How do you solve #5 / (x - 2)^2 = 10 / (3x + 3)#?

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Answer 1 · 2016-08-29T07:36:09

#x=0.5 or x = 5#

#5/(x-2)^2=10/(3x+3)#

#15x+15=10(x^2-4x+4)=10x^2-40x+40#

#:. 10x^2-55x+25=0#

#5(2x^2-11x+5)=0#

#5(2x-1)(x-5)=0#

#(2x-1)(x-5)=0#

graph{(y-5/(x-2)^2)(y-10/(3*x+3))=0 [-1, 6, -1, 3]}

graph{10x^2-55x+25 [-1, 6, -60, 10]}

Answer 2 · 2016-08-29T09:32:52

#x = 1/2 or x =5#

There is one term (a fraction) on each side of the equation.
We can cross-multiply to get rid of the denominators.

#10(x-2)^2 = 5(3x+3) color(white)(...........................) div5#

#2(x-2)^2 = (3x+3)#

#2(x^2 -4x +4) = 3x+3#

#2x^2 -8x +8 -3x -3=0#

#2x^2 -11x +5 =0#

Find factors of 2 and 5 which add up to 11.
(Work towards #1x+10x = 11x#)

#(2x -1)(1 -5)=0#

#x = 1/2 or x =5#