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KS3 Maths · Year 7Chapter 6 · Solving equations and using formulae
Year 7 · Chapter 6

Solving equations and using formulae

By the end of this chapter you will be able to

  • Solve a one-step or two-step equation, by doing the same to both sides
  • Solve an equation with the unknown on both sides, and an equation with brackets or fractions
  • Write an equation from a word problem, and solve it
  • Substitute values into a formula, and rearrange a formula to change its subject
  • Check a solution by putting it back into the original equation

You should already know

  • Substitution and the vocabulary of algebra (Chapter 4)
  • Expanding brackets and collecting like terms (Chapter 5)
  • Inverse operations: addition and subtraction, multiplication and division (Chapter 1)

Where this leads

  • Solving equations with negative and fractional solutions in Year 8
  • Simultaneous equations and quadratics in Years 8 and 9

National Curriculum for England, Key Stage 3 mathematics, Year 7. Content: A5 understand and use standard formulae; rearrange formulae to change the subject, A6 model situations by translating them into algebraic expressions, formulae and equations, A7 solve linear equations in one variable, including equations with the unknown on both sides, brackets and fractions.

Exercise P  Do you remember?

Non-calculator
1
FWork out 2 × 6 + 1.
2
FWhat number goes in the box? □ × 3 = 12
3
FWhat number goes in the box? □ + 8 = 15
4
FWork out 3(2 + 4).
5
FWork out 20 ÷ 4 + 1.
6.1

One-step and two-step equations

Year 7
A7
Video lesson: One-step and two-step equationsComing soon
Ahmed has some money in his bank account. After putting in £15 he has £40. The equation for this is x + 15 = 40, where x is the amount he had at the start. Solving it tells us what x is.

An equation has an equals sign, with the same value on each side. To solve it, do the same thing to both sides until the letter is on its own. Use the inverse (opposite) operation: subtraction undoes addition, and division undoes multiplication.

Key ideas

  • Do the same to both sides to keep the equation balanced.
  • Addition is undone by subtraction; multiplication by division.
  • For two steps, undo the adding or subtracting first, then the multiplying or dividing.
  • Check your solution by putting it back into the original equation.
Worked example 6.1.1
Solve x + 15 = 40.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    x + 15 = 40
  2. Subtract 15 from both sides. This removes the number on the left, so the letter term is on its own.
    x = 25
  3. The letter already has no number in front, so the value is found. Read off the value.
    x = 25
  4. Write the answer. Check it by putting the value back into the original equation.
    x = 25

Check. Put x = 25 into the original equation: both sides give the same value. ✓

Watch out. Do the same to both sides. Doing it to one side only changes the equation.

Worked example 6.1.2
Solve 3x − 5 = 16.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    3x − 5 = 16
  2. Add 5 to both sides. This removes the number on the left, so the letter term is on its own.
    3x = 21
  3. Divide both sides by 3. This leaves the letter on its own.
    x = 7
  4. Write the answer. Check it by putting the value back into the original equation.
    x = 7

Check. Put x = 7 into the original equation: both sides give the same value. ✓

Watch out. Do the same to both sides. Doing it to one side only changes the equation.

Worked example 6.1.3
Solve x/4 = 6.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    x/4 = 6
  2. Multiply both sides by 4. This undoes the division by 4.
    x = 6 × 4
  3. Work out the right-hand side. Multiply 6 by 4.
    x = 24
  4. Write the answer. Check it by putting the value back into the original equation.
    x = 24

Check. Put x = 24 into the original equation: both sides give the same value. ✓

Watch out. Do the same to both sides. Doing it to one side only changes the equation.

Watch out

Watch out: in 2x + 5 = 17, subtract 5 first (2x = 12), then divide by 2 (x = 6). Do not divide by 2 first.

Exam tip

Grown-up tip. Ask the pupil to say what the inverse of each operation is, before they start. The order is always the reverse of the way the equation was built.

Exercise 6A  One-step and two-step equations

Non-calculator
1
FSolve x + 7 = 12.
2
FSolve y − 9 = 4.
3
FSolve 3a = 21.
4
FSolve b/4 = 6.
5
FSolve 2x + 5 = 17.
6
FSolve 4y − 3 = 25.
7
FSolve 5m + 8 = 3.
8
FSolve n/3 − 2 = 5.
9
FSolve 10 − 2p = 4.
10
FSolve: (a) 3x + 1 = 10 (b) 6 − 2y = −4 (c) x/5 + 4 = 6
(a)(a) 3x + 1 = 10
(b)(b) 6 − 2y = −4
(c)(c) x/5 + 4 = 6
11
PSSam thinks of a number. He multiplies it by 4, then adds 3. His answer is 23. What number did he start with?
12
RSam solves 2x = 10 and writes x = 20. Explain his mistake, and find the correct answer.
6.2

Equations with the unknown on both sides

Year 7
A7
Video lesson: Equations with the unknown on both sidesComing soon
Two cards at a fair cost 5x + 2 pence and 3x + 10 pence, where x is the number of rides. The stalls charge the same amount at one number of rides. To find it, set the two costs equal and solve.

When the letter is on both sides, collect the letter terms on one side and the number terms on the other. Choose the side with the bigger letter term to keep the coefficient positive.

Key ideas

  • Collect the letter terms on one side, and the numbers on the other.
  • Subtract the smaller letter term from both sides, so the letter stays positive.
  • Check the answer by putting it into both sides.
  • An equation with no solution has the same letter on both sides, with different numbers left over.
Worked example 6.2.1
Solve 5x + 2 = 3x + 10.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    5x + 2 = 3x + 10
  2. Get the x terms on one side: subtract 3x from both sides. Collect the letter terms on one side, so the letter is in one place.
    2x + 2 = 10
  3. Get the numbers on the other side: subtract 2 from both sides. The numbers go on the other side of the equals sign.
    2x = 8
  4. Divide both sides by 2. This leaves the letter on its own.
    x = 4
  5. Write the answer. Check it by putting the value back into the original equation.
    x = 4

Check. Put x = 4 into the original equation: both sides give the same value. ✓

Watch out. Collect the letters on one side, then the numbers. Do not move a letter to the side with a bigger number.

Worked example 6.2.2
Solve 7y − 4 = 2y + 16.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    7y − 4 = 2y + 16
  2. Get the y terms on one side: subtract 2y from both sides. Collect the letter terms on one side, so the letter is in one place.
    5y − 4 = 16
  3. Get the numbers on the other side: subtract −4 from both sides. The numbers go on the other side of the equals sign.
    5y = 20
  4. Divide both sides by 5. This leaves the letter on its own.
    y = 4
  5. Write the answer. Check it by putting the value back into the original equation.
    y = 4

Check. Put y = 4 into the original equation: both sides give the same value. ✓

Watch out. Collect the letters on one side, then the numbers. Do not move a letter to the side with a bigger number.

Worked example 6.2.3
Solve 2b + 9 = 5b.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    2b + 9 = 5b
  2. Get the b terms on one side: subtract 5b from both sides. Collect the letter terms on one side, so the letter is in one place.
    −3b + 9 = 0
  3. Get the numbers on the other side: subtract 9 from both sides. The numbers go on the other side of the equals sign.
    −3b = −9
  4. Divide both sides by −3. This leaves the letter on its own.
    b = 3
  5. Write the answer. Check it by putting the value back into the original equation.
    b = 3

Check. Put b = 3 into the original equation: both sides give the same value. ✓

Watch out. Collect the letters on one side, then the numbers. Do not move a letter to the side with a bigger number.

Watch out

Watch out: when you move a letter term to the other side, its sign changes. 5x = 3x + 10 becomes 2x = 10, not 8x = 10.

Exam tip

Grown-up tip. Ask the pupil to write the letter terms in one colour and the numbers in another, so they do not get mixed up.

Exercise 6B  Unknowns on both sides

Non-calculator
1
FSolve 5x + 2 = 3x + 10.
2
FSolve 7y − 4 = 2y + 16.
3
FSolve 4a + 3 = a + 12.
4
FSolve 2b + 9 = 5b.
5
FSolve 3x − 8 = x + 4.
6
FSolve 6p + 5 = 2p − 7.
7
FSolve 9m − 2 = 4m + 13.
8
FSolve: (a) 8y + 2 = 3y + 27 (b) 4x = 2x + 10 (c) 5 − t = 2t − 4
(a)(a) 8y + 2 = 3y + 27
(b)(b) 4x = 2x + 10
(c)(c) 5 − t = 2t − 4
9
PSCard A has the rule 3n + 1. Card B has the rule n + 9. For which number n are the two cards equal?
10
PSBox A has 5x + 3 sweets. Box B has 2x + 15 sweets. The boxes have the same number of sweets. Find x.
11
RExplain why 3x + 4 = 3x + 7 has no solution.
6.3

Brackets and fractions in equations

Year 7
A7
Video lesson: Brackets and fractions in equationsComing soon
A school trip costs £x per pupil, and the coach costs a fixed sum. A total of 3 identical groups of 4 pupils each is shared between brackets. Equations with brackets and fractions are solved by removing the brackets or fractions first.

For brackets, expand first: 3(x + 4) = 3x + 12. For a fraction, multiply both sides by the bottom number: if (x + 3)/2 = 5, then x + 3 = 10.

Key ideas

  • Expand brackets first, then solve as before.
  • To remove a fraction, multiply both sides by the number on the bottom.
  • Keep the brackets in the working until you have expanded them.
  • Check the answer in the original equation, including the brackets.
Worked example 6.3.1
Solve 3(x + 4) = 27.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    3(x + 4) = 27
  2. Divide both sides by 3. This removes the bracket without expanding.
    x + 4 = 9
  3. Subtract 4 from both sides. This leaves x on its own.
    x = 5
  4. Write the answer. Check it by putting the value back into the original equation.
    x = 5

Check. Put x = 5 into the original equation: both sides give the same value. ✓

Watch out. Do the same to both sides. Doing it to one side only changes the equation.

Worked example 6.3.2
Solve (x + 3)/2 = 5.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    (x + 3)/2 = 5
  2. Multiply both sides by 2. This removes the division by 2.
    x + 3 = 10
  3. Subtract 3 from both sides. This leaves x on its own.
    x = 7
  4. Write the answer. Check it by putting the value back into the original equation.
    x = 7

Check. Put x = 7 into the original equation: both sides give the same value. ✓

Watch out. Do the same to both sides. Doing it to one side only changes the equation.

Worked example 6.3.3
Solve x/4 + 2 = 5.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    x/4 + 2 = 5
  2. Subtract 2 from both sides. This leaves the fraction on its own.
    x/4 = 3
  3. Multiply both sides by 4. This removes the fraction.
    x = 12
  4. Write the answer. Check it by putting the value back into the original equation.
    x = 12

Check. Put x = 12 into the original equation: both sides give the same value. ✓

Watch out. Do the same to both sides. Doing it to one side only changes the equation.

Watch out

Watch out: (x + 3)/2 is not x/2 + 3. The division by 2 applies to the whole bracket.

Exam tip

Grown-up tip. Ask the pupil to check each answer by putting it back into the brackets: 3(5 + 4) = 27 is quick to check.

Exercise 6C  Brackets and fractions

Non-calculator
1
FSolve 2(x + 3) = 14.
2
FSolve 4(y − 2) = 20.
3
FSolve 3(2a + 1) = 21.
4
FSolve 5(b − 1) = 2b + 7.
5
FSolve 2(x + 5) = 3(x − 1).
6
FSolve (x + 3)/2 = 5.
7
FSolve (2y − 1)/3 = 3.
8
FSolve x/4 + 2 = 5.
9
FSolve (n − 3)/2 = n.
10
FSolve: (a) 3(x − 4) = 18 (b) (2x + 6)/4 = 4 (c) x/3 − 1 = 4
(a)(a) 3(x − 4) = 18
(b)(b) (2x + 6)/4 = 4
(c)(c) x/3 − 1 = 4
11
PSI think of a number. I double it, subtract 6, and then divide by 4. My answer is 5. What number did I start with?
12
RExplain why (x + 2)/2 is not the same as x/2 + 2. Use x = 4 to show it.
6.4

Forming and solving equations from words

Year 7
A6
Video lesson: Forming and solving equations from wordsComing soon
Maya is x years old. Her brother is 4 years older. Together their ages add up to 22. We can write this as an equation: x + (x + 4) = 22. Solving it tells us how old Maya is.

To form an equation from words, choose a letter for the unknown number, write each part of the sentence in algebra, and set the expression equal to the given value. Solve the equation, and then answer the question in words, with units.

Key ideas

  • Choose a letter for the unknown, and say what it stands for.
  • Write each operation in the words in the same order, using brackets where needed.
  • Set the expression equal to the number in the question.
  • Answer the question in words, with the correct units.
Worked example 6.4.1
Sam thinks of a number, multiplies it by 3 and adds 4. The answer is 19. Form and solve an equation.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    3n + 4 = 19
  2. Write the equation. Multiply by 3, then add 4, gives 19.
    3n + 4 = 19
  3. Subtract 4 from both sides. This undoes the adding of 4.
    3n = 15
  4. Divide both sides by 3. This undoes the multiplying by 3.
    n = 5
  5. Write the answer. Check it by putting the value back into the original equation.
    n = 5

Check. Put n = 5 into the original equation: both sides give the same value. ✓

Watch out. Do the same to both sides. Doing it to one side only changes the equation.

Worked example 6.4.2
Maya is x years old and her brother is 4 years older. Their ages add up to 22. Find Maya's age.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    x + (x + 4) = 22
  2. Simplify the left-hand side. Add the two ages.
    2x + 4 = 22
  3. Subtract 4 from both sides. This leaves the 2x term on its own.
    2x = 18
  4. Divide both sides by 2. This gives Maya's age.
    x = 9
  5. Write the answer. Check it by putting the value back into the original equation.
    x = 9

Check. Put x = 9 into the original equation: both sides give the same value. ✓

Watch out. Do the same to both sides. Doing it to one side only changes the equation.

Worked example 6.4.3
A rectangle has length (x + 3) cm and width x cm. Its perimeter is 26 cm. Find x.

Method. Do the same to both sides of the equation, so that it stays balanced, until the letter is on its own.

  1. Write the equation. This is the equation we need to solve.
    2(x + 3) + 2x = 26
  2. Expand the bracket. The perimeter is two lengths and two widths.
    2x + 6 + 2x = 26
  3. Collect the x terms. Add the two x terms.
    4x + 6 = 26
  4. Subtract 6, then divide by 4. This leaves x on its own.
    x = 5
  5. Write the answer. Check it by putting the value back into the original equation.
    x = 5

Check. Put x = 5 into the original equation: both sides give the same value. ✓

Watch out. Do the same to both sides. Doing it to one side only changes the equation.

Watch out

Watch out: "4 years older" means add 4, not multiply. If the brother is 4 years older, his age is x + 4.

Exam tip

Grown-up tip. Ask the pupil to say the sentence in words first, then write the equation. Check the answer makes sense as an age, or a number of sweets.

Exercise 6D  Forming equations

Non-calculator
1
PSI think of a number, multiply it by 5 and subtract 3. The answer is 22. What is my number?
2
PSThree consecutive whole numbers add up to 42. Let the smallest be n. Find the three numbers.
3
PSA ticket costs £x. Four tickets cost £28. Find the cost of one ticket.
4
PSA pencil costs p pence. Three pencils and a 45p ruler cost 180p. Find the cost of one pencil.
5
PSTom has three times as many marbles as Sam. Together they have 36 marbles. Let Sam have m marbles. How many marbles does Sam have?
6
PSThe angles of a triangle are x, 2x and 3x degrees. The angles add up to 180°. Find x.
7
PSA rectangle has length (x + 4) cm and width x cm. Its perimeter is 32 cm. Find x.
8
PSA bike hire costs £10 plus £4 for each hour. The total is £34. Find the number of hours, h.
9
FHalf of a number, plus 7, is 15. Find the number.
10
FWrite and solve an equation for each: (a) Double a number and take away 9. The answer is 13. (b) Three less than a number is 10.
(a)(a) Double a number and take away 9 is 13
(b)(b) Three less than a number is 10
11
PSPriya is 3 years older than her sister. Their ages add up to 33. How old is the sister?
12
RWhen I add 7 to a number and double the answer, I get 30. Write an equation and explain how to solve it.
6.5

Formulae and changing the subject

Year 7
A5
Video lesson: Formulae and changing the subjectComing soon
The formula F = 9C/5 + 32 changes a temperature from degrees Celsius (C) to degrees Fahrenheit (F). To find F for a given C, substitute. To find C for a given F, change the subject of the formula.

A formula links quantities with letters. The subject is the letter on its own on one side, such as F in F = 9C/5 + 32. To change the subject, move the letters and numbers across by doing the inverse operation to both sides.

Key ideas

  • Substitute: put in the values and work out the answer.
  • Change the subject: do the inverse operations to both sides, in the reverse order.
  • Example: P = 2l + 2w. Subtract 2w, then divide by 2: l = (P − 2w)/2.
  • Check a rearranged formula by putting in numbers, both ways.
Worked example 6.5.1
Use v = u + at to find v when u = 2, a = 5 and t = 4.

Method. Change the subject by doing the same to both sides, until the wanted letter is on its own.

  1. Write the formula. This is the starting formula.
    v = u + at
  2. Substitute the values. Put each letter with its value in brackets.
    v = 2 + 5 × 4
  3. Work out the multiplication. Multiplication comes before addition.
    5 × 4 = 20
  4. Work out the addition. Add 2.
    v = 2 + 20 = 22
  5. Write the answer. The wanted letter is now on its own.
    22

Check. Check: 2 + 5 × 4 = 22. ✓

Watch out. Do not add 2 and 5 first.

Worked example 6.5.2
Make l the subject of P = 2l + 2w.

Method. Change the subject by doing the same to both sides, until the wanted letter is on its own.

  1. Write the formula. This is the starting formula.
    P = 2l + 2w
  2. Subtract 2w from both sides. This removes the term with w from the right-hand side.
    P − 2w = 2l
  3. Divide both sides by 2. This leaves l on its own.
    (P − 2w)/2 = l
  4. Write the answer. The wanted letter is now on its own.
    l = (P − 2w)/2

Check. Check with P = 10 and w = 2: l = (10 − 4)/2 = 3, and 2(3) + 2(2) = 10. ✓

Watch out. Write the bracket (P − 2w) on the top, so that the whole of P − 2w is divided by 2.

Worked example 6.5.3
Make t the subject of v = u + at.

Method. Change the subject by doing the same to both sides, until the wanted letter is on its own.

  1. Write the formula. This is the starting formula.
    v = u + at
  2. Subtract u from both sides. This removes u from the right-hand side.
    v − u = at
  3. Divide both sides by a. This leaves t on its own.
    (v − u)/a = t
  4. Write the answer. The wanted letter is now on its own.
    t = (v − u)/a

Check. Check with u = 2, a = 5, t = 4: v = 22, and (22 − 2)/5 = 4. ✓

Watch out. Do not divide only the v. The whole of v − u is divided by a.

Watch out

Watch out: to make x the subject of y = 3x + 4, subtract 4 and then divide by 3: x = (y − 4)/3. Do not write x = y − 4/3.

Exam tip

Grown-up tip. Ask the pupil to pick numbers, substitute them, and check both the original and the rearranged formula.

Exercise 6E  Formulae and changing the subject

Non-calculator
1
FFind A = lw when l = 6 and w = 4.
2
FUse v = u + at to find v when u = 2, a = 5 and t = 4.
3
FThe formula is F = 9C/5 + 32. Find F when C = 25.
4
FThe formula is P = 2l + 2w. Make l the subject.
5
FMake x the subject of y = 3x + 4.
6
FMake a the subject of t = a/5.
7
FMake C the subject of F = 9C/5 + 32.
8
FMake m the subject of p = m − 7.
9
PSUse v = u + at to find t when v = 20, u = 5 and a = 3.
10
FMake the letter in brackets the subject: (a) x = 4y (b) a = 2b + c (in terms of the letter b)
(a)(a) x = 4y, make y the subject
(b)(b) a = 2b + c, make b the subject
11
PSThe cost of hiring a van is C = 40 + 15d pounds, where d is the number of days. Chris pays £130. How many days does he hire it for?
12
RSam says that if y = 3x + 4, then x = (y + 4)/3. Is he right? Explain.

Chapter summary

SkillMethod
One and two stepsUndo the operations in reverse order, doing the same to both sides
Unknown on both sidesCollect the letters on one side and the numbers on the other
Brackets and fractionsExpand the brackets, or multiply both sides by the bottom number
Forming equationsChoose a letter, write each part of the words, set equal to the value, solve, answer in words
Substituting into a formulaPut each value in, then work out using BIDMAS
Changing the subjectUse inverse operations on both sides; the whole of a term on one side is divided or subtracted together

Can you…?

  • I can solve one-step and two-step equations, and check the answer
  • I can solve an equation with the unknown on both sides
  • I can solve equations with brackets and fractions
  • I can write an equation from a word problem, and solve it
  • I can substitute values into a formula
  • I can change the subject of a formula

Key terms

equation
a statement with an equals sign, which is true for only some values of the letter
solve
find the value of the letter that makes an equation true
inverse operation
the operation that undoes another, such as subtraction for addition
formula
a rule that links quantities using letters, such as P = 2l + 2w
subject
the letter on its own on one side of a formula
consecutive
next to each other in order, such as 13, 14 and 15

Exercise 6Z  Check-up

Non-calculator
1
Solve x − 6 = 10.
2
Solve 5y + 2 = 22.
3
Solve 3(a − 2) = 12.
4
Solve 2b + 7 = b + 11.
5
Use s = vt to find s when v = 12 and t = 2.5.
6
Make x the subject of y = 2x + 5.
7
Solve (m + 1)/4 = 3.
8
A number is doubled and then 3 is added. The answer is 17. Find the number.
9
Solve 8 − 3p = 2.
10
Solve 6x + 1 = 4x + 9.

Total 21 marks. Allow about 21 minutes.

Exercise 6X  Practice test questions

Non-calculator
1
FSolve 4x + 3 = 19.[1 mark]
2
FSolve y/5 = 7.[1 mark]
3
FSolve 3(x + 2) = 24.[2 marks]
4
FSolve 7m − 2 = 3m + 10.[2 marks]
5
FThe cost of hiring a van is C = 40 + 15d pounds. Find C when d = 4.[2 marks]
6
FSolve (2x − 5)/3 = 3.[2 marks]
7
FMake x the subject of y = 5x − 2.[2 marks]
8
FThree consecutive whole numbers add up to 60. Let the smallest be n. Find the largest number.[3 marks]
9
FA rectangle has length 3x cm and width (x + 2) cm. Its perimeter is 28 cm. Find x.[3 marks]
10
FAli has £x. Ben has 4 times as much as Ali. Together they have £45. How much does Ben have?[3 marks]
← Chapter 5: Simplifying, expanding and factorisingChapter 7: Sequences and graphs →

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