How do you add $\frac { 12y } { y ^ { 2} + 16y + 48} + \frac { 6} { y + 4}$ ?

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Answer 1 · 2017-02-02T19:55:37

$(6(y+14))/((y+4)(y+12))$

$12/(y^2+16y+48)+6/(y+4)$
Factorise
$12/((y+4)(y+12))+6/(y+4)$

So to get a common denominator multiply numerator and denominator of the second term by y+12

$=12/((y+4)(y+12))+6/(y+4)*(y+12)/(y+12)$

$=(12+6(y+12))/((y+4)(y+12))$
$=(12+6y+72)/((y+4)(y+12))$
$=(6y+84)/((y+4)(y+12))$
$=(6(y+14))/((y+4)(y+12))$

How do you add \frac { 12y } { y ^ { 2} + 16y + 48} + \frac { 6} { y + 4}\frac { 12y } { y ^ { 2} + 16y + 48} + \frac { 6} { y + 4}?