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

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Answer 1 · 2018-05-24T00:29:23

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

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))#