A linear equation in one variable contains only one variable with degree one. We use the method of rearranging equations to make the variable the subject of the equation and hence find its value.

Worked Examples

Example 1

Solve $4x-1=\dfrac{1}{3}(2+2x)$.

Solution

The aim is to collect the $x$ terms on one side and the constants on the other. Begin by multiplying through by $3$ to get rid of the fraction. \begin{align} 3 \times (4x-1) &= 3 \times \frac{1}{3}(2+2x) \\ 12x - 3 &= 2+2x \end{align}

Subtract $2x$ from both sides so that there is no $x$ term on the right. \begin{align} 12x - 3 - 2x &= 2 + 2x - 2x \\ 10x - 3 &= 2 \\ \end{align}

Add $3$ to both sides so that the $x$ term is on its own on the left. \begin{align} 10x - 3 + 3 &= 2 + 3 \\ 10x &= 5 \end{align}

Finally, divide by $10$ to get an expression for $x$. \begin{align} \frac{10x}{10} &= \frac{5}{10} \\ \\ x &= \frac{1}{2} \end{align}

Example 2

Solve $\dfrac{3}{3-a} + \dfrac{4}{2a+1} = 0$.

Solution

Take the second fraction over to the right hand side and then cross multiply. \begin{align} \frac{3}{3-a} &= -\frac{4}{2a+1} \\ 3(2a+1) &= -4(3-a) \\ 6a+3 &=-12+4a \end{align} Subtract $4a$ from both sides. \begin{align} 6a + 3 - 4a &= -12 + 4a - 4a \\ 2a + 3 &= -12 \end{align}

Subtract $3$ from both sides. \begin{align} 2a + 3 - 3 &= -12 - 3 \\ 2a &= -15 \end{align}

Divide by $2$ to get a solution for $a$. \begin{align} \frac{2a}{2} &= - \frac{15}{2} \\ \\ a &= - \frac{15}{2} \end{align}

Video Examples

Example 1

Prof. Robin Johnson solves the linear equation $2x-5 = \dfrac{1}{2}(3-x)$.

Example 2

Prof. Robin Johnson solves the equation $\dfrac{2}{1-x} + \dfrac{3}{x+2} =0$.

Workbook

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.