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

Question

Avery Alexander · Accepted Answer

#x=color(red)5#

First, let's address the initial issue:

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

To make the equation simpler, we'll apply cross multiplication:

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

Then, to simplify things, we'll apply the distributive property to both sides of the equation:

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

We now combine similar terms:

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

Julia Adams · Answer

undefined To solve the equation (3x)/(x-2)=(3x+5)/(x-1) and find any extraneous solutions, we can start by cross-multiplying to eliminate the fractions. This gives us 3x(x-1) = (3x+5)(x-2). Expanding both sides of the equation, we get 3x^2 - 3x = 3x^2 - x - 10. Simplifying further, we have -3x = -x - 10. By combining like terms, we get -2x = -10. Dividing both sides by -2, we find x = 5.

To check for extraneous solutions, we substitute x = 5 back into the original equation. We have (3(5))/(5-2) = (3(5)+5)/(5-1), which simplifies to 15/3 = 20/4. This equation is true, so x = 5 is a valid solution and there are no extraneous solutions.