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

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Answer 1 · 2017-04-03T06:08:56

#x = 8" "# or #" "x = -2#

Given:

#(x-4)/3 = (x+4)/(x+1)#

Multiply both sides by #3(x+1)# to get:

#(x-4)(x+1) = 3(x+4)#

Multiply out to get:

#x^2-3x-4 = 3x+12#

Subtract #3x+12# from both sides to get:

#x^2-6x-16 = 0#

Note that #16 = 8*2# and #6 = 8-2#. So we find:

#0 = x^2-6x-16 = (x-8)(x+2)#

So the solutions are:

#x = 8" "# or #" "x = -2#

Finally note that neither of these values results in a denominator becoming #0# in the original problem, so both are valid solutions.