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

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Answer 1 · 2017-05-14T15:39:31

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

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

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

Then you have a quadratic equation.

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