How do you combine #6/ ( y ^ { 2} - 16) - 5/ ( y + 4)#?

Question

424 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-14T13:21:38

See a solution process below:

First, multiple the fraction on the right by the appropriate form of #1# to put it over a common denominator with the fraction on the left:

#6/(y^2 - 16) - (5/(y + 4) xx (y - 4)/(y - 4)) =>#

#6/(y^2 - 16) - (5(y - 4))/((y + 4) xx (y - 4)) =>#

#6/(y^2 - 16) - (5y - 20)/(y^2 - 16)#

We can now subtract the numerators over the common denominator:

#(6 - (5y - 20))/(y^2 - 16) =>#

#(6 - 5y + 20)/(y^2 - 16) =>#

#(-5y + 20 + 6)/(y^2 - 16) =>#

#(-5y + 26)/(y^2 - 16) =>#