Solving Linear Equations

30 min
0/4 practice checks

A linear equation in one variable can be written in the form ax+b=cax + b = c with a0a \neq 0. The variable appears only to the first power — no x2x^2, no 1x\frac{1}{x}.

The general strategy:

  1. Simplify each side (expand brackets, collect like terms).
  2. Move variable terms to one side, numbers to the other, using inverse operations on both sides.
  3. Divide by the coefficient of the variable.
  4. Check by substitution.

Worked example. Solve 3x+4=193x + 4 = 19.

3x+4=193x + 4 = 19
3x=15(subtract 4)3x = 15 \qquad \text{(subtract 4)}
x=5(divide by 3)x = 5 \qquad \text{(divide by 3)}

Check: 3(5)+4=193(5) + 4 = 19

A second example with a real bill. A minibus taxi charges a R10 flag fee plus R8 per kilometre. A trip costs R50 — how far was it?

10+8k=50    8k=40    k=5 km10 + 8k = 50 \;\Rightarrow\; 8k = 40 \;\Rightarrow\; k = 5 \text{ km}

Translating a sentence into an equation and solving it is the whole game of algebra.

Solve for xx: 3x+4=193x + 4 = 19

Solve and answer in the form x=…: 5x2=3x+85x - 2 = 3x + 8

Solving 4x+7=314x + 7 = 31: which is the best first step?

A minibus taxi charges a R10 flag fee plus R8 per kilometre. Your trip costs R50. How many kilometres was the trip?