How do you solve #1/2(x-3)+3/(2-x)=5 x#?

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Answer 1 · 2016-03-01T01:29:27

First, distribute.

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

The LCD (least common denominator) is #(2)(2 - x)#

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

You can now eliminate the denominators since everything is equivalent,

#-x^2 - 3x + 2x - 6 + 6 = 20x - 10x^2#

#-x^2 - 21x + 10x^2 = 0#

#9x^2 - 21x = 0#

#3x(3x - 7) = 0#

#x = 0 and 7/3#

Hopefully this helps!