## Method For Solving Linear Equations

The general steps for solving linear equations are:

Expand all brackets.

Rearrange the terms so that all terms containing the variable are on one side of the equation and all constant terms are on the other side.

Group like terms together and simplify.

Factorise if necessary.

Find the solution and write down the answer.

Check the answer by substituting the solution back into the original equation.

### Important:

An equation must always be balanced, whatever you do to the left-hand side, you must also do to the right-hand side.

## Example

### Question

Solve for \(x\):

\[4(2x – 9) – 4x = 4 – 6x\]

### Expand the brackets and simplify

\begin{align*} 4(2x – 9) – 4x & = 4 – 6x \\ 8x – 36 – 4x & = 4 – 6x \\ 8x – 4x + 6x & = 4 + 36 \\ 10x & = 40 \end{align*}

### Divide both sides by 10

\[x = 4\]

### Check the answer by substituting the solution back into the original equation

\begin{align*} \text{LHS } & = 4[2(4) – 9] – 4(4) \\ & = 4(8 – 9) – 16 \\ & = 4(-1) – 16 \\ & = -4 – 16 \\ & = -20 \\ \text{RHS } & = 4 – 6(4) \\ & =4 – 24 \\ & = -20 \\ \therefore \text{LHS } = \text{ RHS} \end{align*}

Since both sides are equal, the answer is correct.

## Example

### Question

Solve for \(x\):

\[\cfrac{2 – x}{3x + 1} = 2\]

### Multiply both sides of the equation by \((3x + 1)\)

Division by \(\text{0}\) is undefined so there must be a restriction: \((x\ne -\cfrac{1}{3})\).

\begin{align*} \cfrac{2 – x}{3x + 1} & = 2 \\ (2 – x) & = 2(3x + 1) \end{align*}

### Expand the brackets and simplify

\begin{align*} 2 – x & = 6x + 2 \\ -x – 6x & = 2 – 2 \\ -7x & = 0 \end{align*}

### Divide both sides by \(-\text{7}\)

\begin{align*} x & = \cfrac{0}{-7} \\ x & = 0 \end{align*}

### Check the answer by substituting the solution back into the original equation

\begin{align*} \text{LHS } & = \cfrac{2 – (0)}{3(0) + 1} \\ & = 2 \\ & = \text{ RHS} \end{align*}

Since both sides are equal, the answer is correct.

## Example

### Question

Solve for \(a\): \[\cfrac{2a – 3}{3} – 3a = \cfrac{a}{3}\]

### Multiply the equation by the common denominator \(\text{3}\) and simplify

\begin{align*} 2a – 3 – 9a & = a \\ -7a – 3 & = a \end{align*}

### Rearrange the terms and simplify

\begin{align*} -7a – a & = 3 \\ -8a & = 3 \end{align*}

### Divide both sides by \(-\text{8}\)

\[a= -\cfrac{3}{8}\]

### Check the answer by substituting the solution back into the original equation

\begin{align*} \text{LHS } & = \cfrac{2(-\cfrac{3}{8}) – 3}{3} – 3(-\cfrac{3}{8}) \\ & = \cfrac{(-\cfrac{3}{4}) – \cfrac{12}{4}}{3} + \cfrac{9}{8} \\ & = [-\cfrac{15}{4}\times \cfrac{1}{3}] + \cfrac{9}{8} \\ & = -\cfrac{5}{4} + \cfrac{9}{8} \\ & = -\cfrac{10}{8} + \cfrac{9}{8} \\ & = -\cfrac{1}{8} \\ \text{RHS } & = \cfrac{-\cfrac{3}{8}}{3} \\ & = \cfrac{-\cfrac{3}{8}}{3} \\ & = -\cfrac{3}{8}\times \cfrac{1}{3} \\ & = -\cfrac{1}{8} \\ \therefore \text{LHS } = \text{ RHS} \end{align*}

Since both sides are equal, the answer is correct.