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Improve Tuition · Foundation Maths · Year 6–10
Algebra · Linear Equations

Solving Linear Equations

From one-step equations up to forming and solving from words.

6 lessons · 35 questions (A–G) · 20 calculation + 15 word problems · no time pressure

Solving a linear equation means finding the value of the letter that makes both sides balance, by doing the same thing to each side until the letter is on its own. Pupils most often slip by changing one side but not the other, mishandling a negative when moving a term across, or expanding a bracket wrongly before solving.

At Improve Tuition, a qualified teacher uses the balance idea to make every step make sense, in small worked steps — with read-aloud, comfort spacing and a reading tint for pupils who take in maths more easily that way.

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Lesson 1

One-step equations

Aim: Solve an equation in one step using the inverse operation.

An equation is a balance: whatever you do to one side, do to the other. To get x on its own, undo what is done to it with the inverse — add undoes subtract, multiply undoes divide.

x + 712
The scales balance because the sides are equal. Take 7 off each side to leave x = 5.
Worked example
Solve x + 7 = 12.
  1. We want x on its own. At the moment 7 is being added to it.
  2. To undo ‘+ 7’, do the opposite — subtract 7. Do it to both sides so the scales stay balanced.
  3. x + 7 − 7 = 12 − 7.
  4. On the left the +7 and −7 cancel, leaving x = 5.
  5. Check: 5 + 7 = 12. ✓
Watch out
Do the same to both sides — that keeps the balance.

Try one

Solve x − 4 = 9.

Show answer
x − 4 = 9
What is happening to x? 4 has been subtracted.
The opposite of − 4 is + 4, so add 4 to both sides:
x − 4 + 4 = 9 + 4
x = 13
Check: 13 − 4 = 9. ✓
Practise this now — Section A →
Lesson 2

Two-step equations

Aim: Solve an equation in two steps, undoing in reverse order.

When x is multiplied and something is added, undo in reverse: deal with the + or − first, then the multiply or divide.

2x + 311
Take 3 off both pans, then split the 8 between the two x’s: x = 4.
Worked example
Solve 2x + 3 = 11.
  1. Two things are happening to x: it is multiplied by 2, then 3 is added.
  2. Undo them in reverse, so deal with the ‘+ 3’ first: subtract 3 from both sides.
  3. 2x + 3 − 3 = 11 − 3, which leaves 2x = 8.
  4. Now undo the ‘× 2’ by dividing both sides by 2.
  5. x = 8 ÷ 2 = 4.
  6. Check: 2 × 4 + 3 = 8 + 3 = 11. ✓
Watch out
When solving, undo the last operation first. This is usually the opposite order to BIDMAS.

Try one

Solve 3x − 5 = 16.

Show answer
3x − 5 = 16
What is happening to x? It is multiplied by 3, then 5 is subtracted.
Undo the last step first. The opposite of − 5 is + 5, so add 5 to both sides:
3x − 5 + 5 = 16 + 5
3x = 21
Now undo × 3. The opposite is ÷ 3:
x = 21 ÷ 3
x = 7
Check: 3 × 7 − 5 = 21 − 5 = 16. ✓
Practise this now — Section B →
Lesson 3

Equations with brackets

Aim: Solve equations that contain brackets.

Two ways in: divide both sides by the number outside first, or expand the bracket and then solve. Dividing first is often quicker.

Worked example
Solve 2(x + 3) = 14.
  1. The bracket (x + 3) is all multiplied by 2, so undo that first.
  2. Divide both sides by 2: x + 3 = 7.
  3. Now only ‘+ 3’ is left on x. Undo it by subtracting 3 from both sides.
  4. x = 7 − 3 = 4.
  5. Check: 2 × (4 + 3) = 2 × 7 = 14. ✓
Watch out
If you expand instead, multiply every term inside the bracket.

Try one

Solve 3(x − 2) = 12.

Show answer
3(x − 2) = 12
The whole bracket is multiplied by 3.
Undo × 3 first by dividing both sides by 3:
x − 2 = 4
Now undo − 2 by adding 2 to both sides:
x = 6
Check: 3(6 − 2) = 3 × 4 = 12. ✓
Practise this now — Section C →
Lesson 4

Unknown on both sides

Aim: Solve equations with the unknown on both sides.

Collect the x terms on one side and the numbers on the other. Subtract the smaller x term from both sides so x stays positive, then finish as a two-step equation.

5x2x + 12
x’s sit on both pans. Take 2x off each side, leaving 3x = 12, so x = 4.
Worked example
Solve 5x = 2x + 12.
  1. x appears on both sides, so first gather all the x’s on one side.
  2. Take 2x off both sides: 5x − 2x = 12, which leaves 3x = 12.
  3. Now undo the ‘× 3’ by dividing both sides by 3.
  4. x = 12 ÷ 3 = 4.
  5. Check: 5 × 4 = 20, and 2 × 4 + 12 = 20. ✓
Watch out
Move the smaller x term to avoid a negative coefficient.

Try one

Solve 7x − 2 = 4x + 13.

Show answer
7x − 2 = 4x + 13
x is on both sides. Move the smaller x term first.
Subtract 4x from both sides:
3x − 2 = 13
Add 2 to both sides:
3x = 15
Divide both sides by 3:
x = 5
Check: 7 × 5 − 2 = 33 and 4 × 5 + 13 = 33. ✓
Practise this now — Section D →
Lesson 5

Forming equations from words

Aim: Turn a worded situation into an equation, then solve it.

Let the unknown be x. Translate the words step by step: “multiply by 3” → 3x, “then subtract 4” → 3x − 4, “equals 20” → = 20. Then solve.

Worked example
I think of a number, multiply by 3, subtract 4, and get 20. Find the number.
  1. Call the number x. ‘Multiply by 3’ gives 3x; ‘subtract 4’ gives 3x − 4; ‘gets 20’ means = 20.
  2. So the equation is 3x − 4 = 20.
  3. Undo the ‘− 4’ first: add 4 to both sides, giving 3x = 24.
  4. Undo the ‘× 3’: divide both sides by 3.
  5. x = 24 ÷ 3 = 8.
  6. Check: 3 × 8 − 4 = 24 − 4 = 20. ✓
Watch out
Read in order and build the equation one phrase at a time.

Try one

A number add 6 gives 15. Find it.

Show answer
Let the number be x. ‘Add 6’ means x + 6.
Form the equation: x + 6 = 15.
Undo + 6 by subtracting 6 from both sides:
x = 9
Check: 9 + 6 = 15. ✓
Practise this now — Section E →
Lesson 6

Forming from shapes & harder problems

Aim: Form and solve equations from perimeters, angles and multi-step problems.

Shapes give equations too: add the sides for a perimeter, or the angles to 180° in a triangle. Build the equation, then solve as before.

Worked example
A rectangle has length x and width 5. Its perimeter is 26 cm. Find x.
  1. Perimeter means all four sides added. A rectangle has two lengths and two widths, so that is 2(x + 5).
  2. Set it equal to the perimeter given: 2(x + 5) = 26.
  3. Undo the ‘× 2’ by dividing both sides by 2: x + 5 = 13.
  4. Undo the ‘+ 5’ by subtracting 5: x = 8.
  5. Check: 2 × (8 + 5) = 2 × 13 = 26 cm. ✓
Watch out
Perimeter adds all the sides — a rectangle has two of each.

Try one

An equilateral triangle has sides (x + 2) and perimeter 21 cm. Find x.

Show answer
An equilateral triangle has 3 equal sides, so perimeter = 3(x + 2).
Form the equation: 3(x + 2) = 21.
Undo × 3 by dividing both sides by 3:
x + 2 = 7
Undo + 2 by subtracting 2:
x = 5
Check: 3(5 + 2) = 3 × 7 = 21. ✓
Practise this now — Section F →
↑ Back to the lessons
Section A · One-step equations
LEARNING OBJECTIVE · GRADE 1 · CALCULATION

Solve one-step linear equations.

Success criteria — I can:
  • spot the operation on x
  • apply the inverse to both sides
  • state x
1.x + 7 = 12
Answer:
2.x − 4 = 9
Answer:
3.5x = 35
Answer:
4.x ÷ 3 = 6
Answer:
5.x + 13 = 8
Answer:
↑ Back to the lessons
Section B · Two-step equations
LEARNING OBJECTIVE · GRADE 2 · CALCULATION

Solve two-step linear equations.

Success criteria — I can:
  • undo the + or − first
  • then undo the × or ÷
  • state x
6.2x + 3 = 11
Answer:
7.3x − 5 = 16
Answer:
8.5x + 2 = 27
Answer:
9.4x − 9 = 7
Answer:
10.3x + 14 = 5
Answer:
↑ Back to the lessons
Section C · Equations with brackets
LEARNING OBJECTIVE · GRADE 3 · CALCULATION

Solve linear equations containing brackets.

Success criteria — I can:
  • divide by the number outside, or expand
  • solve the equation that remains
  • state x
11.2(x + 3) = 14
Answer:
12.3(x − 2) = 12
Answer:
13.4(x + 1) = 24
Answer:
14.5(2x − 1) = 35
Answer:
15.2(3x + 4) = 26
Answer:
↑ Back to the lessons
Section D · Unknown on both sides
LEARNING OBJECTIVE · GRADE 3 · CALCULATION

Solve equations with the unknown on both sides.

Success criteria — I can:
  • collect x terms on one side
  • collect numbers on the other
  • state x
16.5x = 3x + 8
Answer:
17.7x − 2 = 4x + 13
Answer:
18.9x + 1 = 5x + 25
Answer:
19.6x − 7 = 2x + 5
Answer:
20.8x + 4 = 3x + 24
Answer:
↑ Back to the lessons
Section E · Forming equations from words
LEARNING OBJECTIVE · GRADE 2 · WORD PROBLEM

Form and solve equations from worded situations.

Success criteria — I can:
  • let the unknown be x
  • translate the words into an equation
  • solve and answer in context
21.I think of a number, add 6, and the answer is 15. What is the number?
Answer:
22.A taxi costs £3 plus £2 for every mile. A journey cost £17. How many miles was it?
Answer:
23.Five identical books cost £40 in total. What is the price of one book?
Answer:
24.I think of a number, multiply it by 3, subtract 4, and get 20. What is the number?
Answer:
25.A plumber charges a £25 call-out plus £30 per hour. The bill was £145. How many hours did the job take?
Answer:
↑ Back to the lessons
Section F · Equations from shapes
LEARNING OBJECTIVE · GRADE 3 · WORD PROBLEM

Form and solve equations from perimeters and angles.

Success criteria — I can:
  • write an expression for the total
  • set it equal to the given value
  • solve for x
26.A rectangle has length x and width 5 cm. Its perimeter is 26 cm. Find x.
Answer:
27.An equilateral triangle has sides of length (x + 2) cm and a perimeter of 21 cm. Find x.
Answer:
28.A rectangle is twice as long as it is wide. Its perimeter is 30 cm. Find the width.
Answer:
29.A square has sides of length (2x − 1) cm and a perimeter of 28 cm. Find x.
Answer:
30.The three angles in a triangle are x, x and (x + 30) degrees. Find x.
Answer:
↑ Back to the lessons
Section G · Multi-step problems
LEARNING OBJECTIVE · GRADE 4 · WORD PROBLEM

Form and solve multi-step equations.

Success criteria — I can:
  • define the unknown clearly
  • build the equation from the conditions
  • solve and check it is sensible
31.I think of a number, double it and add 5. The result equals the number plus 12. Find the number.
Answer:
32.Sam is 4 years older than Tom. Together they are 26. How old is Tom?
Answer:
33.A father is 3 times as old as his son. Their ages add to 48. How old is the son?
Answer:
34.I think of a number, multiply by 4 and add 3. This equals the number multiplied by 6 then subtract 7. Find the number.
Answer:
35.A rectangle's perimeter is 40 cm. Its length is 4 cm more than its width. Find the width.
Answer:
0 of 35 answered
Guide

How do you solve a linear equation?

A linear equation has a letter to find and an equals sign, such as 3x + 4 = 19. Solving it means working out the value of the letter that makes both sides equal — here, x = 5.

The method is to keep the equation balanced: whatever you do to one side, do to the other. Undo it step by step — first remove anything added or subtracted, then anything multiplying or dividing — until the letter stands alone. With brackets, expand first; with the letter on both sides, collect it onto one side.

Common mistakes. Changing one side but not the other, getting a sign wrong when moving a term across, expanding a bracket incorrectly, or dividing only part of a side.

Common questions
What does it mean to solve an equation?
It means finding the value of the unknown letter that makes both sides of the equation equal. For 2x = 10, the solution is x = 5.
What is the balance method?
You treat the equation like a balanced scale: whatever you do to one side, you do the same to the other, so it stays equal while you work towards the letter on its own.
How do you solve an equation with brackets?
Expand the bracket first by multiplying everything inside by the term outside, then solve the resulting equation as normal.
How do you solve an equation with the letter on both sides?
Collect the letter terms onto one side and the numbers onto the other by adding or subtracting the same thing from both sides, then solve.
How can you check the solution to an equation?
Substitute your answer back into the original equation. If both sides come out equal, the solution is correct.
Stuck on this topic?

A teacher can find the exact gap

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