Linear Equations

1. Solve one- and two-step equations

   Undo what has been done to x, in reverse order, doing the same to both sides. For 3x + 5 = 20: subtract 5 from both sides (3x = 15), then divide by 3 (x = 5). Every step keeps the equation balanced - that is the whole game.

   Watch this bit

2. Solve equations with brackets

   Expand the bracket first, then solve as normal: 4(x − 2) = 20 becomes 4x − 8 = 20, so 4x = 28 and x = 7. If everything on the other side divides neatly, you can divide first instead: 4(x − 2) = 20 gives x − 2 = 5 straight away.

   Watch this bit

3. Solve equations with the unknown on both sides

   Collect the x terms on one side and the numbers on the other. For 5x − 4 = 2x + 11: subtract 2x from both sides (3x − 4 = 11), add 4 (3x = 15), divide (x = 5). Moving the smaller x term keeps the coefficient positive and the signs friendly.

   Watch this bit

4. Solve equations involving fractions

   Clear the fractions by multiplying every term by the denominator (or the lowest common denominator). For $\frac{x+3}{4}$ = 5, multiply both sides by 4 to get x + 3 = 20, so x = 17. With two fractions, multiplying by both bottoms - or cross-multiplying - does the same job.

   Watch this bit

5. Set up an equation from a worded problem or a diagram, then solve it

   Name the unknown, translate the story into symbols, then solve. "I think of a number, double it, add 7, and get 31" becomes 2x + 7 = 31. Angle and perimeter diagrams work the same way - the shape gives you the total, the expressions give you the left-hand side.

   Watch this bit

6. Rearrange a formula to change the subject

   Use the same inverse-operation moves as equation solving, but the answer is an expression rather than a number. Making u the subject of v = u + at: subtract at, giving u = v − at. If the new subject appears twice, collect its terms and factorise it out first.

   Watch this bit

• Moving a term across the equals sign without changing its sign, so 5x − 4 = 2x + 11 becomes 3x = 7.

  Do the same thing to both sides: add 4 and subtract 2x, giving 3x = 15, so x = 5. Check it: 5 × 5 − 4 = 21 and 2 × 5 + 11 = 21.

• Multiplying only the first term inside the bracket, so 4(x − 2) = 20 becomes 4x − 2 = 20.

  The 4 multiplies everything inside the bracket: 4x − 8 = 20, so 4x = 28 and x = 7.

• Dividing upside down at the last step, so 6x = 3 gives x = 2 instead of x = $\frac{1}{2}$.

  Divide the number by the coefficient of x, whichever is bigger: 6x = 3 gives x = 3 ÷ 6 = $\frac{1}{2}$. Answers do not have to be whole numbers.

• Multiplying only some of the terms by the denominator, so $\frac{x}{4}$ + 3 = 5 becomes x + 3 = 20.

  Multiply every term on both sides by 4: x + 12 = 20, so x = 8. Or take away the 3 first: $\frac{x}{4}$ = 2, so x = 8.

• Losing a bracket when changing the subject, so v = u + at becomes a = v − $\frac{u}{t}$.

  Subtract u from both sides, then divide the whole of v − u by t: a = $\frac{v-u}{t}$. Keep a bracket round anything you divide as one piece.

More Algebra topics

• Expanding Brackets
• Graphs
• Quadratic Equations
• Sequences
• Simultaneous Equations
• Inequalities

Keep reading

• How do I revise for GCSE maths?
• What's on the GCSE maths syllabus?
• What are the GCSE Maths grade boundaries?
• The GCSE Maths syllabus, explained