Solving Equations with Algebraic Fractions (SQA National 5 Maths): Revision Note

When multiplying by an algebraic expression, use brackets around the expression, e.g. $\left(2x+3\right)$.

Multiplying by both denominators at once can speed up the process, but take care if choosing this technique in the exam!

• And remember to multiply all terms on both sides of the equation

Solve the equation $\frac{5}{x-4}+\frac{7}{x+2}=6$.

Answer:

To clear the fractions, multiply both sides of the equation by each denominator

Start by multiplying all terms in the equation by the denominator $(x-4)$

• The $(x-4)$ on top and bottom will cancel in the first term

$\begin{array}{rcl}\frac{5\cancel{\left(x-4\right)}}{\cancel{x-4}}+\frac{7\left(x-4\right)}{x+2}& =& 6\left(x-4\right)\\ 5+\frac{7\left(x-4\right)}{x+2}& =& 6\left(x-4\right)\end{array}$

Now multiply all terms on both sides by the next denominator, $\left(x+2\right)$

• The $\left(x+2\right)$ on top and bottom will cancel in the second term

$\begin{array}{rcl}5\left(x+2\right)+\frac{7\left(x-4\right)\cancel{\left(x+2\right)}}{\cancel{x+2}}& =& 6\left(x-4\right)\left(x+2\right)\\ 5\left(x+2\right)+7\left(x-4\right)& =& 6\left(x-4\right)\left(x+2\right)\end{array}$

Expand brackets on both sides

• Use FOIL or another method to expand the double brackets on the right

$\begin{array}{rcl}5x+10+7x-28& =& 6\left(x^2-2x-8\right)\\ 5x+10+7x-28& =& 6x^2-12x-48\end{array}$

Collect like terms on the left

$\begin{array}{rcl}12x-18& =& 6x^2-12x-48\end{array}$

Subtract $12x$ and add 18 to both sides of the equation to get zero on one side

$0=6x^2-24x-30$

Divide both sides of the equation by 6 to simplify the coefficients

$0=x^2-4x-5$

That is a quadratic equation that can be solved by factorising

$0=\left(x+1\right)\left(x-5\right)$

$x=-1,x=5$

Those are the solutions you are looking for

• You can substitute them back into the original equation to check that they are right

$x=-1,x=5$