GCSE Maths · Quadratic and Simultaneous Equations · Higher · Lesson 14.11

GCSE Maths: Linear and quadratic simultaneous equations (Higher)

Part of the GCSE Maths quadratic and simultaneous equations chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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14.11

Linear and quadratic simultaneous equations

Edexcel 1MA1: A19
Grade 7-9

So far both simultaneous equations have been straight lines. Now one equation can be a curve, such as y = x² − 4, or a circle, such as x² + y² = 25. Solving the pair means finding the points where the line meets the curve. A line can cross a parabola in two points, touch it at one point, or miss it completely.

The line y = x + 2 meets the curve y = x² − 4 at two points, (−2, 0) and (3, 5). Each point is one solution of the simultaneous equations.

The graph shows what the algebra must find, but graphs are only approximate. The algebraic method is substitution:

Rearrange the linear equation to make one letter the subject (for example y = x + 2).
Substitute it into the quadratic equation, in brackets. This gives an equation with only one letter. For y = x² − 4 and y = x + 2 we get x² − 4 = x + 2.
Rearrange to equal 0, factorise (or use the formula) and solve. This gives two values: x² − x − 6 = 0 so x = −2 or x = 3.
Put each value into the linear equation to find its y. This gives the pairs (−2, 0) and (3, 5).
Write the solutions as pairs.

Always use the linear equation to find the second value. If you use the quadratic you may get extra “answers” that are not solutions. For example in x² + y² = 25, x = 3 gives y = 4 or y = −4, but only one of them lies on the line.

Circles. The equation x² + y² = 25 is a circle with centre the origin and radius 5 (you will study circles in Chapter 25). The method is the same: substitute the linear equation into it. Be careful to square the whole bracket: (x + 1)² = x² + 2x + 1, not x² + 1.

How many solutions? After substitution you get a quadratic equation. If it has two solutions the line cuts the curve twice. If it has exactly one (a repeated root) the line is a tangent (it touches the curve). If the discriminant is negative there are no solutions and the line misses the curve.

The line y = x + 1 meets the circle x² + y² = 25 at (3, 4) and (−4, −3).

Key ideas

One equation is linear, one is quadratic (a parabola or a circle). The solutions are the points where they meet.
Make one letter the subject of the linear equation, substitute into the quadratic (in brackets), and solve.
There are usually two solutions. Find the second coordinate of each using the linear equation.
Write each solution as a pair (x, y), keeping the matching values together.
One repeated root means the line touches the curve; a negative discriminant means no solution.

Worked example 14.11.1

Solve the simultaneous equations y = x² − 4 and y = x + 2.

Key words. A linear equation gives a straight line. A quadratic equation involves a squared letter and gives a curve. Solving them simultaneously finds the points where the line meets the curve.

Method. Linear and quadratic simultaneous equations: rearrange the linear equation to make one letter the subject, substitute it into the quadratic equation, solve the resulting quadratic, then use the linear equation to find the other letter for each solution.

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² − 4
(2) y = x + 2
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = x + 2
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x + 2:
x² − 4 = x + 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 x from both sides: x² − 4 − x = 2
Subtract 2 from both sides: x² − 4 − x − 2 = 0
Collect like terms: x² − x − 6 = 0
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 = -1 and c = -6.
We need two numbers that multiply to -6 (the last number) and add to -1 (the middle number). List the pairs that multiply to -6:
1 × -6 = -6, add: 1 + -6 = -5 ✗
-1 × 6 = -6, add: -1 + 6 = 5 ✗
2 × -3 = -6, add: 2 + -3 = -1 ✓ this pair works
Put the pair into two brackets: (x + 2)(x − 3).
Check by expanding: x² − 3x + 2x − 6 = x² − x − 6. ✓
x² − x − 6 = (x + 2)(x − 3)
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 − 3) = 0
So (x − 3) = 0 or (x + 2) = 0.
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 + 2 = 0
Subtract 2 from both sides: x = −2
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −2 or x = 3
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x + 2 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −2: y = (−2) + 2 = 0
When x = 3: y = 3 + 2 = 5
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −2, y = 0 and x = 3, y = 5

Check. (−2, 0) in equation (1): x-part (−2)² − 4 = 4 − 4 = 0; this equals 0, the y value. ✓
(3, 5) in equation (1): x-part 3² − 4 = 9 − 4 = 5; this equals 5, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Worked example 14.11.2

Solve y = x² + x − 6 and 2x − y = 4.

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² + x − 6
(2) 2x − y = 4
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2): 2x − y = 4
Subtract 2x from both sides: −y = −2x + 4
Multiply both sides by −1 (change every sign): y = 2x − 4
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with 2x − 4:
x² + x − 6 = 2x − 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 2x from both sides: x² + x − 6 − 2x = −4
Add 4 to both sides: x² + x − 6 − 2x + 4 = 0
Collect like terms: x² − x − 2 = 0
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 = -1 and c = -2.
We need two numbers that multiply to -2 (the last number) and add to -1 (the middle number). List the pairs that multiply to -2:
1 × -2 = -2, add: 1 + -2 = -1 ✓ this pair works
Put the pair into two brackets: (x + 1)(x − 2).
Check by expanding: x² − 2x + 1x − 2 = x² − x − 2. ✓
x² − x − 2 = (x + 1)(x − 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 + 1)(x − 2) = 0
So (x − 2) = 0 or (x + 1) = 0.
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 + 1 = 0
Subtract 1 from both sides: x = −1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −1 or x = 2
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = 2x − 4 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −1: y = 2 × (−1) − 4 = −6
When x = 2: y = 2 × 2 − 4 = 0
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −1, y = −6 and x = 2, y = 0

Check. (−1, −6) in equation (1): x-part (−1)² + (−1) − 6 = 1 − 1 − 6 = −6; this equals −6, the y value. ✓
(2, 0) in equation (1): x-part 2² + 2 − 6 = 4 + 2 − 6 = 0; this equals 0, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Worked example 14.11.3

Solve x² + y² = 25 and y = x + 1.

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 25
(2) y = x + 1
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = x + 1
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x + 1. Put the expression in brackets:
x² + (x + 1)² = 25
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 + 1)(x + 1):
First: x × x = x²
Outer: x × 1 = x
Inner: 1 × x = x
Last: 1 × 1 = 1
so (x + 1)(x + 1) = x² + 2x + 1
The equation is now: 2x² + 2x + 1 = 25
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 25 from both sides: 2x² + 2x + 1 − 25 = 0
Collect like terms: 2x² + 2x − 24 = 0
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 2x², 2x, −24. The biggest number and the highest power of x that divide into all of them is 2.
Take it outside a bracket, and inside the bracket put each term divided by it:
2x² ÷ 2 = x²
2x ÷ 2 = x
−24 ÷ 2 = −12
So 2x² + 2x − 24 = 2(x² + x − 12).
Check by expanding: 2 × (x² + x − 12) = 2 × x², 2 × x, 2 × (−12) → 2x² + 2x − 24. ✓
Can the bracket be factorised again? Look at (x² + x − 12).
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 1 and c = -12.
We need two numbers that multiply to -12 (the last number) and add to 1 (the middle number). List the pairs that multiply to -12:
1 × -12 = -12, add: 1 + -12 = -11 ✗
-1 × 12 = -12, add: -1 + 12 = 11 ✗
2 × -6 = -12, add: 2 + -6 = -4 ✗
-2 × 6 = -12, add: -2 + 6 = 4 ✗
3 × -4 = -12, add: 3 + -4 = -1 ✗
-3 × 4 = -12, add: -3 + 4 = 1 ✓ this pair works
Put the pair into two brackets: (x − 3)(x + 4).
Check by expanding: x² + 4x − 3x − 12 = x² + x − 12. ✓
2x² + 2x − 24 = 2(x − 3)(x + 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.
2(x − 3)(x + 4) = 0
So (x − 3) = 0 or (x + 4) = 0.
The number 2 in front can never be 0, so it gives no solution. Ignore it.
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 + 4 = 0
Subtract 4 from both sides: x = −4
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −4 or x = 3
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x + 1 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −4: y = (−4) + 1 = −3
When x = 3: y = 3 + 1 = 4
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −4, y = −3 and x = 3, y = 4

Check. (−4, −3) in equation (1): (−4)² + (−3)² = 16 + 9 = 25. ✓
(3, 4) in equation (1): 3² + 4² = 9 + 16 = 25. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Worked example 14.11.4

Solve x² + y² = 25 and x = 2y − 5. (This time x is already the subject.)

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 25
(2) x = 2y − 5
Make x the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives x: x = 2y − 5
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace x in equation (1) with 2y − 5. Put the expression in brackets:
(2y − 5)² + y² = 25
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 (2y − 5)(2y − 5):
First: 2y × 2y = 4y²
Outer: 2y × (−5) = −10y
Inner: (−5) × 2y = −10y
Last: (−5) × (−5) = 25
so (2y − 5)(2y − 5) = 4y² − 20y + 25
The equation is now: 5y² − 20y + 25 = 25
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 25 from both sides: 5y² − 20y + 25 − 25 = 0
Collect like terms: 5y² − 20y = 0
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 5y², −20y. The biggest number and the highest power of y that divide into all of them is 5y.
Take it outside a bracket, and inside the bracket put each term divided by it:
5y² ÷ 5y = y
−20y ÷ 5y = −4
So 5y² − 20y = 5y(y − 4).
Check by expanding: 5y × (y − 4) = 5y × y, 5y × (−4) → 5y² − 20y. ✓
5y² − 20y = 5y(y − 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.
5y(y − 4) = 0
So (y − 4) = 0 or y = 0.
The number 5 in front can never be 0, so it gives no solution. Ignore it.
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.
y − 4 = 0
Add 4 to both sides: y = 4

y = 0
y = 0 is a solution.
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
y = 0 or y = 4
Find the other coordinate for each solution. We have the y values. Put each one into the linear equation x = 2y − 5 (never into the curve, which could give extra wrong pairs) to find the matching x value.
When y = 0: x = 2 × 0 − 5 = −5
When y = 4: x = 2 × 4 − 5 = 3
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −5, y = 0 and x = 3, y = 4

Check. (−5, 0) in equation (1): (−5)² + 0² = 25 + 0 = 25. ✓
(3, 4) in equation (1): 3² + 4² = 9 + 16 = 25. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Worked example 14.11.5

Show that the line y = 2x + 1 touches the curve y = x² + 2, and find the point where it touches.

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² + 2
(2) y = 2x + 1
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = 2x + 1
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with 2x + 1:
x² + 2 = 2x + 1
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 + 2 − 2x = 1
Subtract 1 from both sides: x² + 2 − 2x − 1 = 0
Collect like terms: x² − 2x + 1 = 0
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 = 1.
We need two numbers that multiply to 1 (the last number) and add to -2 (the middle number). List the pairs that multiply to 1:
1 × 1 = 1, add: 1 + 1 = 2 ✗
-1 × -1 = 1, add: -1 + -1 = -2 ✓ this pair works
Put the pair into two brackets: (x − 1)(x − 1).
Check by expanding: x² − 1x − 1x + 1 = x² − 2x + 1. ✓
x² − 2x + 1 = (x − 1)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 − 1)² = 0
So (x − 1) = 0.
A bracket that appears twice gives the same answer twice, so there is only one solution from it (a repeated root).
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 − 1 = 0
Add 1 to both sides: x = 1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = 1
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = 2x + 1 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = 1: y = 2 × 1 + 1 = 3
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = 1, y = 3

Check. (1, 3) in equation (1): x-part 1² + 2 = 1 + 2 = 3; this equals 3, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Watch out

The solutions are pairs. Do not write the two x values and then two unrelated y values. Also remember to find y from the linear equation, and to expand a squared bracket properly, for example (x − 3)² = x² − 6x + 9.

Exercise 14K Linear and quadratic simultaneous equations

Non-calculator

FSolve each pair of simultaneous equations.

(a)y = x² and y = x + 6

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x²
(2) y = x + 6
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = x + 6
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x + 6:
x² = x + 6
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 x from both sides: x² − x = 6
Subtract 6 from both sides: x² − x − 6 = 0
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 = -1 and c = -6.
We need two numbers that multiply to -6 (the last number) and add to -1 (the middle number). List the pairs that multiply to -6:
1 × -6 = -6, add: 1 + -6 = -5 ✗
-1 × 6 = -6, add: -1 + 6 = 5 ✗
2 × -3 = -6, add: 2 + -3 = -1 ✓ this pair works
Put the pair into two brackets: (x + 2)(x − 3).
Check by expanding: x² − 3x + 2x − 6 = x² − x − 6. ✓
x² − x − 6 = (x + 2)(x − 3)
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 − 3) = 0
So (x − 3) = 0 or (x + 2) = 0.
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 + 2 = 0
Subtract 2 from both sides: x = −2
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −2 or x = 3
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x + 6 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −2: y = (−2) + 6 = 4
When x = 3: y = 3 + 6 = 9
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −2, y = 4 and x = 3, y = 9

Check. (−2, 4) in equation (1): x-part (−2)² = 4; this equals 4, the y value. ✓
(3, 9) in equation (1): x-part 3² = 9; this equals 9, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

(b)y = x² + 2x and y = 3x + 2

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² + 2x
(2) y = 3x + 2
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = 3x + 2
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with 3x + 2:
x² + 2x = 3x + 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 3x from both sides: x² + 2x − 3x = 2
Subtract 2 from both sides: x² + 2x − 3x − 2 = 0
Collect like terms: x² − x − 2 = 0
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 = -1 and c = -2.
We need two numbers that multiply to -2 (the last number) and add to -1 (the middle number). List the pairs that multiply to -2:
1 × -2 = -2, add: 1 + -2 = -1 ✓ this pair works
Put the pair into two brackets: (x + 1)(x − 2).
Check by expanding: x² − 2x + 1x − 2 = x² − x − 2. ✓
x² − x − 2 = (x + 1)(x − 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 + 1)(x − 2) = 0
So (x − 2) = 0 or (x + 1) = 0.
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 + 1 = 0
Subtract 1 from both sides: x = −1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −1 or x = 2
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = 3x + 2 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −1: y = 3 × (−1) + 2 = −1
When x = 2: y = 3 × 2 + 2 = 8
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −1, y = −1 and x = 2, y = 8

Check. (−1, −1) in equation (1): x-part (−1)² + 2 × (−1) = 1 − 2 = −1; this equals −1, the y value. ✓
(2, 8) in equation (1): x-part 2² + 2 × 2 = 4 + 4 = 8; this equals 8, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

(c)y = x² − 5x + 4 and y = x − 1

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² − 5x + 4
(2) y = x − 1
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = x − 1
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x − 1:
x² − 5x + 4 = x − 1
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 x from both sides: x² − 5x + 4 − x = −1
Add 1 to both sides: x² − 5x + 4 − x + 1 = 0
Collect like terms: x² − 6x + 5 = 0
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 = 5.
We need two numbers that multiply to 5 (the last number) and add to -6 (the middle number). List the pairs that multiply to 5:
1 × 5 = 5, add: 1 + 5 = 6 ✗
-1 × -5 = 5, add: -1 + -5 = -6 ✓ this pair works
Put the pair into two brackets: (x − 1)(x − 5).
Check by expanding: x² − 5x − 1x + 5 = x² − 6x + 5. ✓
x² − 6x + 5 = (x − 1)(x − 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 − 1)(x − 5) = 0
So (x − 5) = 0 or (x − 1) = 0.
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 − 1 = 0
Add 1 to both sides: x = 1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = 1 or x = 5
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x − 1 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = 1: y = 1 − 1 = 0
When x = 5: y = 5 − 1 = 4
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = 1, y = 0 and x = 5, y = 4

Check. (1, 0) in equation (1): x-part 1² − 5 × 1 + 4 = 1 − 5 + 4 = 0; this equals 0, the y value. ✓
(5, 4) in equation (1): x-part 5² − 5 × 5 + 4 = 25 − 25 + 4 = 4; this equals 4, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

(d)y = x² + x − 2 and y = 2x + 4

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² + x − 2
(2) y = 2x + 4
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = 2x + 4
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with 2x + 4:
x² + x − 2 = 2x + 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 2x from both sides: x² + x − 2 − 2x = 4
Subtract 4 from both sides: x² + x − 2 − 2x − 4 = 0
Collect like terms: x² − x − 6 = 0
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 = -1 and c = -6.
We need two numbers that multiply to -6 (the last number) and add to -1 (the middle number). List the pairs that multiply to -6:
1 × -6 = -6, add: 1 + -6 = -5 ✗
-1 × 6 = -6, add: -1 + 6 = 5 ✗
2 × -3 = -6, add: 2 + -3 = -1 ✓ this pair works
Put the pair into two brackets: (x + 2)(x − 3).
Check by expanding: x² − 3x + 2x − 6 = x² − x − 6. ✓
x² − x − 6 = (x + 2)(x − 3)
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 − 3) = 0
So (x − 3) = 0 or (x + 2) = 0.
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 + 2 = 0
Subtract 2 from both sides: x = −2
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −2 or x = 3
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = 2x + 4 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −2: y = 2 × (−2) + 4 = 0
When x = 3: y = 2 × 3 + 4 = 10
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −2, y = 0 and x = 3, y = 10

Check. (−2, 0) in equation (1): x-part (−2)² + (−2) − 2 = 4 − 2 − 2 = 0; this equals 0, the y value. ✓
(3, 10) in equation (1): x-part 3² + 3 − 2 = 9 + 3 − 2 = 10; this equals 10, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

(e)y = x² − 4x + 3 and x + y = 3

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² − 4x + 3
(2) x + y = 3
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2): x + y = 3
Subtract x from both sides: y = −x + 3
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with −x + 3:
x² − 4x + 3 = −x + 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.
Add x to both sides: x² − 4x + 3 + x = 3
Subtract 3 from both sides: x² − 4x + 3 + x − 3 = 0
Collect like terms: x² − 3x = 0
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, −3x. 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
−3x ÷ x = −3
So x² − 3x = x(x − 3).
Check by expanding: x × (x − 3) = x × x, x × (−3) → x² − 3x. ✓
x² − 3x = x(x − 3)
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 − 3) = 0
So (x − 3) = 0 or x = 0.
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 = 0
x = 0 is a solution.
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = 0 or x = 3
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = −x + 3 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = 0: y = −0 + 3 = 3
When x = 3: y = −3 + 3 = 0
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = 0, y = 3 and x = 3, y = 0

Check. (0, 3) in equation (1): x-part 0² − 4 × 0 + 3 = 0 + 0 + 3 = 3; this equals 3, the y value. ✓
(3, 0) in equation (1): x-part 3² − 4 × 3 + 3 = 9 − 12 + 3 = 0; this equals 0, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

(f)y = 2x² − 3 and y = x

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = 2x² − 3
(2) y = x
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = x
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x:
2x² − 3 = x
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 x from both sides: 2x² − 3 − x = 0
Collect like terms: 2x² − x − 3 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. A quadratic where the number in front of x² is 2, not 1. We use the split the middle term method.
Multiply the first and last numbers: 2 × (-3) = -6. We need two numbers that multiply to -6 and add to -1 (the middle number).
1 × -6 = -6, add: -5 ✗
-1 × 6 = -6, add: 5 ✗
2 × -3 = -6, add: -1 ✓ this pair works
Split the middle term (-1x) into 2x and -3x: 2x² + 2x − 3x − 3.
Factorise each pair separately. First pair: 2x² + 2x = 2x(x + 1). Second pair: −3x − 3 = −3(x + 1).
The two brackets (x + 1) are the same. That is how we know it has worked. Write: 2x(x + 1) − 3(x + 1).
Take the common bracket out: (x + 1) multiplied by (2x − 3).
Result: (x + 1)(2x − 3).
Check by expanding: (x + 1)(2x − 3) = 2x² − x − 3. ✓
2x² − x − 3 = (x + 1)(2x − 3)
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 + 1)(2x − 3) = 0
So (x + 1) = 0 or (2x − 3) = 0.
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 + 1 = 0
Subtract 1 from both sides: x = −1

2x − 3 = 0
Add 3 to both sides: 2x = 3
Divide both sides by 2: x = 3 ÷ 2 = 32
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −1 or x = 32
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −1: y = (−1) = −1
When x = 32: y = 32 = 32
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −1, y = −1 and x = 32, y = 32

Check. (−1, −1) in equation (1): x-part 2 × (−1)² − 3 = 2 − 3 = −1; this equals −1, the y value. ✓
(32, 32) in equation (1): x-part 2 × 32² − 3 = 92 − 3 = 32; this equals 32, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Done

FSolve the simultaneous equations
y = x² − 2x − 8
y = x + 2

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² − 2x − 8
(2) y = x + 2
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = x + 2
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x + 2:
x² − 2x − 8 = x + 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 x from both sides: x² − 2x − 8 − x = 2
Subtract 2 from both sides: x² − 2x − 8 − x − 2 = 0
Collect like terms: x² − 3x − 10 = 0
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 ✓ 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)
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 − 5) = 0 or (x + 2) = 0.
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 + 2 = 0
Subtract 2 from both sides: x = −2
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −2 or x = 5
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x + 2 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −2: y = (−2) + 2 = 0
When x = 5: y = 5 + 2 = 7
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −2, y = 0 and x = 5, y = 7

Check. (−2, 0) in equation (1): x-part (−2)² − 2 × (−2) − 8 = 4 + 4 − 8 = 0; this equals 0, the y value. ✓
(5, 7) in equation (1): x-part 5² − 2 × 5 − 8 = 25 − 10 − 8 = 7; this equals 7, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Done

FSolve the simultaneous equations
y = x² − 3x − 4
3x − y = 4

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² − 3x − 4
(2) 3x − y = 4
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2): 3x − y = 4
Subtract 3x from both sides: −y = −3x + 4
Multiply both sides by −1 (change every sign): y = 3x − 4
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with 3x − 4:
x² − 3x − 4 = 3x − 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 3x from both sides: x² − 3x − 4 − 3x = −4
Add 4 to both sides: x² − 3x − 4 − 3x + 4 = 0
Collect like terms: x² − 6x = 0
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, −6x. 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
−6x ÷ x = −6
So x² − 6x = x(x − 6).
Check by expanding: x × (x − 6) = x × x, x × (−6) → x² − 6x. ✓
x² − 6x = x(x − 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(x − 6) = 0
So (x − 6) = 0 or x = 0.
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 = 0
x = 0 is a solution.
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = 0 or x = 6
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = 3x − 4 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = 0: y = 3 × 0 − 4 = −4
When x = 6: y = 3 × 6 − 4 = 14
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = 0, y = −4 and x = 6, y = 14

Check. (0, −4) in equation (1): x-part 0² − 3 × 0 − 4 = 0 + 0 − 4 = −4; this equals −4, the y value. ✓
(6, 14) in equation (1): x-part 6² − 3 × 6 − 4 = 36 − 18 − 4 = 14; this equals 14, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Done

FSolve the simultaneous equations
y = x² − 6
x + y = 0

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² − 6
(2) x + y = 0
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2): x + y = 0
Subtract x from both sides: y = −x
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with −x:
x² − 6 = −x
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.
Add x to both sides: x² − 6 + x = 0
Collect like terms: x² + x − 6 = 0
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 = 1 and c = -6.
We need two numbers that multiply to -6 (the last number) and add to 1 (the middle number). List the pairs that multiply to -6:
1 × -6 = -6, add: 1 + -6 = -5 ✗
-1 × 6 = -6, add: -1 + 6 = 5 ✗
2 × -3 = -6, add: 2 + -3 = -1 ✗
-2 × 3 = -6, add: -2 + 3 = 1 ✓ this pair works
Put the pair into two brackets: (x − 2)(x + 3).
Check by expanding: x² + 3x − 2x − 6 = x² + x − 6. ✓
x² + x − 6 = (x − 2)(x + 3)
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 + 3) = 0
So (x − 2) = 0 or (x + 3) = 0.
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 + 3 = 0
Subtract 3 from both sides: x = −3
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −3 or x = 2
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = −x (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −3: y = −(−3) = 3
When x = 2: y = −2 = −2
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −3, y = 3 and x = 2, y = −2

Check. (−3, 3) in equation (1): x-part (−3)² − 6 = 9 − 6 = 3; this equals 3, the y value. ✓
(2, −2) in equation (1): x-part 2² − 6 = 4 − 6 = −2; this equals −2, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Done

FSolve each pair of simultaneous equations. Each is a circle and a straight line.

(a)x² + y² = 25 and y = x + 1

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 25
(2) y = x + 1
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = x + 1
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x + 1. Put the expression in brackets:
x² + (x + 1)² = 25
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 + 1)(x + 1):
First: x × x = x²
Outer: x × 1 = x
Inner: 1 × x = x
Last: 1 × 1 = 1
so (x + 1)(x + 1) = x² + 2x + 1
The equation is now: 2x² + 2x + 1 = 25
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 25 from both sides: 2x² + 2x + 1 − 25 = 0
Collect like terms: 2x² + 2x − 24 = 0
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 2x², 2x, −24. The biggest number and the highest power of x that divide into all of them is 2.
Take it outside a bracket, and inside the bracket put each term divided by it:
2x² ÷ 2 = x²
2x ÷ 2 = x
−24 ÷ 2 = −12
So 2x² + 2x − 24 = 2(x² + x − 12).
Check by expanding: 2 × (x² + x − 12) = 2 × x², 2 × x, 2 × (−12) → 2x² + 2x − 24. ✓
Can the bracket be factorised again? Look at (x² + x − 12).
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 1 and c = -12.
We need two numbers that multiply to -12 (the last number) and add to 1 (the middle number). List the pairs that multiply to -12:
1 × -12 = -12, add: 1 + -12 = -11 ✗
-1 × 12 = -12, add: -1 + 12 = 11 ✗
2 × -6 = -12, add: 2 + -6 = -4 ✗
-2 × 6 = -12, add: -2 + 6 = 4 ✗
3 × -4 = -12, add: 3 + -4 = -1 ✗
-3 × 4 = -12, add: -3 + 4 = 1 ✓ this pair works
Put the pair into two brackets: (x − 3)(x + 4).
Check by expanding: x² + 4x − 3x − 12 = x² + x − 12. ✓
2x² + 2x − 24 = 2(x − 3)(x + 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.
2(x − 3)(x + 4) = 0
So (x − 3) = 0 or (x + 4) = 0.
The number 2 in front can never be 0, so it gives no solution. Ignore it.
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 + 4 = 0
Subtract 4 from both sides: x = −4
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −4 or x = 3
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x + 1 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −4: y = (−4) + 1 = −3
When x = 3: y = 3 + 1 = 4
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −4, y = −3 and x = 3, y = 4

Check. (−4, −3) in equation (1): (−4)² + (−3)² = 16 + 9 = 25. ✓
(3, 4) in equation (1): 3² + 4² = 9 + 16 = 25. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

(b)x² + y² = 25 and x + y = 7

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 25
(2) x + y = 7
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2): x + y = 7
Subtract x from both sides: y = −x + 7
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with −x + 7. Put the expression in brackets:
x² + (−x + 7)² = 25
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 + 7)(−x + 7):
First: (−x) × (−x) = x²
Outer: (−x) × 7 = −7x
Inner: 7 × (−x) = −7x
Last: 7 × 7 = 49
so (−x + 7)(−x + 7) = x² − 14x + 49
The equation is now: 2x² − 14x + 49 = 25
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 25 from both sides: 2x² − 14x + 49 − 25 = 0
Collect like terms: 2x² − 14x + 24 = 0
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 2x², −14x, 24. The biggest number and the highest power of x that divide into all of them is 2.
Take it outside a bracket, and inside the bracket put each term divided by it:
2x² ÷ 2 = x²
−14x ÷ 2 = −7x
24 ÷ 2 = 12
So 2x² − 14x + 24 = 2(x² − 7x + 12).
Check by expanding: 2 × (x² − 7x + 12) = 2 × x², 2 × (−7x), 2 × 12 → 2x² − 14x + 24. ✓
Can the bracket be factorised again? Look at (x² − 7x + 12).
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 ✗
-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. ✓
2x² − 14x + 24 = 2(x − 4)(x − 3)
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.
2(x − 4)(x − 3) = 0
So (x − 4) = 0 or (x − 3) = 0.
The number 2 in front can never be 0, so it gives no solution. Ignore it.
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 − 4 = 0
Add 4 to both sides: x = 4

x − 3 = 0
Add 3 to both sides: x = 3
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = 3 or x = 4
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = −x + 7 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = 3: y = −3 + 7 = 4
When x = 4: y = −4 + 7 = 3
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = 3, y = 4 and x = 4, y = 3

Check. (3, 4) in equation (1): 3² + 4² = 9 + 16 = 25. ✓
(4, 3) in equation (1): 4² + 3² = 16 + 9 = 25. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

(c)x² + y² = 10 and y = 3x

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 10
(2) y = 3x
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = 3x
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with 3x. Put the expression in brackets:
x² + (3x)² = 10
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 3x3x:
3x × 3x = 9x²
so 3x3x = 9x²
The equation is now: 10x² = 10
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: 10x² − 10 = 0
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 10x², −10. The biggest number and the highest power of x that divide into all of them is 10.
Take it outside a bracket, and inside the bracket put each term divided by it:
10x² ÷ 10 = x²
−10 ÷ 10 = −1
So 10x² − 10 = 10(x² − 1).
Check by expanding: 10 × (x² − 1) = 10 × x², 10 × (−1) → 10x² − 10. ✓
Can the bracket be factorised again? Look at (x² − 1).
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 1 = 1², so b = 1.
Put them into the rule: (x − 1)(x + 1).
Check by expanding: (x − 1)(x + 1) = x² + 1x − 1x − 1 = x² − 1. ✓ (the middle terms cancel)
10x² − 10 = 10(x − 1)(x + 1)
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.
10(x − 1)(x + 1) = 0
So (x − 1) = 0 or (x + 1) = 0.
The number 10 in front can never be 0, so it gives no solution. Ignore it.
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 − 1 = 0
Add 1 to both sides: x = 1

x + 1 = 0
Subtract 1 from both sides: x = −1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −1 or x = 1
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = 3x (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −1: y = 3 × (−1) = −3
When x = 1: y = 3 × 1 = 3
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −1, y = −3 and x = 1, y = 3

Check. (−1, −3) in equation (1): (−1)² + (−3)² = 1 + 9 = 10. ✓
(1, 3) in equation (1): 1² + 3² = 1 + 9 = 10. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

(d)x² + y² = 13 and y = x + 1

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 13
(2) y = x + 1
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = x + 1
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x + 1. Put the expression in brackets:
x² + (x + 1)² = 13
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 + 1)(x + 1):
First: x × x = x²
Outer: x × 1 = x
Inner: 1 × x = x
Last: 1 × 1 = 1
so (x + 1)(x + 1) = x² + 2x + 1
The equation is now: 2x² + 2x + 1 = 13
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 13 from both sides: 2x² + 2x + 1 − 13 = 0
Collect like terms: 2x² + 2x − 12 = 0
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 2x², 2x, −12. The biggest number and the highest power of x that divide into all of them is 2.
Take it outside a bracket, and inside the bracket put each term divided by it:
2x² ÷ 2 = x²
2x ÷ 2 = x
−12 ÷ 2 = −6
So 2x² + 2x − 12 = 2(x² + x − 6).
Check by expanding: 2 × (x² + x − 6) = 2 × x², 2 × x, 2 × (−6) → 2x² + 2x − 12. ✓
Can the bracket be factorised again? Look at (x² + x − 6).
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 1 and c = -6.
We need two numbers that multiply to -6 (the last number) and add to 1 (the middle number). List the pairs that multiply to -6:
1 × -6 = -6, add: 1 + -6 = -5 ✗
-1 × 6 = -6, add: -1 + 6 = 5 ✗
2 × -3 = -6, add: 2 + -3 = -1 ✗
-2 × 3 = -6, add: -2 + 3 = 1 ✓ this pair works
Put the pair into two brackets: (x − 2)(x + 3).
Check by expanding: x² + 3x − 2x − 6 = x² + x − 6. ✓
2x² + 2x − 12 = 2(x − 2)(x + 3)
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.
2(x − 2)(x + 3) = 0
So (x − 2) = 0 or (x + 3) = 0.
The number 2 in front can never be 0, so it gives no solution. Ignore it.
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 + 3 = 0
Subtract 3 from both sides: x = −3
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −3 or x = 2
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x + 1 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −3: y = (−3) + 1 = −2
When x = 2: y = 2 + 1 = 3
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −3, y = −2 and x = 2, y = 3

Check. (−3, −2) in equation (1): (−3)² + (−2)² = 9 + 4 = 13. ✓
(2, 3) in equation (1): 2² + 3² = 4 + 9 = 13. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

(e)x² + y² = 20 and y = 2x

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 20
(2) y = 2x
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = 2x
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with 2x. Put the expression in brackets:
x² + (2x)² = 20
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 2x2x:
2x × 2x = 4x²
so 2x2x = 4x²
The equation is now: 5x² = 20
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 20 from both sides: 5x² − 20 = 0
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 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)
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.
5(x − 2)(x + 2) = 0
So (x − 2) = 0 or (x + 2) = 0.
The number 5 in front can never be 0, so it gives no solution. Ignore it.
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 + 2 = 0
Subtract 2 from both sides: x = −2
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −2 or x = 2
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = 2x (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −2: y = 2 × (−2) = −4
When x = 2: y = 2 × 2 = 4
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −2, y = −4 and x = 2, y = 4

Check. (−2, −4) in equation (1): (−2)² + (−4)² = 4 + 16 = 20. ✓
(2, 4) in equation (1): 2² + 4² = 4 + 16 = 20. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

(f)x² + y² = 5 and x + y = 3

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 5
(2) x + y = 3
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2): x + y = 3
Subtract x from both sides: y = −x + 3
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with −x + 3. Put the expression in brackets:
x² + (−x + 3)² = 5
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 + 3)(−x + 3):
First: (−x) × (−x) = x²
Outer: (−x) × 3 = −3x
Inner: 3 × (−x) = −3x
Last: 3 × 3 = 9
so (−x + 3)(−x + 3) = x² − 6x + 9
The equation is now: 2x² − 6x + 9 = 5
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 5 from both sides: 2x² − 6x + 9 − 5 = 0
Collect like terms: 2x² − 6x + 4 = 0
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 2x², −6x, 4. The biggest number and the highest power of x that divide into all of them is 2.
Take it outside a bracket, and inside the bracket put each term divided by it:
2x² ÷ 2 = x²
−6x ÷ 2 = −3x
4 ÷ 2 = 2
So 2x² − 6x + 4 = 2(x² − 3x + 2).
Check by expanding: 2 × (x² − 3x + 2) = 2 × x², 2 × (−3x), 2 × 2 → 2x² − 6x + 4. ✓
Can the bracket be factorised again? Look at (x² − 3x + 2).
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -3 and c = 2.
We need two numbers that multiply to 2 (the last number) and add to -3 (the middle number). List the pairs that multiply to 2:
1 × 2 = 2, add: 1 + 2 = 3 ✗
-1 × -2 = 2, add: -1 + -2 = -3 ✓ this pair works
Put the pair into two brackets: (x − 1)(x − 2).
Check by expanding: x² − 2x − 1x + 2 = x² − 3x + 2. ✓
2x² − 6x + 4 = 2(x − 2)(x − 1)
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.
2(x − 2)(x − 1) = 0
So (x − 2) = 0 or (x − 1) = 0.
The number 2 in front can never be 0, so it gives no solution. Ignore it.
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 − 1 = 0
Add 1 to both sides: x = 1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = 1 or x = 2
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = −x + 3 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = 1: y = −1 + 3 = 2
When x = 2: y = −2 + 3 = 1
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = 1, y = 2 and x = 2, y = 1

Check. (1, 2) in equation (1): 1² + 2² = 1 + 4 = 5. ✓
(2, 1) in equation (1): 2² + 1² = 4 + 1 = 5. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Done

FSolve the simultaneous equations
x² + y² = 25
x = 2y − 5

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 25
(2) x = 2y − 5
Make x the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives x: x = 2y − 5
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace x in equation (1) with 2y − 5. Put the expression in brackets:
(2y − 5)² + y² = 25
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 (2y − 5)(2y − 5):
First: 2y × 2y = 4y²
Outer: 2y × (−5) = −10y
Inner: (−5) × 2y = −10y
Last: (−5) × (−5) = 25
so (2y − 5)(2y − 5) = 4y² − 20y + 25
The equation is now: 5y² − 20y + 25 = 25
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 25 from both sides: 5y² − 20y + 25 − 25 = 0
Collect like terms: 5y² − 20y = 0
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 5y², −20y. The biggest number and the highest power of y that divide into all of them is 5y.
Take it outside a bracket, and inside the bracket put each term divided by it:
5y² ÷ 5y = y
−20y ÷ 5y = −4
So 5y² − 20y = 5y(y − 4).
Check by expanding: 5y × (y − 4) = 5y × y, 5y × (−4) → 5y² − 20y. ✓
5y² − 20y = 5y(y − 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.
5y(y − 4) = 0
So (y − 4) = 0 or y = 0.
The number 5 in front can never be 0, so it gives no solution. Ignore it.
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.
y − 4 = 0
Add 4 to both sides: y = 4

y = 0
y = 0 is a solution.
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
y = 0 or y = 4
Find the other coordinate for each solution. We have the y values. Put each one into the linear equation x = 2y − 5 (never into the curve, which could give extra wrong pairs) to find the matching x value.
When y = 0: x = 2 × 0 − 5 = −5
When y = 4: x = 2 × 4 − 5 = 3
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −5, y = 0 and x = 3, y = 4

Check. (−5, 0) in equation (1): (−5)² + 0² = 25 + 0 = 25. ✓
(3, 4) in equation (1): 3² + 4² = 9 + 16 = 25. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Done

FSolve the simultaneous equations
x² + y² = 74
x − y = 2

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) x² + y² = 74
(2) x − y = 2
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2): x − y = 2
Subtract x from both sides: −y = −x + 2
Multiply both sides by −1 (change every sign): y = x − 2
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x − 2. Put the expression in brackets:
x² + (x − 2)² = 74
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 − 2):
First: x × x = x²
Outer: x × (−2) = −2x
Inner: (−2) × x = −2x
Last: (−2) × (−2) = 4
so (x − 2)(x − 2) = x² − 4x + 4
The equation is now: 2x² − 4x + 4 = 74
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 74 from both sides: 2x² − 4x + 4 − 74 = 0
Collect like terms: 2x² − 4x − 70 = 0
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 2x², −4x, −70. The biggest number and the highest power of x that divide into all of them is 2.
Take it outside a bracket, and inside the bracket put each term divided by it:
2x² ÷ 2 = x²
−4x ÷ 2 = −2x
−70 ÷ 2 = −35
So 2x² − 4x − 70 = 2(x² − 2x − 35).
Check by expanding: 2 × (x² − 2x − 35) = 2 × x², 2 × (−2x), 2 × (−35) → 2x² − 4x − 70. ✓
Can the bracket be factorised again? Look at (x² − 2x − 35).
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -2 and c = -35.
We need two numbers that multiply to -35 (the last number) and add to -2 (the middle number). List the pairs that multiply to -35:
1 × -35 = -35, add: 1 + -35 = -34 ✗
-1 × 35 = -35, add: -1 + 35 = 34 ✗
5 × -7 = -35, add: 5 + -7 = -2 ✓ this pair works
Put the pair into two brackets: (x + 5)(x − 7).
Check by expanding: x² − 7x + 5x − 35 = x² − 2x − 35. ✓
2x² − 4x − 70 = 2(x − 7)(x + 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.
2(x − 7)(x + 5) = 0
So (x − 7) = 0 or (x + 5) = 0.
The number 2 in front can never be 0, so it gives no solution. Ignore it.
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 − 7 = 0
Add 7 to both sides: x = 7

x + 5 = 0
Subtract 5 from both sides: x = −5
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −5 or x = 7
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x − 2 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −5: y = (−5) − 2 = −7
When x = 7: y = 7 − 2 = 5
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −5, y = −7 and x = 7, y = 5

Check. (−5, −7) in equation (1): (−5)² + (−7)² = 25 + 49 = 74. ✓
(7, 5) in equation (1): 7² + 5² = 49 + 25 = 74. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Done

RShow that the line y = 2x + 1 meets the curve y = x² + 2 at exactly one point, and find the coordinates of that point.

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² + 2
(2) y = 2x + 1
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = 2x + 1
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with 2x + 1:
x² + 2 = 2x + 1
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 + 2 − 2x = 1
Subtract 1 from both sides: x² + 2 − 2x − 1 = 0
Collect like terms: x² − 2x + 1 = 0
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 = 1.
We need two numbers that multiply to 1 (the last number) and add to -2 (the middle number). List the pairs that multiply to 1:
1 × 1 = 1, add: 1 + 1 = 2 ✗
-1 × -1 = 1, add: -1 + -1 = -2 ✓ this pair works
Put the pair into two brackets: (x − 1)(x − 1).
Check by expanding: x² − 1x − 1x + 1 = x² − 2x + 1. ✓
x² − 2x + 1 = (x − 1)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 − 1)² = 0
So (x − 1) = 0.
A bracket that appears twice gives the same answer twice, so there is only one solution from it (a repeated root).
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 − 1 = 0
Add 1 to both sides: x = 1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = 1
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = 2x + 1 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = 1: y = 2 × 1 + 1 = 3
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = 1, y = 3

Check. (1, 3) in equation (1): x-part 1² + 2 = 1 + 2 = 3; this equals 3, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Done

RShow that the line y = 2x − 3 does not meet the curve y = x² + 1.

Key words. The discriminant is b² − 4ac. A surd is a square root that is not a whole number, like √5. An exact answer keeps the surd; a decimal answer is rounded.

Method. Rearrange to ax² + bx + c = 0, read off a, b, c, substitute them (in brackets) into the quadratic formula, work out the discriminant first, then give the answer in the form asked.

Write the equations and substitute. The line is already y = 2x − 3, so replace y in the curve’s equation by 2x − 3. If the line meets the curve, the y values are equal.
x² + 1 = 2x − 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 2x from both sides: x² + 1 − 2x = −3
Add 3 to both sides: x² + 1 − 2x + 3 = 0
Collect like terms: x² − 2x + 4 = 0
Read off a, b and c. a is the number in front of x², b is the number in front of x, and c is the number on its own. Each number keeps its sign (a minus sign belongs to the number after it). If there is no number in front of a letter, it is 1.
x² − 2x + 4 = 0
a = 1, b = −2, c = 4
Write down the quadratic formula. This formula works for every quadratic equation. (It is proved by completing the square in general; see the start of this section.)
x = −b ± √(b² − 4ac)2a
Substitute a, b and c, using brackets. Brackets stop sign mistakes, especially when a number is negative, because (−5)² = 25 but −5² would be −25.
x = −(−2) ± √((−2)² − 4 × 1 × 4)2 × 1
Work out the pieces one at a time. The part under the square root, b² − 4ac, is called the discriminant. Work it out separately and carefully; it decides what kind of answer we get.
b² = (−2)² = 4
4ac = 4 × 1 × 4 = 16
b² − 4ac = 4 − 16 = −12
−b = −(−2) = 2
2a = 2 × 1 = 2
Look at the discriminant. You cannot take the square root of a negative number and get an ordinary (real) number, because a number multiplied by itself is never negative.
−12 is negative, so there are no real solutions.
Conclude. The equation has no real solutions, so there is no point that is on both the line and the curve.
The line does not meet the curve.

Check. The line y = 2x − 3 is always below the curve: at x = 1 the curve is at 2 and the line is at −1. ✓

Watch out. Put every number in brackets when you substitute. The whole top line is divided by 2a, not just the square root.

Done

PSFind the coordinates of the points where the line and the curve meet.
y = x² − x − 5
y = x + 3

Write the equations and number them. One equation is a curve (it has a squared term) and one is a straight line. The solutions are the points where they meet.
(1) y = x² − x − 5
(2) y = x + 3
Make y the subject of the linear equation. Rearrange the straight-line equation (the easier one) so that one letter is on its own. Then we can substitute it into the curve.
Equation (2) already gives y: y = x + 3
Substitute into equation (1). Now there is only one letter. The result is a quadratic equation, which we can solve by factorising.
Replace y in equation (1) with x + 3:
x² − x − 5 = x + 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 x from both sides: x² − x − 5 − x = 3
Subtract 3 from both sides: x² − x − 5 − x − 3 = 0
Collect like terms: x² − 2x − 8 = 0
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 = -8.
We need two numbers that multiply to -8 (the last number) and add to -2 (the middle number). List the pairs that multiply to -8:
1 × -8 = -8, add: 1 + -8 = -7 ✗
-1 × 8 = -8, add: -1 + 8 = 7 ✗
2 × -4 = -8, add: 2 + -4 = -2 ✓ this pair works
Put the pair into two brackets: (x + 2)(x − 4).
Check by expanding: x² − 4x + 2x − 8 = x² − 2x − 8. ✓
x² − 2x − 8 = (x + 2)(x − 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 − 4) = 0
So (x − 4) = 0 or (x + 2) = 0.
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 − 4 = 0
Add 4 to both sides: x = 4

x + 2 = 0
Subtract 2 from both sides: x = −2
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −2 or x = 4
Find the other coordinate for each solution. We have the x values. Put each one into the linear equation y = x + 3 (never into the curve, which could give extra wrong pairs) to find the matching y value.
When x = −2: y = (−2) + 3 = 1
When x = 4: y = 4 + 3 = 7
Write the solutions as pairs. Each solution is a pair: an x value with its own y value. Do not mix them up.
x = −2, y = 1 and x = 4, y = 7

Check. (−2, 1) in equation (1): x-part (−2)² − (−2) − 5 = 4 + 2 − 5 = 1; this equals 1, the y value. ✓
(4, 7) in equation (1): x-part 4² − 4 − 5 = 16 − 4 − 5 = 7; this equals 7, the y value. ✓

Watch out. Give the answers as pairs. Find each second value from the linear equation: if you use the curve you may produce pairs that are not solutions.

Done

PSA rectangle has a perimeter of 26 cm and an area of 40 cm². Its length is l cm and its width is w cm. Form a pair of simultaneous equations and solve them to find the dimensions of the rectangle.

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.

Write the two equations. The perimeter is 2l + 2w = 26, so l + w = 13. The area is lw = 40. One equation is linear and one is quadratic.
(1) l + w = 13
(2) lw = 40
Substitute. From (1), w = 13 − l. Put this into (2) in brackets.
l(13 − l) = 40
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 l(−l + 13):
l × (−l) = −l²
l × 13 = 13l
so l(−l + 13) = −l² + 13l
The equation is now: −l² + 13l = 40
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.
Add l² to both sides: 13l = 40 + l²
Subtract 13l from both sides: 0 = 40 + l² − 13l
Collect like terms: 0 = l² − 13l + 40
Write it the other way round: l² − 13l + 40 = 0
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 l² + bl + c, with b = -13 and c = 40.
We need two numbers that multiply to 40 (the last number) and add to -13 (the middle number). List the pairs that multiply to 40:
1 × 40 = 40, add: 1 + 40 = 41 ✗
-1 × -40 = 40, add: -1 + -40 = -41 ✗
2 × 20 = 40, add: 2 + 20 = 22 ✗
-2 × -20 = 40, add: -2 + -20 = -22 ✗
4 × 10 = 40, add: 4 + 10 = 14 ✗
-4 × -10 = 40, add: -4 + -10 = -14 ✗
5 × 8 = 40, add: 5 + 8 = 13 ✗
-5 × -8 = 40, add: -5 + -8 = -13 ✓ this pair works
Put the pair into two brackets: (l − 5)(l − 8).
Check by expanding: l² − 8l − 5l + 40 = l² − 13l + 40. ✓
l² − 13l + 40 = (l − 5)(l − 8)
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.
(l − 5)(l − 8) = 0
So (l − 8) = 0 or (l − 5) = 0.
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.
l − 8 = 0
Add 8 to both sides: l = 8

l − 5 = 0
Add 5 to both sides: l = 5
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
l = 5 or l = 8
Find w for each solution. Use the linear equation w = 13 − l.
When l = 5: w = 13 − 5 = 8
When l = 8: w = 13 − 8 = 5
Answer the question. Both pairs are possible, and they describe the same rectangle.
The rectangle is 8 cm by 5 cm.

Check. Perimeter = 2 × (8 + 5) = 26 cm and area = 8 × 5 = 40 cm². ✓

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.

Done

RThe line y = 2x + k is a tangent to the curve y = x². Find the value of k.

Done