How do you add #3/(x +2) + 4/(x-7)#?

1 Answer
May 24, 2018

#(7x-13)/((x-7)(x +2))#

Explanation:

First you need a common denominator to add any fraction:

yours would be #(x+2)(x-7)#

so we need to multiply the first term by:

#(x-7)/(x-7)#

and the second by:

#(x+2)/(x+2)#

#3/(x +2) + 4/(x-7)#

#(x-7)/(x-7)*3/(x +2) + 4/(x-7)#

#(3(x-7))/((x-7)(x +2)) + 4/(x-7)#

#(3(x-7))/((x-7)(x +2)) + 4/(x-7)*(x+2)/(x+2)#

#(3(x-7))/((x-7)(x +2)) + (4(x+2))/((x-7)(x +2))#

#(3(x-7) + 4(x+2))/((x-7)(x +2))#

#(3x-21 + 4x+8)/((x-7)(x +2))#

#(7x-13)/((x-7)(x +2))#