How to Solve Algebra Equations

This guide shows you how to solve algebra equations clearly and systematically. You'll learn how to balance equations (do the same to both sides), see how multi-step equations with brackets are tackled, and practise neat, vertical working with one equals sign per line.

Prior Knowledge Make sure you are comfortable with Simplifying Algebraic Expressions and Expanding Single Brackets before starting.
⇩ Jump to Practice Questions ⇩

What is an equation?

An equation is a statement that two expressions are equal. It always contains an equals sign \(=\). The aim in solving is to find the value of the variable (e.g. \(x\)) that makes the statement true.

  • Variables represent unknown numbers, e.g. \(x, y\).
  • Constants are fixed numbers, e.g. \(7, -3, \tfrac{1}{2}\).
  • We solve by performing the same operation on both sides, keeping the equation balanced at every step.

The job of = is to keep balance

Balance scale showing x + 2 on the left and 10 on the right
Whatever you do to one side, do to the other: keep the equation balanced.

Inverse operations

Every operation has an inverse that undoes it. Apply the inverse to both sides to keep the equation balanced.

Add: subtract from both sides

\[\begin{aligned} x+7 &= 19 \\ x &= 12 \end{aligned}\]

subtract 7 from both sides ✓

Subtract: add to both sides

\[\begin{aligned} x-9 &= 4 \\ x &= 13 \end{aligned}\]

add 9 to both sides ✓

Multiply: divide both sides

\[\begin{aligned} 6x &= 42 \\ x &= 7 \end{aligned}\]

divide both sides by 6 ✓

Divide: multiply both sides

\[\begin{aligned} \dfrac{x}{5} &= 9 \\ x &= 45 \end{aligned}\]

multiply both sides by 5 ✓

Square: square root both sides

\[\begin{aligned} x^2 &= 25 \\ x &= \pm5 \end{aligned}\]

square root both sides (\(\pm\)) ✓

Square root: square both sides

\[\begin{aligned} \sqrt{x} &= 7 \\ x &= 49 \end{aligned}\]

square both sides ✓

Step-by-step method for solving equations

Follow these steps in order every time; the golden rule is one operation per line, always applied to both sides.

  1. Expand any brackets if present, distributing carefully and watching the signs.
  2. Collect like terms on each side so each side is as simple as possible.
  3. Add or subtract from both sides to move all constants to one side.
  4. Add or subtract from both sides to move all \(x\) terms to the other side.
  5. Divide both sides by the coefficient of \(x\) to isolate it.
  6. Check by substituting your answer back into the original equation.

BIDMAS vs solving order (reverse BIDMAS)

When evaluating you follow BIDMAS. To solve, first tidy up, then reverse it: undo the operations on \(x\) from the outside in.

Step 0 (tidy up): expand any brackets and collect like terms first, so each side is as simple as possible.
Evaluate (BIDMAS)▼ work top to bottom
Solve (reverse BIDMAS)▲ work bottom to top
1Brackets: simplify inside brackets first.
B
0Brackets: expand first, before you undo (the tidy-up step above).
2Indices: powers and roots.
I
3Indices: deal with powers or roots if present.
3Divide / Multiply: left to right.
D/M
2Multiply / Divide: remove the coefficient of \(x\).
4Add / Subtract: left to right.
A/S
1Add / Subtract: move constants first.

Shortcut: for a single bracket like \(4(x+3)=20\) you can divide both sides first and undo the bracket last. Expanding first always works, so use it whenever \(x\) appears more than once.

Worked Examples

💡 Example 1: Expand, then solve

What's happening?
\( 2(x+3)+4 \)
\(=\)
\( x+19 \)
\( 2x+6+4 \)
\(=\)
\( x+19 \)
expand the bracket first
\( 2x+10 \)
\(=\)
\( x+19 \)
collect like terms
\( x+10 \)
\(=\)
\( 19 \)
subtract \(x\) from both sides
\( x \)
\(=\)
\( 9 \)
subtract 10 from both sides ✓

💡 Example 2: Fractions and brackets

What's happening?
\( \dfrac{x}{3}+2 \)
\(=\)
\( \dfrac{x-1}{2}+5 \)
\( \dfrac{x}{3} \)
\(=\)
\( \dfrac{x-1}{2}+3 \)
subtract 2 from both sides
\( 2x \)
\(=\)
\( 3(x-1)+18 \)
multiply both sides by 6
\( 2x \)
\(=\)
\( 3x+15 \)
expand the bracket
\( -x \)
\(=\)
\( 15 \)
subtract \(3x\) from both sides
\( x \)
\(=\)
\( -15 \)
divide both sides by \(-1\) ✓

💡 Example 3: Variable on both sides

What's happening?
\( 5(2x-3)-4(x+1) \)
\(=\)
\( 2 \)
\( 10x-15-4x-4 \)
\(=\)
\( 2 \)
expand both brackets
\( 6x-19 \)
\(=\)
\( 2 \)
collect like terms
\( 6x \)
\(=\)
\( 21 \)
add 19 to both sides
\( x \)
\(=\)
\( \dfrac{7}{2} \)
divide both sides by 6 ✓

💡 Example 4: Nested minus and brackets

What's happening?
\( 4-(x-5) \)
\(=\)
\( 2(3-x)-1 \)
\( 4-x+5 \)
\(=\)
\( 6-2x-1 \)
expand both brackets
\( 9-x \)
\(=\)
\( 5-2x \)
collect like terms
\( 9+x \)
\(=\)
\( 5 \)
add \(2x\) to both sides
\( x \)
\(=\)
\( -4 \)
subtract 9 from both sides ✓

🔑 Key Points

  • Expand brackets and collect like terms first, then reverse BIDMAS on what is left.
  • Reverse BIDMAS order: undo + / − (constants) first, then × / ÷ (the coefficient), then indices.
  • Always do the same operation to both sides.
  • Shortcut: for a single bracket like \(4(x+3)=20\) you may divide first and undo the bracket last.
  • One equals sign per line; check your answer by substituting back in.

⚠️ Common Pitfalls

  • Trying to move terms before expanding the brackets, then losing track of the bracketed parts.
  • Dropping a negative sign when moving terms across the equals sign.
  • Forgetting \(x^2 = 25\) gives \(x = \pm 5\), not just \(x = 5\).
  • Only multiplying one term inside a bracket instead of all terms.
⇩ Jump to Practice Questions ⇩

Ready for more practice? Work through the rooms below ↓

Next: Simplifying Algebraic Fractions →

Solving Linear Equations: Practice Rooms

Practise solving linear equations with auto-marked, randomly generated questions. Difficulty rises left to right across each room. Write answers including the variable, e.g. x = 3, and fractions as a = 5/2. Answers without the variable, or the wrong letter, are marked wrong.

✓ Correct 0
✗ Re-attempts 0
🔥 Streak 0
🏆 Best 0