Solve for x?

Question

#(4x)/(x+3) = (37)/(x^2 -9) -3#

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Answer 1 · 2018-06-17T17:21:29

#color(blue)(x=-16/7 " and " x=4)#

Move all terms to the #LHS#:

#(4x)/(x+3)-37/(x^2-9)+3=0#

Factor denominator of second term:

#(x^2-9)=(x^2-3^2)=(x+3)(x-3)# ( difference of two squares )

#(4x)/(x+3)-37/((x+3)(x-3))+3=0#

Add #LHS#:

#(4x(x-3)-37+3(x+3)(x-3))/((x+3)(x-3))=0#

Since the denominator can't be zero, we concentrate on the numerator:

#4x(x-3)-37+3(x+3)(x-3)=0#

Expand this:

#4x^2-12x-37+3x^2-27=0#

Collecting like terms:

#7x^2-12x-64=0#

Factor:

#(7x+16)(x-4)=0=>x=-16/7 and x=4#