SOLUTION: Solve and explain if there is an extraneous solution. (2x-5)/(3x-6) =2/5

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Question 821299: Solve and explain if there is an extraneous solution.
(2x-5)/(3x-6) =2/5
Found 2 solutions by ankor@dixie-net.com, KMST:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
%282x-5%29%2F%283x-6%29 = 2%2F5
Cross multiply
5(2x-5) = 2(3x-6)
10x - 25 = 6x - 12
10x - 6x = -12 + 25
4x = 13
x = 13%2F4
:
You can check this in the original equation using decimal equiv; x=3.25

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
%282x-5%29%2F%283x-6%29+=2%2F5
5%282x-5%29%2F%283x-6%29+=2
5%282x-5%29+=2%283x-6%29
10x-25=6x-12
10x-6x-25=-12
4x=25-12
4x=13
highlight%28x=13%2F4%29

Verification:
I know that the only step where I could have introduced an extraneous solution is hen I multiplied both sides of the equal sign times the 2x-5 denominator.
If I had found 1%2F2 for a solution, it would have been an extraneous solution that would make the denominator zero.
The easiest way to check your work for extraneous solutions, and also for mistakes is to substitute the value found for the solution in the original equation, and see if that value makes it true.

The solution found is not an extraneous solution,
and I did not make a mistake in the calculations either.
If I had found 1%2F2 for a solution, it would have been an extraneous solution that would make the denominator zero.