How do you solve #(3x)/(x-2)=(3x+5)/(x-1)# and find any extraneous solutions?

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Answer 1 · 2017-06-07T23:18:50

#x=color(red)5#

Let's start with the original problem:

#(3x)/(x-2)=(3x+5)/(x-1)#

We'll use cross multiplication to simplify the equation:

#3x(x-1)=(x-2)(3x+5)#

We'll then use the distributive property on both sides of the equation to make things easier:

#(3x)(x)-(3x)(1)=(x-2)(3x)+(x-2)(5)#
#3x^2-3x=3x^2-6x+5x-10#

Now, we combine like terms:

#3x^2-3x=3x^2-6x+5x-10#
#-3x=-6x+5x-10#
#-2x=-10#
#x=(-10)/-2=color(red)5#