Solving Equations with Algebraic Fractions (Edexcel IGCSE Maths A (Modular))

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Solving Algebraic Fractions

How do I solve an equation that contains algebraic fractions?

There are two methods for solving equations that contain algebraic fractions
One method is to add or subtract the algebraic fractions first and then solve as usual
- For example, to solve $\frac{8}{x + 1} - \frac{5}{x + 2} = 1$
- First subtract the fractions and simplify, $\frac{3 x + 11}{(x + 1) (x + 2)} = 1$
- Then cross-multiply, expand and solve
  $\begin{array}{rcl} 3 x + 11 & = & 1 (x + 1) (x + 2) \\ 3 x + 11 & = & x^{2} + 3 x + 2 \\ 0 & = & x^{2} - 9 \\ 0 & = & (x - 3) (x + 3) \\ x & = & 3 or x = - 3 \end{array}$
Alternatively, you can remove the fractions first by multiplying everything on both sides of the equation by each expression in the denominators and then solve
- For example, to solve the equation $\frac{4}{x - 3} + \frac{5}{x + 1} = 5$
- First multiply every term in the equation by both $(x - 3)$ and $(x + 1)$ and cancel common factors where possible
  - Multiply every term by $(x - 3)$ (this bracket goes in the numerator of any fractions)
    $\begin{array}{rcl} \frac{4}{(x - 3)} (x - 3) + \frac{5 (x - 3)}{x + 1} & = & 5 (x - 3) \\ 4 + \frac{5 (x - 3)}{x + 1} & = & 5 (x - 3) \end{array}$
  - Then multiply every term by $\begin{array}{rcl} (x + 1) \end{array}$
    $\begin{array}{rcl} 4 (x + 1) + \frac{5 (x - 3)}{(x + 1)} (x + 1) & = & 5 (x - 3) (x + 1) \\ 4 (x + 1) + 5 (x - 3) & = & 5 (x - 3) (x + 1) \end{array}$
- Then solve
  $\begin{array}{rcl} 4 x + 4 + 5 x - 15 & = & 5 (x^{2} - 2 x - 3) \\ 9 x - 11 & = & 5 x^{2} - 10 x - 15 \\ 0 & = & 5 x^{2} - 19 x - 4 \\ 0 & = & (5 x + 1) (x - 4) \\ x & = & - \frac{1}{5} or x = 4 \end{array}$

Exam Tip

When multiplying by an algebraic expression, use brackets around the expression, e.g. $(2 x + 3)$
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 either side of the equation

Worked Example

$\frac{2}{p + 3} - \frac{5}{p} = 6 p$

Show that this equation can be written as $6 p^{3} + 18 p^{2} + 3 p + 15 = 0$ .

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

Start by multiplying all terms in the equation by the denominator $(p + 3)$
The $(p + 3)$ on top and bottom will cancel in the first term

$2 - \frac{5 (p + 3)}{p} = 6 p (p + 3)$

Now multiply all terms on both sides by the next denominator, $p$
The $p$ on top and bottom will cancel in the second term

$2 (p) - 5 (p + 3) = 6 p (p + 3) (p)$

Expand brackets
Be careful with negative signs

$\begin{array}{rcl} 2 p - 5 (p + 3) & = & 6 p^{2} (p + 3) \\ 2 p - 5 p - 15 & = & 6 p^{3} + 18 p^{2} \end{array}$

Collect like terms

$- 3 p - 15 = 6 p^{3} + 18 p^{2}$

Add $3 p$ and 15 to both sides of the equation

$0 = 6 p^{3} + 18 p^{2} + 3 p + 15$

$6 p^{3} + 18 p + 3 p + 15 = 0$

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