Solving Linear Equations

Solving linear equations with one or two steps, the unknown on both sides, brackets, and one or two fractions.

In the exam for Edexcel IGCSE Maths 4MA1, Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024 · 173 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

What Solving an Equation Means

Solving means finding the number x stands for. An equation is a balance: the two sides are equal.

Rule

Whatever you do to one side, do to the other. Then the balance stays level.

Example

Think of wrapping a present: box, then paper, then ribbon. To open it you undo in reverse: ribbon first, box last. Solving undoes what was done to x, last thing first.

Key idea

Answers are often fractions or decimals, like x = 5.5. That is fine. Don't round unless you are told to.

One and Two Steps

Undo each operation with its opposite: $+$ with $-$ , $\times$ with $\div$ .

Worked example

1
$4 x + 5 = 27$ — The question: x was multiplied by 4, then 5 was added.
2
$4 x + 5 - 5 = 27 - 5$ — Undo the +5 first: take 5 from both sides.
3
$4 x = 22$ — 27 − 5 = 22. The 5 is gone.
4
$\frac{4 x}{4} = \frac{22}{4}$ — Undo the ×4: divide both sides by 4.
5
$x = \frac{11}{2} = 5.5$ — 22 ÷ 4 = 5.5. A decimal answer is fine.

One step

\frac{x}{6} = 3

x = 18

Undo ÷6: multiply by 6.

Watch the minus

19 - k = 4

19 = 4 + k

k = 15

Add k to both sides, then take 4.

Common mistake

Writing $x = 22 - 4 = 18$ . The 4 multiplies x, so you divide by 4.

x on Both Sides

Many 3-mark questions have x on both sides, often with a bracket. Expand first, then gather the x terms on one side.

Worked example

1
$5 (x + 3) = 3 x - 4$ — The question. x is on both sides, and the left has a bracket.
2
$5 x + 15 = 3 x - 4$ — Expand the bracket first: 5 × x = 5x and 5 × 3 = 15.
3
$2 x + 15 = - 4$ — Take 3x from both sides. 3x is the smaller x term, so the x's stay on the left only.
4
$2 x = - 19$ — Take 15 from both sides: −4 − 15 = −19.
5
$x = - \frac{19}{2} = - 9.5$ — Divide both sides by 2. A negative answer is fine.

Rule

Take the smaller x term away from both sides. If the smaller one has a minus, like −3x, taking it away means adding 3x.

Common mistake

Changing a sign on one side only. Every move is “do the same to both sides”.

A Minus in Front of x

Exams often give x a minus sign, like 13 − 2x = 6. Add the x term to both sides, so x becomes positive.

Worked example

1
$13 - 2 x = 6$ — The question. The x term −2x has a minus.
2
$13 - 2 x + 2 x = 6 + 2 x$ — Add 2x to both sides. This clears the minus.
3
$13 = 6 + 2 x$ — −2x + 2x = 0. Now x is on the right, and it is positive.
4
$7 = 2 x$ — Take 6 from both sides: 13 − 6 = 7.
5
$x = 3.5$ — Divide both sides by 2: 7 ÷ 2 = 3.5.
6
$13 - 2 (3.5) = 13 - 7 = 6$ — Check: put 3.5 back in. We get 6, the same as the right side, so it is correct.

x on both sides, one with a minus

Worked example

1
$14 - 3 p = 2 + 5 p$ — The question. The smaller p term is −3p, because −3 is less than 5.
2
$14 = 2 + 8 p$ — Add 3p to both sides: −3p + 3p = 0 and 5p + 3p = 8p.
3
$12 = 8 p$ — Take 2 from both sides: 14 − 2 = 12.
4
$p = 1.5$ — Divide both sides by 8: 12 ÷ 8 = 1.5.

Common mistake

Stopping at $- 2 x = - 7$ . Divide both sides by −2: $(- 7) \div (- 2) = 3.5$ . A minus divided by a minus is a plus.

One Fraction

If x is inside a fraction, clear the fraction first: multiply both sides by the bottom number.

Worked example

1
$\frac{5 x - 1}{3} = 2 x + 4$ — The question. 5x − 1 is divided by 3, so clear the 3 first.
2
$3 \times \frac{5 x - 1}{3} = 3 (2 x + 4)$ — Multiply both sides by 3. Keep the right side in a bracket, so the 3 multiplies all of it.
3
$5 x - 1 = 6 x + 12$ — The 3s cancel on the left. On the right, 3 × 2x = 6x and 3 × 4 = 12.
4
$- 1 = x + 12$ — Take 5x, the smaller x term, from both sides.
5
$x = - 13$ — Take 12 from both sides: −1 − 12 = −13.
6
$\frac{5 ( - 13 ) - 1}{3} = \frac{- 66}{3} = - 22$ — Check the left side. The right side is 2(−13) + 4 = −22 too, so both sides match.

Key idea

Always check by putting your answer back into both sides. If they match, you are right. It is the quickest way to catch a sign slip.

Common mistake

Writing $3 (2 x + 4)$ as $6 x + 4$ . The 3 multiplies every term on that side.

Two Fractions (Extended/Higher)

Two fractions are worth 3–4 marks. Multiply every term by a number both bottoms go into.

Worked example

1
$\frac{x + 1}{3} - \frac{x - 2}{4} = 2$ — The question. 3 and 4 both go into 12, so 12 clears both bottoms.
2
$4 (x + 1) - 3 (x - 2) = 24$ — Multiply every term by 12: 12 ÷ 3 = 4, 12 ÷ 4 = 3, and 2 × 12 = 24. Keep the brackets.
3
$4 x + 4 - 3 x + 6 = 24$ — Expand. The minus in front multiplies too: (−3) × (−2) = +6.
4
$x + 10 = 24$ — Collect: 4x − 3x = x and 4 + 6 = 10.
5
$x = 14$ — Take 10 from both sides.

When the right side is a decimal

Worked example

1
$\frac{x + 2}{4} - \frac{5 - x}{5} = 1.3$ — The question. 4 and 5 both go into 20. The 1.3 is a term too, so it gets multiplied.
2
$5 (x + 2) - 4 (5 - x) = 26$ — Multiply every term by 20: 1.3 × 20 = 26, a whole number.
3
$5 x + 10 - 20 + 4 x = 26$ — Expand. Watch the minus: (−4) × (−x) = +4x.
4
$9 x - 10 = 26$ — Collect: 5x + 4x = 9x and 10 − 20 = −10.
5
$9 x = 36$ — Add 10 to both sides.
6
$x = 4$ — Divide both sides by 9.

Common mistake

Forgetting the right side. The 2 must become 24, and the 1.3 must become 26.

x on the Bottom

When x is on the bottom of a fraction, do the same as with a number: multiply both sides by the bottom.

Worked example

1
$\frac{36}{x} = 8$ — The question. x is on the bottom.
2
$36 = 8 x$ — Multiply both sides by x. On the left, the x's cancel.
3
$x = 4.5$ — Divide both sides by 8: 36 ÷ 8 = 4.5.

A fraction on each side: cross-multiply

Worked example

1
$\frac{3 x}{x + 4} = \frac{5}{2}$ — The question. There is a fraction on each side.
2
$2 \times 3 x = 5 (x + 4)$ — Cross-multiply: each top times the other bottom. Put x + 4 in a bracket.
3
$6 x = 5 x + 20$ — Work out 2 × 3x = 6x, and expand 5(x + 4).
4
$x = 20$ — Take 5x from both sides.
5
$\frac{3 \times 20}{20 + 4} = \frac{60}{24} = 2.5$ — Check: 5 ÷ 2 = 2.5 as well, so both sides match.

Common mistake

Leaving out the bracket: writing 5(x + 4) as 5x + 4. The 5 multiplies the 4 too.

Key idea

If clearing the fractions leaves an $x^{2}$ that does not cancel, the equation is a quadratic. See the lesson Equations That Become Quadratics.

When x² Cancels

Sometimes both sides have $x^{2}$ after you expand. If the $x^{2}$ terms are the same, they cancel and you are left with a linear equation.

Worked example

1
$(2 x + 3)^{2} = (2 x + 1) (2 x + 7)$ — The question. It looks like a quadratic, but expand first.
2
$(2 x + 3) (2 x + 3) = (2 x + 1) (2 x + 7)$ — A square means the bracket times itself.
3
$4 x^{2} + 12 x + 9 = 4 x^{2} + 16 x + 7$ — Expand both sides. Each side starts with 4x².
4
$12 x + 9 = 16 x + 7$ — Take 4x² from both sides. The x² terms cancel.
5
$9 = 4 x + 7$ — Take 12x, the smaller x term, from both sides.
6
$2 = 4 x$ — Take 7 from both sides.
7
$x = 0.5$ — Divide both sides by 4: 2 ÷ 4 = 0.5.

Common mistake

Writing $(2 x + 3)^{2}$ as $4 x^{2} + 9$ . You lose the middle term, 12x.

Which Method? Read the Question

Look at where x is before you start.

You see	You do
One thing done to x	Undo it
Two things done to x	Undo the adding first, then the multiplying
A minus in front of x	Add the x term to both sides
x on both sides	Take the smaller x term from both sides
A bracket	Expand it first
x² on both sides	Expand; the x² terms may cancel
One fraction	Multiply both sides by the bottom
x on the bottom	Multiply both sides by it
A fraction on each side	Cross-multiply
Two fractions	Multiply every term by the common bottom

Key idea

Show every line. Most of the marks are for the working, and when a question says “show clear algebraic working”, a right answer with no lines can score 0.

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