GCSE Maths Quadratic and Simultaneous Equations (Foundation): check-up and exam-style questions

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? (3) is it a quadratic with three terms?
   x² + 9x + 20
2. Factorise. Follow the method.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 9 and c = 20.
   We need two numbers that multiply to 20 (the last number) and add to 9 (the middle number). List the pairs that multiply to 20:
   1 × 20 = 20,   add: 1 + 20 = 21 ✗
   -1 × -20 = 20,   add: -1 + -20 = -21 ✗
   2 × 10 = 20,   add: 2 + 10 = 12 ✗
   -2 × -10 = 20,   add: -2 + -10 = -12 ✗
   4 × 5 = 20,   add: 4 + 5 = 9 ✓ this pair works
   Put the pair into two brackets: (x + 4)(x + 5).
   Check by expanding: x² + 5x + 4x + 20 = x² + 9x + 20. ✓

Check. Expand your answer: (x + 4)(x + 5) should give x² + 9x + 20 again. ✓

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

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? (3) is it a quadratic with three terms?
   x² − 81
2. Factorise. Follow the method.
   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)

Check. Expand your answer: (x − 9)(x + 9) should give x² − 81 again. ✓

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

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

Check. When x = −2: left side = (−2)² − 3 × (−2) − 10 = 4 + 6 − 10 = 0. It equals 0. ✓
When x = 5: left side = 5² − 3 × 5 − 10 = 25 − 15 − 10 = 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.

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 + 4x = 21
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 21 from both sides: x² + 4x − 21 = 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 = 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)
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 + 7) = 0
   So (x − 3) = 0 or (x + 7) = 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
   Add 3 to both sides: x = 3
   
   x + 7 = 0
   Subtract 7 from both sides: x = −7
6. 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.

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)   x + y = 11
   (2)   x − 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 1 and 1: they are already the same size, so no multiplying is needed.
3. Subtract (2) from (1). The x terms are x and x. They have the same sign, so subtract the equations: x − x = 0, so x disappears. Do the same thing to the left sides and to the right sides.
   (1)   x + y = 11
   (2)   x − y = 3
   x terms: x − x = 0 (they cancel)
   y terms: y − (−y) = 2y
   Numbers: 11 − 3 = 8
   2y = 8
4. Solve for y. 2y means 2 multiplied by y, so divide both sides by 2.
   Divide both sides by 2: y = 8 ÷ 2 = 4
5. Substitute y = 4 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): x + y = 11
   x + 4 = 11
   x + 4 = 11
   Subtract 4 from both sides: x = 7
6. Write the solution. Give both values. They belong together as a pair.
   x = 7, y = 4

Check. Put x = 7, y = 4 into the other equation (equation (2), which we did not use to find x): 7 − 4 = 7 − 4 = 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.

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 − y = 10
2. Choose a letter to eliminate: x, and make the numbers match. The numbers in front of x are 2 and 4. The lowest common multiple of 2 and 4 is 4, so we will make both equal to 4. 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 x 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 × 3y on the left and 2 × 12 on the right.
   (3)   4x + 6y = 24
4. Subtract (2) from (3). The x terms are 4x and 4x. They have the same sign, so subtract the equations: 4x − 4x = 0, so x disappears. Do the same thing to the left sides and to the right sides.
   (3)   4x + 6y = 24
   (2)   4x − y = 10
   x terms: 4x − 4x = 0 (they cancel)
   y terms: 6y − (−y) = 7y
   Numbers: 24 − 10 = 14
   7y = 14
5. Solve for y. 7y means 7 multiplied by y, so divide both sides by 7.
   Divide both sides by 7: y = 14 ÷ 7 = 2
6. Substitute y = 2 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): 2x + 3y = 12
   2x + 3 × 2 = 12
   2x + 6 = 12
   Subtract 6 from both sides: 2x = 6
   Divide both sides by 2: x = 6 ÷ 2 = 3
7. 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 x): 4 × 3 − 2 = 12 − 2 = 10. This equals 10, 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.

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)   x + y = 8
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): x + y = 8
   Subtract x from both sides: y = 8 − x
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): −2x + y = −1
   −2x + (8 − x) = −1
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: (8 − x) = −x + 8
   The equation becomes: −x + 8 − 2x = −1
   Collect like terms: −3x + 8 = −1
5. Solve the equation for x. It is now a one-letter equation. Use inverse operations on both sides.
   Subtract 8 from both sides: −3x = −9
   Divide both sides by −3: x = −9 ÷ (−3) = 3
6. Substitute x = 3 into the rearranged equation to find y. The rearranged equation y = 8 − x is the easiest to use.
   y = 8 − x
   y = 8 − 3
   y = 5
7. Write the solution. Give both values as a pair.
   x = 3, y = 5

Check. Put x = 3, y = 5 into equation (2): 3 + 5 = 3 + 5 = 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.

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 + 4) cm.
2. Form an equation from the information. Area of a rectangle = length × width.
   x(x + 4) = 45
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 = 45
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 45 from both sides: x² + 4x − 45 = 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 = -45.
   We need two numbers that multiply to -45 (the last number) and add to 4 (the middle number). List the pairs that multiply to -45:
   1 × -45 = -45,   add: 1 + -45 = -44 ✗
   -1 × 45 = -45,   add: -1 + 45 = 44 ✗
   3 × -15 = -45,   add: 3 + -15 = -12 ✗
   -3 × 15 = -45,   add: -3 + 15 = 12 ✗
   5 × -9 = -45,   add: 5 + -9 = -4 ✗
   -5 × 9 = -45,   add: -5 + 9 = 4 ✓ this pair works
   Put the pair into two brackets: (x − 5)(x + 9).
   Check by expanding: x² + 9x − 5x − 45 = x² + 4x − 45. ✓
   x² + 4x − 45 = (x − 5)(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 − 5)(x + 9) = 0
   So (x − 5) = 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 − 5 = 0
   Add 5 to both sides: x = 5
   
   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 = 5
9. Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −9.
   x = 5 cm
10. Find the length and answer the question. The length is x + 4.
    Length = 5 + 4 = 9 cm
    Width = 5 cm, length = 9 cm

Check. Check in the words: width 5 cm and length 9 cm differ by 4 cm and the area is 5 × 9 = 45 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).

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, 5).
3. Write the solution. The x-coordinate is the value of x and the y-coordinate is the value of y.
   x = 3, y = 5

Check. Put x = 3 into both equations. y = x + 2: y = 3 + 2 = 5. y = 8 − x: y = −3 + 8 = 5. Both give y = 5. ✓

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.

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 £24.
   Total cost in pounds = (2 × a) + (3 × c).
   2a + 3c = 24
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   3 adult tickets and 1 child ticket cost £22.
   Total cost in pounds = (3 × a) + (1 × c).
   3a + c = 22
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   2a + 3c = 24
   (2)   3a + c = 22
5. Choose a letter to eliminate: c, and make the numbers match. The numbers in front of c are 3 and 1. The lowest common multiple of 3 and 1 is 3, so we will make both equal to 3. So multiply equation (2) by 3. Multiply every term, including the number on the right-hand side.
6. Multiply to make the numbers in front of c match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
   Equation (2) × 3: 3 × 3a, 3 × c on the left and 3 × 22 on the right.
   (3)   9a + 3c = 66
7. Subtract (1) from (3). The c terms are 3c and 3c. They have the same sign, so subtract the equations: 3c − 3c = 0, so c disappears. Do the same thing to the left sides and to the right sides.
   (3)   9a + 3c = 66
   (1)   2a + 3c = 24
   c terms: 3c − 3c = 0 (they cancel)
   a terms: 9a − 2a = 7a
   Numbers: 66 − 24 = 42
   7a = 42
8. Solve for a. 7a means 7 multiplied by a, so divide both sides by 7.
   Divide both sides by 7: a = 42 ÷ 7 = 6
9. Substitute a = 6 into equation (2) to find c. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (2): 3a + c = 22
   3 × 6 + c = 22
   c + 18 = 22
   Subtract 18 from both sides: c = 4
10. Write the solution. Give both values. They belong together as a pair.
    a = 6, c = 4
11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Adult ticket: a = 6, so one adult ticket costs £6.
    Child ticket: c = 4, so one child ticket costs £4.
    One adult ticket costs £6 and one child ticket costs £4.

Check. Put a = 6, c = 4 into the other equation (equation (1), which we did not use to find c): 2 × 6 + 3 × 4 = 12 + 12 = 24. This equals 24, 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.

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? (3) is it a quadratic with three terms?
   x² + x − 20
2. Factorise. Follow the method.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 1 and c = -20.
   We need two numbers that multiply to -20 (the last number) and add to 1 (the middle number). List the pairs that multiply to -20:
   1 × -20 = -20,   add: 1 + -20 = -19 ✗
   -1 × 20 = -20,   add: -1 + 20 = 19 ✗
   2 × -10 = -20,   add: 2 + -10 = -8 ✗
   -2 × 10 = -20,   add: -2 + 10 = 8 ✗
   4 × -5 = -20,   add: 4 + -5 = -1 ✗
   -4 × 5 = -20,   add: -4 + 5 = 1 ✓ this pair works
   Put the pair into two brackets: (x − 4)(x + 5).
   Check by expanding: x² + 5x − 4x − 20 = x² + x − 20. ✓

Check. Expand your answer: (x − 4)(x + 5) should give x² + x − 20 again. ✓

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

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? (3) is it a quadratic with three terms?
   4x² − 25
2. Factorise. Follow the method.
   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. 4x² = (2x)², so a = 2x. And 25 = 5², so b = 5.
   Put them into the rule: (2x − 5)(2x + 5).
   Check by expanding: (2x − 5)(2x + 5) = 4x² + 52x − 52x − 25 = 4x² − 25. ✓ (the middle terms cancel)

Check. Expand your answer: (2x − 5)(2x + 5) should give 4x² − 25 again. ✓

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

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 − 14 = 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 = -5 and c = -14.
   We need two numbers that multiply to -14 (the last number) and add to -5 (the middle number). List the pairs that multiply to -14:
   1 × -14 = -14,   add: 1 + -14 = -13 ✗
   -1 × 14 = -14,   add: -1 + 14 = 13 ✗
   2 × -7 = -14,   add: 2 + -7 = -5 ✓ this pair works
   Put the pair into two brackets: (x + 2)(x − 7).
   Check by expanding: x² − 7x + 2x − 14 = x² − 5x − 14. ✓
   x² − 5x − 14 = (x + 2)(x − 7)
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 − 7) = 0
   So (x − 7) = 0 or (x + 2) = 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 − 7 = 0
   Add 7 to both sides: x = 7
   
   x + 2 = 0
   Subtract 2 from both sides: x = −2
6. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
   x = −2 or x = 7

Check. When x = −2: left side = (−2)² − 5 × (−2) − 14 = 4 + 10 − 14 = 0. It equals 0. ✓
When x = 7: left side = 7² − 5 × 7 − 14 = 49 − 35 − 14 = 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.

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 = 28
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 28 from both sides: x² + 3x − 28 = 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 = -28.
   We need two numbers that multiply to -28 (the last number) and add to 3 (the middle number). List the pairs that multiply to -28:
   1 × -28 = -28,   add: 1 + -28 = -27 ✗
   -1 × 28 = -28,   add: -1 + 28 = 27 ✗
   2 × -14 = -28,   add: 2 + -14 = -12 ✗
   -2 × 14 = -28,   add: -2 + 14 = 12 ✗
   4 × -7 = -28,   add: 4 + -7 = -3 ✗
   -4 × 7 = -28,   add: -4 + 7 = 3 ✓ this pair works
   Put the pair into two brackets: (x − 4)(x + 7).
   Check by expanding: x² + 7x − 4x − 28 = x² + 3x − 28. ✓
   x² + 3x − 28 = (x − 4)(x + 7)
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 − 4)(x + 7) = 0
   So (x − 4) = 0 or (x + 7) = 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 − 4 = 0
   Add 4 to both sides: x = 4
   
   x + 7 = 0
   Subtract 7 from both sides: x = −7
6. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
   x = −7 or x = 4

Check. When x = −7: left side = (−7)² + 3 × (−7) = 49 − 21 = 28; right side = 28 = 28. Both sides equal. ✓
When x = 4: left side = 4² + 3 × 4 = 16 + 12 = 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.

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 + 5) = 24
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 + 5):
   x × x = x²
   x × 5 = 5x
   so x(x + 5) = x² + 5x
   The equation is now: x² + 5x = 24
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 24 from both sides: x² + 5x − 24 = 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 = 5 and c = -24.
   We need two numbers that multiply to -24 (the last number) and add to 5 (the middle number). List the pairs that multiply to -24:
   1 × -24 = -24,   add: 1 + -24 = -23 ✗
   -1 × 24 = -24,   add: -1 + 24 = 23 ✗
   2 × -12 = -24,   add: 2 + -12 = -10 ✗
   -2 × 12 = -24,   add: -2 + 12 = 10 ✗
   3 × -8 = -24,   add: 3 + -8 = -5 ✗
   -3 × 8 = -24,   add: -3 + 8 = 5 ✓ this pair works
   Put the pair into two brackets: (x − 3)(x + 8).
   Check by expanding: x² + 8x − 3x − 24 = x² + 5x − 24. ✓
   x² + 5x − 24 = (x − 3)(x + 8)
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 + 8) = 0
   So (x − 3) = 0 or (x + 8) = 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 + 8 = 0
   Subtract 8 from both sides: x = −8
7. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
   x = −8 or x = 3

Check. When x = −8: left side = (−8)² + 5 × (−8) = 64 − 40 = 24; right side = 24 = 24. Both sides equal. ✓
When x = 3: left side = 3² + 5 × 3 = 9 + 15 = 24; right side = 24 = 24. 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.

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 = 13
   (2)   5x − 2y = 11
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 2 and −2: they are already the same size, so no multiplying is needed.
3. Add (1) 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.
   (1)   3x + 2y = 13
   (2)   5x − 2y = 11
   y terms: 2y + (−2y) = 0 (they cancel)
   x terms: 3x + 5x = 8x
   Numbers: 13 + 11 = 24
   8x = 24
4. Solve for x. 8x means 8 multiplied by x, so divide both sides by 8.
   Divide both sides by 8: x = 24 ÷ 8 = 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): 3x + 2y = 13
   3 × 3 + 2y = 13
   2y + 9 = 13
   Subtract 9 from both sides: 2y = 4
   Divide both sides by 2: y = 4 ÷ 2 = 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): 5 × 3 − 2 × 2 = 15 − 4 = 11. This equals 11, 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.

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 = 1
   (2)   3x + 2y = 4
2. Choose a letter to eliminate: x, and make the numbers match. The numbers in front of x 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 x 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 × 2x, 3 × 3y on the left and 3 × 1 on the right.
   (3)   6x + 9y = 3
   
   Equation (2) × 2: 2 × 3x, 2 × 2y on the left and 2 × 4 on the right.
   (4)   6x + 4y = 8
4. Subtract (4) from (3). The x terms are 6x and 6x. They have the same sign, so subtract the equations: 6x − 6x = 0, so x disappears. Do the same thing to the left sides and to the right sides.
   (3)   6x + 9y = 3
   (4)   6x + 4y = 8
   x terms: 6x − 6x = 0 (they cancel)
   y terms: 9y − 4y = 5y
   Numbers: 3 − 8 = −5
   5y = −5
5. Solve for y. 5y means 5 multiplied by y, so divide both sides by 5.
   Divide both sides by 5: y = −5 ÷ 5 = −1
6. Substitute y = −1 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): 2x + 3y = 1
   2x + 3 × (−1) = 1
   2x − 3 = 1
   Add 3 to both sides: 2x = 4
   Divide both sides by 2: x = 4 ÷ 2 = 2
7. Write the solution. Give both values. They belong together as a pair.
   x = 2, y = −1

Check. Put x = 2, y = −1 into the other equation (equation (2), which we did not use to find x): 3 × 2 + 2 × (−1) = 6 − 2 = 4. This equals 4, 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.

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 + 5) cm.
2. Form an equation from the information. Area of a rectangle = length × width.
   x(x + 5) = 84
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 + 5):
   x × x = x²
   x × 5 = 5x
   so x(x + 5) = x² + 5x
   The equation is now: x² + 5x = 84
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 84 from both sides: x² + 5x − 84 = 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 = 5 and c = -84.
   We need two numbers that multiply to -84 (the last number) and add to 5 (the middle number). List the pairs that multiply to -84:
   1 × -84 = -84,   add: 1 + -84 = -83 ✗
   -1 × 84 = -84,   add: -1 + 84 = 83 ✗
   2 × -42 = -84,   add: 2 + -42 = -40 ✗
   -2 × 42 = -84,   add: -2 + 42 = 40 ✗
   3 × -28 = -84,   add: 3 + -28 = -25 ✗
   -3 × 28 = -84,   add: -3 + 28 = 25 ✗
   4 × -21 = -84,   add: 4 + -21 = -17 ✗
   -4 × 21 = -84,   add: -4 + 21 = 17 ✗
   6 × -14 = -84,   add: 6 + -14 = -8 ✗
   -6 × 14 = -84,   add: -6 + 14 = 8 ✗
   7 × -12 = -84,   add: 7 + -12 = -5 ✗
   -7 × 12 = -84,   add: -7 + 12 = 5 ✓ this pair works
   Put the pair into two brackets: (x − 7)(x + 12).
   Check by expanding: x² + 12x − 7x − 84 = x² + 5x − 84. ✓
   x² + 5x − 84 = (x − 7)(x + 12)
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 − 7)(x + 12) = 0
   So (x − 7) = 0 or (x + 12) = 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 − 7 = 0
   Add 7 to both sides: x = 7
   
   x + 12 = 0
   Subtract 12 from both sides: x = −12
8. Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
   x = −12 or x = 7
9. Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −12.
   x = 7 cm
10. Find the length and answer the question. The length is x + 5.
    Length = 7 + 5 = 12 cm
    Width = 7 cm, length = 12 cm

Check. Check in the words: width 7 cm and length 12 cm differ by 5 cm and the area is 7 × 12 = 84 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).

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 £33.
   Total cost in pounds = (2 × a) + (3 × c).
   2a + 3c = 33
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   1 adult ticket and 4 child tickets cost £29.
   Total cost in pounds = (1 × a) + (4 × c).
   a + 4c = 29
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   2a + 3c = 33
   (2)   a + 4c = 29
5. Choose a letter to eliminate: a, and make the numbers match. The numbers in front of a 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 a 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 × a, 2 × 4c on the left and 2 × 29 on the right.
   (3)   2a + 8c = 58
7. Subtract (1) from (3). The a terms are 2a and 2a. They have the same sign, so subtract the equations: 2a − 2a = 0, so a disappears. Do the same thing to the left sides and to the right sides.
   (3)   2a + 8c = 58
   (1)   2a + 3c = 33
   a terms: 2a − 2a = 0 (they cancel)
   c terms: 8c − 3c = 5c
   Numbers: 58 − 33 = 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 (2) 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 (2): a + 4c = 29
   a + 4 × 5 = 29
   a + 20 = 29
   Subtract 20 from both sides: a = 9
10. Write the solution. Give both values. They belong together as a pair.
    a = 9, c = 5
11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Adult ticket: a = 9, so one adult ticket costs £9.
    Child ticket: c = 5, so one child ticket costs £5.
    One adult ticket costs £9 and one child ticket costs £5.

Check. Put a = 9, c = 5 into the other equation (equation (1), which we did not use to find a): 2 × 9 + 3 × 5 = 18 + 15 = 33. This equals 33, 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.

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 (4, 5).
3. Write the solution. The x-coordinate is the value of x and the y-coordinate is the value of y.
   x = 4, y = 5

Check. Put x = 4 into both equations. y = x + 1: y = 4 + 1 = 5. y = 9 − x: y = −4 + 9 = 5. Both give y = 5. ✓

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.

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? (3) is it a quadratic with three terms?
   x² + 7x + 10
2. Factorise. Follow the method.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 7 and c = 10.
   We need two numbers that multiply to 10 (the last number) and add to 7 (the middle number). List the pairs that multiply to 10:
   1 × 10 = 10,   add: 1 + 10 = 11 ✗
   -1 × -10 = 10,   add: -1 + -10 = -11 ✗
   2 × 5 = 10,   add: 2 + 5 = 7 ✓ this pair works
   Put the pair into two brackets: (x + 2)(x + 5).
   Check by expanding: x² + 5x + 2x + 10 = x² + 7x + 10. ✓

Check. Expand your answer: (x + 2)(x + 5) should give x² + 7x + 10 again. ✓

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

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 − x − 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 = -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 ✓ this pair works
   Put the pair into two brackets: (x + 3)(x − 4).
   Check by expanding: x² − 4x + 3x − 12 = x² − x − 12. ✓
   x² − x − 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 − 4) = 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 − 4 = 0
   Add 4 to both sides: x = 4
   
   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 = 4

Check. When x = −3: left side = (−3)² − (−3) − 12 = 9 + 3 − 12 = 0. It equals 0. ✓
When x = 4: left side = 4² − 4 − 12 = 16 − 4 − 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.

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)   x + y = 9
   (2)   x − y = 5
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 1 and 1: they are already the same size, so no multiplying is needed.
3. Subtract (2) from (1). The x terms are x and x. They have the same sign, so subtract the equations: x − x = 0, so x disappears. Do the same thing to the left sides and to the right sides.
   (1)   x + y = 9
   (2)   x − y = 5
   x terms: x − x = 0 (they cancel)
   y terms: y − (−y) = 2y
   Numbers: 9 − 5 = 4
   2y = 4
4. Solve for y. 2y means 2 multiplied by y, so divide both sides by 2.
   Divide both sides by 2: y = 4 ÷ 2 = 2
5. Substitute y = 2 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): x + y = 9
   x + 2 = 9
   x + 2 = 9
   Subtract 2 from both sides: x = 7
6. Write the solution. Give both values. They belong together as a pair.
   x = 7, y = 2

Check. Put x = 7, y = 2 into the other equation (equation (2), which we did not use to find x): 7 − 2 = 7 − 2 = 5. This equals 5, 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.