# How do you solve the rational equation 5 / (y - 2) = y + 2 ?

Mar 7, 2018

color(magenta)(therefore y=3 or y=-3

#### Explanation:

$\frac{5}{y - 2} = y + 2$

$\left(y + 2\right) \left(y - 2\right) = 5$

Identity:
color(red)((a+b)(a-b)=a^2-b^2

${y}^{2} - 4 = 5$

${y}^{2} - 4 - 5 = 0$

${y}^{2} - 9 = 0$

Identity:
color(red)((a+b)(a-b)=a^2-b^2

$\left(y + 3\right) \left(y - 3\right) = 0$

$\therefore y - 3 = 0 \mathmr{and} y + 3 = 0$

color(magenta)(therefore y=3 or y=-3

Mar 7, 2018

$y = \pm 3$

#### Explanation:

$\text{multiply both sides by } \left(y - 2\right)$

$\cancel{\left(y - 2\right)} \times \frac{5}{\cancel{\left(y - 2\right)}} = \left(y - 2\right) \left(y + 2\right)$

$\Rightarrow \left(y - 2\right) \left(y + 2\right) = 5 \leftarrow \textcolor{b l u e}{\text{expand factors}}$

$\Rightarrow {y}^{2} - 4 = 5 \leftarrow \textcolor{b l u e}{\text{add 4 to both sides}}$

$\Rightarrow {y}^{2} = 9$

$\textcolor{b l u e}{\text{take square root of both sides}}$

$\Rightarrow y = \pm 3 \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\textcolor{b l u e}{\text{As a check}}$

$y = 3 \to \frac{5}{3 - 2} = 3 + 2 \to 5 = 5 \text{ True}$

$y = - 3 \to \frac{5}{- 3 - 2} = - 3 + 2 \to - 1 = - 1 \text{ True}$

$\Rightarrow y = \pm 3 \text{ are the solutions}$