How do you solve #15/(x^2-1)=4/(x-1)-3/(x+1)#?

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Answer 1 · 2018-03-30T18:24:44

#x = 8#

We can combine like fractions:

#4/(x-1) - 3/(x+1) = (4(x+1))/((x+1)(x-1)) - (3(x-1))/((x-1)(x+1)) #
# = (4x + 4 - 3x + 3)/(x^2-1) = (x+7)/(x^2+1) #

Setting this equal to the first equation, we get
#15/(x^2-1) = (x+7) / (x^2-1) #

Let #x ne pm 1# so there aren't infinities we will deal with.

This means that
#x+7 = 15 #
#x = 8 #

If we plug in #x=8# to the original, we can check:

#15/(8^2 - 1) = 4/(8-1) - 3/(8+1) #
#15/63 = 4/7 - 3/9 = 4/7 - 1/3#
#5/21 = 12/21 - 7/21 = 5/21. #
As we hope.