 + bx + c = (x + b/2)2 − (b/2)2 + c; turning point (−p, q).
Linear/quadratic simultaneous (Higher)Substitute the linear equation into the quadratic, solve, find each second value from the linear equation, write pairs.
Quadratics with fractions (Higher)Multiply every term by the common denominator, rearrange to 0, solve, reject any value that makes a denominator 0.

Can you…?

  • I can factorise quadratics of the form x2 + bx + c and the difference of two squares.
  • I can solve quadratic equations by factorising, after rearranging so that one side is 0.
  • I can form a quadratic equation from an area or consecutive-number problem and reject impossible solutions.
  • I can solve simultaneous linear equations by elimination and by substitution.
  • I can solve simultaneous equations and estimate roots of quadratics using graphs.
  • I can form and solve simultaneous equations from a word problem.
  • Higher: I can factorise ax2 + bx + c, use the quadratic formula, complete the square, solve linear/quadratic simultaneous equations and solve quadratic equations with fractions.

Key terms

Quadratic expression
An expression whose highest power is 2, such as x2 + 5x + 6.
Quadratic equation
An equation of the form ax2 + bx + c = 0.
Factorise
Write an expression as a product of brackets.
Null factor law
If a product is 0, at least one of its factors is 0.
Root
A solution of an equation. The roots of a quadratic are where its graph crosses the x-axis.
Difference of two squares
An expression a2 − b2, which factorises as (a − b)(a + b).
Simultaneous equations
Two or more equations that must be true at the same time.
Elimination
Adding or subtracting equations to remove one letter.
Substitution
Replacing a letter by an expression that equals it.
Discriminant
The expression b2 − 4ac; its sign tells how many real solutions a quadratic has.
Completing the square
Writing x2 + bx + c as (x + p)2 + q.
Turning point
The lowest (minimum) or highest (maximum) point of a quadratic curve.
Surd
A square root that is not a whole number, such as √5; an exact answer keeps the surd.

Exercise 14Z  Check-up (non-calculator)

Non-calculator
1
Factorise x2 + 9x + 20.
2
Factorise x2 − 81.
3
Solve x2 − 3x − 10 = 0.
4
Solve x2 + 4x = 21.
5
Solve the simultaneous equations
x + y = 11
x − y = 3
6
Solve the simultaneous equations
2x + 3y = 12
4x − y = 10
7
Solve by substitution
y = 2x − 1
x + y = 8
8
A rectangle has width x cm and length (x + 4) cm. Its area is 45 cm2. Find the width and the length.
9
Use the graph to solve the simultaneous equations
y = x + 2
y = 8 − x
xy-1123456-112345678910y = x + 2y = -x + 8
10
2 adult tickets and 3 child tickets cost £24. 3 adult tickets and 1 child ticket cost £22. Work out the cost of one adult ticket and the cost of one child ticket.

Total 40 marks. Allow about 40 minutes.

Exercise 14X  Exam-style questions

Non-calculator
1
FFactorise x2 + x − 20.[2 marks]
2
FFactorise 4x2 − 25.[2 marks]
3
FSolve x2 − 5x − 14 = 0.[3 marks]
4
FSolve x2 + 3x = 28.[3 marks]
5
FSolve x(x + 5) = 24.[3 marks]
6
FSolve the simultaneous equations
3x + 2y = 13
5x − 2y = 11[3 marks]
7
FSolve the simultaneous equations
2x + 3y = 1
3x + 2y = 4[4 marks]
8
FA rectangle has width x cm and length (x + 5) cm. Its area is 84 cm2. Find the width and the length.[5 marks]
9
F2 adult tickets and 3 child tickets cost £33. 1 adult ticket and 4 child tickets cost £29. Work out the cost of one adult ticket and the cost of one child ticket.[4 marks]
10
FThe diagram shows the lines y = x + 1 and y = 9 − x. Use the diagram to solve the simultaneous equations y = x + 1 and y = 9 − x.[3 marks]
xy-11234567-11234567891011y = x + 1y = -x + 9
11
FFactorise x2 + 7x + 10.[2 marks]
12
FSolve x2 − x − 12 = 0.[3 marks]
13
FSolve the simultaneous equations
x + y = 9
x − y = 5[3 marks]
← Chapter 13: Quadratic and other graphsChapter 15: Ratio and proportion →

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