How do you solve $\frac{1}{y} + \frac{1}{y + 8} = \frac{1}{11}$ $\frac { 1} { y } + \frac { 1} { y + 8} = \frac { 1} { 11}$ ?

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How do you solve $\frac{1}{y} + \frac{1}{y + 8} = \frac{1}{11}$ $\frac { 1} { y } + \frac { 1} { y + 8} = \frac { 1} { 11}$ ?

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Answer 1 · 2017-05-14T15:39:31

First, note that $y$ $y$ cannot be equal to $0$ $0$ or $- 8$ $-8$

Then, multiply both sides by $11 \cdot y \cdot (y + 8)$ $11*y*(y+8)$ , you get :

$11 (y + 8) + 11 y = y (y + 8)$ $11(y+8) + 11y = y(y+8)$
$\Leftrightarrow$ $<=>$
$22 y + 88 = y^{2} + 8 y$ $22y+88 = y^2 + 8y$
$\Leftrightarrow$ $<=>$
$y^{2} - 14 y - 88 = 0$ $y^2 - 14y - 88 = 0$

Then you have a quadratic equation.

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How do you solve $\frac{1}{y} + \frac{1}{y + 8} = \frac{1}{11}$ $\frac { 1} { y } + \frac { 1} { y + 8} = \frac { 1} { 11}$ ?

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