An equation is a balance
A linear equation in one variable has the form ax + b = 0 with a ≠ 0. The highest power of x is 1, which is exactly why the graph is a straight line and why there is exactly one solution.
Solving is a sequence of moves that keep both sides equal. Add the same thing to both sides, subtract the same thing, multiply both sides by the same non-zero number, divide both sides by the same non-zero number. Nothing else is permitted, and every legitimate step you have ever taken is one of those four.
Multiplying by an expression is not always safe
You may multiply both sides by a number. You may only multiply by an expression such as (x − 2) if you check afterwards that the value making it zero is not one of your answers. Multiplying by zero turns any false statement into a true one, which is where "extra" solutions come from.
Equations that need clearing first
Most exam equations are not handed to you in the form ax + b = 0. They arrive with brackets, fractions or an unknown on both sides. The order of operations is fixed: clear fractions, expand brackets, collect the x terms on one side, collect the numbers on the other, divide.
Solve (2x − 1)/3 + (x + 4)/2 = 5.
- The LCM of 3 and 2 is 6. Multiply every term by 6:
2(2x − 1) + 3(x + 4) = 30.Multiply the whole equation, including the right-hand side. Forgetting the 5 × 6 is the classic error here. - Expand:
4x − 2 + 3x + 12 = 30.The 2 and the 3 multiply everything inside their brackets, not just the first term. - Collect:
7x + 10 = 30, so7x = 20. x = 20/7.Check by substitution: (40/7 − 1)/3 + (20/7 + 4)/2 = 11/7 + 24/7 = 5. ✓
x = 20/7
Absolute value equations have two cases
The modulus |x| is the distance of x from zero, so it is never negative. That single fact settles every question in this part of the chapter.
To solve |expression| = k: if k < 0 there is no solution, because a distance cannot be negative. If k ≥ 0, write two equations — the expression equals k, and the expression equals −k — and solve both.
Solve |3x − 5| = 7.
- 7 is positive, so there are two cases.Check the sign of the right-hand side before doing anything else — a negative would end the question immediately.
- Case 1:
3x − 5 = 7gives3x = 12, sox = 4. - Case 2:
3x − 5 = −7gives3x = −2, sox = −2/3.The second case is the one students forget; it is worth a mark on its own. - Check both:
|12 − 5| = 7✓ and|−2 − 5| = 7✓.
x = 4 or x = −2/3
Inequalities: one rule is different
Everything you do to solve an equation works for an inequality too, with a single exception that the paper tests every year.
When you multiply or divide both sides by a negative number, the inequality sign reverses. Adding and subtracting never reverse it; multiplying or dividing by a positive number never reverses it.
The reason is easy to see on a number line. 3 < 5 is true. Multiply both sides by −1 and you get −3 and −5 — and −3 is now to the right of −5, so the correct statement is −3 > −5. The order flipped because the whole line was reflected.
| Operation | Effect on the sign |
|---|---|
| Add or subtract anything | unchanged |
| Multiply or divide by a positive number | unchanged |
| Multiply or divide by a negative number | reverses |
| Take reciprocals of both sides (both positive) | reverses |
Solve 4 − 3x ≤ 19 and represent the solution on a number line.
- Subtract 4 from both sides:
−3x ≤ 15.Subtraction never affects the direction of the sign. - Divide both sides by −3 — and reverse the sign:
x ≥ −5.This is the only step where the direction changes, and it is the whole point of the question. - On the number line, draw a filled circle at −5 with an arrow pointing right.Filled because ≥ includes the endpoint; an open circle would mean strictly greater.
x ≥ −5
Double inequalities and modulus inequalities
A double inequality such as −3 < 2x + 1 ≤ 9 is solved by doing the same thing to all three parts at once. Subtract 1 from each: −4 < 2x ≤ 8. Divide each by 2: −2 < x ≤ 4.
Modulus inequalities reduce to that same shape, and there are only two patterns to remember. Read them as distances and they stop needing memorisation at all.
Why the two patterns differ
|x| is the distance from 0. "Distance less than k" means you are inside a band around zero — one interval. "Distance greater than k" means you are further out than k in either direction — two separate intervals, and the answer must be joined by or, never written as a single chain like k < x < −k, which is impossible.
Before you leave this chapter
- Clear fractions by multiplying every term by the LCM — including the constant on the right.
- |x| is never negative, so |anything| = negative has no solution at all.
- |expression| = k with k ≥ 0 gives two equations to solve.
- Multiplying or dividing an inequality by a negative number reverses the sign. Nothing else does.
- |x| < k gives one interval; |x| > k gives two intervals joined by "or".
Showing the answer on a number line
An inequality has infinitely many solutions, so the answer is a region rather than a number, and the paper expects to see it drawn. Two details carry the marks.
A filled circle means the endpoint is included, which is the case for ≤ and ≥. An open circle means it is excluded, for < and >. Then shade the part of the line that satisfies the inequality and put an arrowhead on it if it continues for ever.
A modulus inequality is drawn the same way, and the picture makes the difference between the two patterns obvious: less-than gives one band around zero, greater-than gives two rays heading in opposite directions.
Compare |x| < k with |x| > k. One is a single band in the middle; the other is two separate pieces. That is why the second answer must be joined by the word "or" and can never be written as one chain.