Solving Linear Equations

≈ 25 min

A linear equation has the variable to the first power only — no $x^2$ , no $\frac{1}{x}$ . Solving means isolating the variable using inverse operations, always applied to both sides so the equation stays balanced.

The routine:

Expand any brackets and clear fractions.
Collect variable terms on one side, numbers on the other.
Divide by the coefficient of the variable.
Check by substituting back.

Worked example. Solve $2(x - 3) = x + 4$ .

2x - 6 = x + 4 \qquad \text{(expand)}

x - 6 = 4 \qquad \text{(subtract } x \text{)}

x = 10 \qquad \text{(add 6)}

Check: LHS $= 2(10 - 3) = 14$ ; RHS $= 10 + 4 = 14$ ✓.

Which is the solution of $2x + 1 = 9$ ?

$x = 3$

$x = 4$

$x = 5$

$x = 8$

Solve for $x$ : $4x - 3 = 17$

Solve $2(x - 3) = x + 4$ . Answer in the form x=….

A learner pays R8 per minibus-taxi trip plus a R5 weekly rank fee. In one week she spends R61 in total. Solving $8t + 5 = 61$ , how many trips $t$ did she take?