GCSE Maths · Algebra

Linear Equations flashcards

13 cards on Linear Equations: the key points, worked examples and the slips to watch. Swipe through them, or print them.

Print these cards

Card 1 of 13

Linear Equations

Solving one- and two-step equations

How do you solve an equation like 3x + 5 = 20?

Answer it in your head, then turn the card over.

Solving one- and two-step equations

Key points

An equation is true only for particular values of x: solving finds them.
Undo what was done to x in reverse order: the last step comes off first.
Do the same to both sides every time, so the equation stays balanced.
Check by putting your answer back into the original equation.

3x + 5 = 20

Subtract 5 from both sides: 3x = 15

Divide both sides by 3: x = 5

Check: 3 × 5 + 5 = 20

Watch out

Undo the + 5 first: dividing only the 3x by 3 gives x + 5 = 20, which is a different equation.

Linear Equations

Answers that are fractions or negative

How do you finish an equation like 6x = 3, where the answer is not a whole number?

Answer it in your head, then turn the card over.

Answers that are fractions or negative

Key points

Divide the number by the coefficient of x (the number in front of x), whichever is bigger.
Answers can be fractions, decimals or negative numbers: that is fine.
Leave a fraction in its simplest form, unless the question asks for a decimal.

Example

6x = 3, so x = 3 ÷ 6 = $\frac{1}{2}$

4x − 7 = 3, so 4x = 10 and x = 2.5

5x + 12 = 2, so 5x = −10 and x = −2

Watch out

Dividing upside down turns the answer over: 6x = 3 gives x = $\frac{1}{2}$ , not x = 2.

Linear Equations

Equations with brackets

How do you solve an equation with a bracket, like 4(x − 2) = 20?

Answer it in your head, then turn the card over.

Linear Equations

Dividing before expanding

When can you solve an equation like 5(x − 3) = 35 without expanding the bracket?

Answer it in your head, then turn the card over.

Dividing before expanding

Key points

If the other side divides exactly by the number outside the bracket, divide first.
That removes the bracket in one step and keeps the numbers small.
If it does not divide neatly, expand instead: both ways give the same answer.

Example

5(x − 3) = 35: divide by 5, x − 3 = 7, so x = 10

3(x + 4) = 27: divide by 3, x + 4 = 9, so x = 5

Watch out

If you expand first, divide every term: 5x − 15 = 35 gives x − 3 = 7, not x − 15 = 7.

Linear Equations

The unknown on both sides

How do you solve an equation with x on both sides, like 5x − 4 = 2x + 11?

Answer it in your head, then turn the card over.

The unknown on both sides

Key points

Collect the x terms on one side and the numbers on the other.
Take away the smaller x term from both sides, so the x term left is positive.
Then solve the two-step equation that is left.

5x − 4 = 2x + 11

Subtract 2x: 3x − 4 = 11

Add 4: 3x = 15, so x = 5

Check: 5 × 5 − 4 = 21 and 2 × 5 + 11 = 21

Watch out

Moving − 4 across without changing its sign gives 3x = 7: adding 4 to both sides gives 3x = 15.

Linear Equations

Brackets on both sides

How do you solve an equation like 2(x + 5) = 3(x − 1)?

Answer it in your head, then turn the card over.

Linear Equations

Equations with a fraction

How do you clear the fraction in an equation like $\frac{x + 3}{4}$ = 5?

Answer it in your head, then turn the card over.

Equations with a fraction

Key points

Multiply both sides by the denominator: the fraction disappears.
Multiply every term on both sides, not just the fraction.
Or, if a number sits outside the fraction, move it away first.

Example

$\frac{x + 3}{4}$ = 5: × 4 gives x + 3 = 20, so x = 17

$\frac{x}{4}$ + 3 = 5: × 4 gives x + 12 = 20, so x = 8

or take 3 first: $\frac{x}{4}$ = 2, so x = 8

Watch out

Multiplying only some terms unbalances it: $\frac{x}{4}$ + 3 = 5 becomes x + 12 = 20, not x + 3 = 20.

Linear Equations

Equations with two fractions

How do you solve an equation with two fractions, like $\frac{x}{3}$ + $\frac{x}{4}$ = 7?

Answer it in your head, then turn the card over.

Equations with two fractions

Key points

Multiply every term by the lowest common multiple of the denominators.
With one fraction on each side, you can cross-multiply instead.
Put brackets round each top before you multiply it, so nothing is lost.

Example

$\frac{x}{3}$ + $\frac{x}{4}$ = 7: multiply every term by 12

4x + 3x = 84, so 7x = 84 and x = 12

$\frac{x + 1}{2}$ = $\frac{x + 4}{3}$ : cross-multiply

3(x + 1) = 2(x + 4), so 3x + 3 = 2x + 8 and x = 5

Watch out

Multiply the right-hand side too: leaving the 7 as 7 gives 7x = 7 and x = 1, not 12.

Linear Equations

Equations from words

How do you turn a worded problem into an equation and solve it?

Answer it in your head, then turn the card over.

Linear Equations

Equations from diagrams

How do you set up an equation from a shape whose angles or sides are given as expressions?

Answer it in your head, then turn the card over.

Equations from diagrams

Key points

Use a fact about the shape for the total: a triangle's angles make 180°, a perimeter adds every side.
Add the expressions and set them equal to that total.
Solve for x, then use it to find what the question actually asks for.

A triangle's angles are x, 2x and 3x

x + 2x + 3x = 180, so 6x = 180

x = 30, so the angles are 30°, 60° and 90°

Watch out

Stopping at x = 30 when the question asks for the largest angle: substitute back to get 90°.

Linear Equations

Changing the subject of a formula

How do you rearrange a formula like y = 3x + 2 to make x the subject?

Answer it in your head, then turn the card over.

Changing the subject of a formula

Key points

The subject is the letter standing on its own on one side.
Use the same moves as solving an equation: undo the steps around the new subject in reverse order.
The answer is an expression, not a number.

Make x the subject of y = 3x + 2

Subtract 2: y − 2 = 3x

Divide by 3: x = $\frac{y - 2}{3}$

Test with x = 4: y = 14, and $\frac{14 - 2}{3}$ = 4

Watch out

Undo + 2 by subtracting 2: x = $\frac{y - 2}{3}$ , not $\frac{y + 2}{3}$ .

Linear Equations

Keeping brackets when you rearrange

Why is v = u + at rearranged to a = $\frac{v - u}{t}$ with a bracket, and not a = v − $\frac{u}{t}$ ?

Answer it in your head, then turn the card over.

Keeping brackets when you rearrange

Key points

Subtract u from both sides first: v − u = at.
Then divide the whole of v − u by t, so it goes in a bracket.
Without the bracket, only the u would be divided by t: a different formula.

Test it: u = 3, a = 2, t = 5

v = 3 + 2 × 5 = 13

$\frac{v - u}{t}$ = $\frac{13 - 3}{5}$ = 2, which is a

but v − $\frac{u}{t}$ = 13 − $\frac{3}{5}$ = 12.4, which is not

Watch out

Put a bracket round anything you divide as one piece: $\frac{v - u}{t}$ , never v − $\frac{u}{t}$ .

Linear Equations

Higher

The subject appearing twice

How do you make x the subject when x appears in two terms?

Answer it in your head, then turn the card over.

The subject appearing twice

Key points

Collect every term with x on one side and everything else on the other.
Factorise x out of those terms, so x appears once.
Divide by the bracket that is left.

Make x the subject of ax + 3 = bx + 7

Collect: ax − bx = 7 − 3 = 4

Factorise: x(a − b) = 4, so x = $\frac{4}{a - b}$

Test a = 5, b = 3: x = 2, and both sides are 13

Watch out

An answer with x on both sides is not finished: x = $\frac{b x + 4}{a}$ still needs the x terms collected.

Print these cards

Four cards to an A4 sheet, each the size of a standard A6 revision card.

One-sided

Each card prints on one face: the question at the top, the notes under it.

Two-sided

Questions on the front, notes on the back. Print double-sided and flip on the long edge, then cut along the lines.

Put it into practice

Practise Linear EquationsExam-style questions that adapt to how you are doing, with feedback on every answer. Free to start.Learn Linear EquationsEvery syllabus point explained, with worked examples and the video lesson where there is one.

More Algebra flashcards

All Maths flashcards