Solve for an unknown by keeping the equation balanced — one-step, two-step, brackets and word problems
Explanation & Worked Examples
Keep the Equation Balanced
A linear equation is a statement that two expressions are equal, containing an unknown (usually x) raised only to the power 1 — no x², no x under a root.
Whatever you do to one side, you must do to the other side.
Think of the = sign as the middle of a balance scale. To find x, you peel away everything attached to it using inverse operations, always applying the same step to both sides so the scale stays level.
You see
Inverse (undo with)
+ a number
subtract it
− a number
add it
× a number
divide by it
÷ a number
multiply by it
Goal: rearrange until you have x by itself on one side, e.g. x = 5. You can always check your answer by substituting it back into the original equation.
The Solving Steps
For a typical two-step equation like 2x + 3 = 11, work in this order:
Step 1 — Undo addition/subtraction
2x + 3 = 11 → subtract 3 → 2x = 8
Step 2 — Undo multiplication/division
2x = 8 → divide by 2 → x = 4
Step 3 — Check
2(4) + 3 = 11 ✓
When x is on both sides
Collect all the x terms on one side and the plain numbers on the other:
Expand (multiply out) the brackets first, then solve as usual:
3(x − 2) = 12 → 3x − 6 = 12 → 3x = 18 → x = 6
Tip: move the smaller x term to avoid a negative coefficient. In 3x + 4 = 5x − 6, subtract 3x (not 5x) so you keep +2x rather than −2x.
Easy Examples
One-step and simple two-step equations — undo the operations in order.
Example 1 — One Step (add/subtract)
Solve x + 7 = 12.
1x is increased by 7, so subtract 7 from both sides
2x = 12 − 7 = 5
Example 2 — One Step (multiply/divide)
Solve 4x = 20.
1x is multiplied by 4, so divide both sides by 4
2x = 20 ÷ 4 = 5
Example 3 — Two Steps
Solve 3x + 2 = 14.
1Subtract 2: 3x = 12
2Divide by 3: x = 4
3Check: 3(4) + 2 = 14 ✓
Medium Examples
x appears on both sides, or there are brackets or fractions to clear first.
Example 1 — x on Both Sides
Solve 5x − 7 = 3x + 9.
1Subtract 3x from both sides: 2x − 7 = 9
2Add 7: 2x = 16
3Divide by 2: x = 8
Example 2 — Brackets
Solve 5(x + 2) = 3x + 16.
1Expand the bracket: 5x + 10 = 3x + 16
2Subtract 3x: 2x + 10 = 16
3Subtract 10, then divide by 2: 2x = 6 → x = 3
Example 3 — A Fraction
Solve x/2 + 3 = 8.
1Subtract 3: x/2 = 5
2Multiply both sides by 2: x = 10
Strategy: always expand brackets and clear fractions (multiply through by the denominator) before collecting x terms — it keeps the working tidy.
Complex Examples
Forming an equation from a word problem, then solving it.
Example 1 — Consecutive Integers
The sum of three consecutive integers is 72. Find the smallest.
1Let the smallest be n. The three are n, n+1, n+2
2n + (n+1) + (n+2) = 72 → 3n + 3 = 72
33n = 69 → n = 23
Example 2 — Ages
A father is 4 times as old as his son. In 20 years he will be twice as old. How old is the son now?
1Let the son be s, so the father is 4s
2In 20 years: 4s + 20 = 2(s + 20)
34s + 20 = 2s + 40 → 2s = 20 → s = 10
Example 3 — Rectangle Perimeter
A rectangle has perimeter 40 cm. The length is 4 cm more than the width. Find the width.
1Let width = w, length = w + 4
2Perimeter: 2(w + w + 4) = 40 → 2(2w + 4) = 40
34w + 8 = 40 → 4w = 32 → w = 8 cm
Example 4 — Fractions Both Sides
Solve (x + 2)/3 = (x − 4)/2.
1Cross-multiply: 2(x + 2) = 3(x − 4)
22x + 4 = 3x − 12
3Subtract 2x, add 12: x = 16
Word-problem checklist: (1) define a letter for the unknown, (2) write each fact as an expression, (3) form the equation, (4) solve, (5) re-read the question — make sure you answer what was actually asked.