Question 821299
{{{(2x-5)/(3x-6) =2/5}}}
{{{5(2x-5)/(3x-6) =2}}}
{{{5(2x-5) =2(3x-6)}}}
{{{10x-25=6x-12}}}
{{{10x-6x-25=-12}}}
{{{4x=25-12}}}
{{{4x=13}}}
{{{highlight(x=13/4)}}}
 
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 {{{1/2}}} 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.
{{{(2(13/4)-5)/(3(13/4)-6) =(13/2+10/2)/(39/4-24/4)=(3/2)/(15/4)=(3/2)*(4/15)=12/30=2/5}}}
The solution found is not an extraneous solution,
and I did not make a mistake in the calculations either.
If I had found {{{1/2}}} for a solution, it would have been an extraneous solution that would make the denominator zero.