# How do you solve y/(y-3) +6/(y+3)= 1?

Oct 12, 2015

The solution is $y = 1$.

#### Explanation:

First of all, turn the left member to a single fraction:

$\frac{y}{y - 3} + \frac{6}{y + 3} = \frac{y \left(y + 3\right) + 6 \left(y - 3\right)}{\left(y + 3\right) \left(y - 3\right)}$

Simplify the numerator, and use the $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ formula for the denominator:

• Numerator:

$y \left(y + 3\right) + 6 \left(y - 3\right) = {y}^{2} + 3 y + 6 y - 18 = {y}^{2} + 9 y - 18$

• Denominator:

$\left(y + 3\right) \left(y - 3\right) = {y}^{2} - 9$

So, the left member (and the whole equation) become

$\frac{{y}^{2} + 9 y - 18}{{y}^{2} - 9} = 1$

Multiply both members for the denominator:

$\cancel{{y}^{2}} + 9 y - 18 = \cancel{{y}^{2}} - 9$

Isolate $y$ terms and constants on the two sides:

$9 y = 9$

Solve for $y$: $y = \frac{9}{9} = 1$.