# How do you solve the rational equation (y +2)/ y = 1 /( y - 5)?

Jun 21, 2018

$x = 2 + \sqrt{14}$

and

$x = 2 - \sqrt{14}$

#### Explanation:

Since there are only two terms you can X-multiply:

$\frac{y + 2}{y} = \frac{1}{y - 5}$

X-multiply:

$\left(y + 2\right) \left(y - 5\right) = y \cdot 1$

${y}^{2} - 3 y - 10 = y$

${y}^{2} - 4 y - 10 = 0$

Now to solve we have to complete the square:

$a {x}^{2} + b x + c$

$a$ must equal one (it does in your function).

$c = {\left(\frac{b}{2}\right)}^{2}$

the completed square is ${\left(x + \frac{b}{2}\right)}^{2}$

${y}^{2} - 4 y = 10$

your $b = - 4$

$c = {\left(- \frac{4}{2}\right)}^{2} = 4$ and the square is ${\left(x - 2\right)}^{2}$

So let's complete the square, remember we need to add $c$ to both sides so we don't alter the equation:

${y}^{2} - 4 y + c = 10 + c$

${y}^{2} - 4 y + 4 = 10 + 4$

$\left(x - 2\right) \left(x - 2\right) = 14$

${\left(x - 2\right)}^{2} = 14$

now solve for $x$:

$\sqrt{{\left(x - 2\right)}^{2}} = \pm \sqrt{14}$

$x - 2 = \pm \sqrt{14}$

$x = 2 \pm \sqrt{14}$

$y = 2 \setminus \pm \setminus \sqrt{14}$

#### Explanation:

Given rational equation:
$\setminus \frac{y + 2}{y} = \setminus \frac{1}{y - 5}$
will have solution such that $y \setminus \ne 0 , y \setminus \ne 5 \setminus \mathmr{and} \setminus y \setminus \notin \left\{0 , 5\right\}$
$\left(y + 2\right) \left(y - 5\right) = y$
${y}^{2} + 2 y - 5 y - 10 = y$
${y}^{2} - 4 y - 10 = 0$
$y = \setminus \frac{- \left(- 4\right) \setminus \pm \setminus \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(- 10\right)}}{2 \left(1\right)}$
y=\frac{4\pm2\sqrt{14}}{2)}
$y = 2 \setminus \pm \setminus \sqrt{14}$