How do you simplify $\frac{4 y}{y^{2} - 9} - \frac{12}{y^{2} - 9}$ $\frac { 4y } { y ^ { 2} - 9} - \frac { 12} { y ^ { 2} - 9}$ ?

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Answer 1 · 2018-04-22T08:36:04

$\Rightarrow \frac{4}{y + 3}$ $=> 4/(y+3 )$

The denominator is the same...

$\Rightarrow \frac{4 y - 12}{y^{2} - 9}$ $=> (4y-12)/(y^2-9)$

Factor $4 y - 12 \to 4 (y - 3)$ $4y-12 -> 4(y-3)$

Use difference in two squares : $a^{2} - b^{2} = (a + b) (a - b)$ $a^2 - b^2 = (a+b)(a-b)$

$\Rightarrow y^{2} - 9 \to (y + 3) (y - 3)$ $=> y^2 - 9 -> (y+3)(y-3)$

$\Rightarrow \frac{4 (y - 3)}{(y + 3) (y - 3)}$ $=> ( 4(y-3) ) / ( (y+3)(y-3) )$

Cancel:

$=> ( 4cancel((y-3)) ) / ( (y+3)cancel((y-3)) )$

$=> 4/(y+3 )$

How do you simplify \frac { 4y } { y ^ { 2} - 9} - \frac { 12} { y ^ { 2} - 9}4yy2−9−12y2−9\frac { 4y } { y ^ { 2} - 9} - \frac { 12} { y ^ { 2} - 9}?