How do you solve $\frac { 1} { y + 3} = \frac { 7} { y - 3} - \frac { 2} { y ^ { 2} - 9}$ ?

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Answer 1 · 2018-05-11T04:18:45

$y=-11/3$

$1/(y+3)=7/(y-3)-2/(y^2-9)$

$1/(y+3)=7/(y-3)-2/((y-3)(y+3))$

$(y-3)/((y-3)(y+3))=(7(y+3))/((y-3)(y+3))-2/((y-3)(y+3))$

$y-3=7y+21-2$

$6y=-3-21+2$

$6y=-22$

$y=-11/3$

How do you solve \frac { 1} { y + 3} = \frac { 7} { y - 3} - \frac { 2} { y ^ { 2} - 9}\frac { 1} { y + 3} = \frac { 7} { y - 3} - \frac { 2} { y ^ { 2} - 9}?