Question 53925
{{{sqrt(2x-1)-sqrt(x-5)=3}}}
{{{sqrt(2x-1)-sqrt(x-5)+sqrt(x-5)=sqrt(x-5)+3}}}
{{{sqrt(2x-1)=sqrt(x-5)+3}}}
{{{(sqrt(2x-1))^2=(sqrt(x-5)+3)^2}}}
{{{2x-1=(sqrt(x-5)+3)(sqrt(x-5)+3)}}}
{{{2x-1=(sqrt(x-5))^2+3sqrt(x-5)+3sqrt(x-5)+9}}}
{{{2x-1=x-5+6sqrt(x-5)+9}}}
{{{2x-1=x+4+6sqrt(x-5)}}}
{{{2x-x-1-4=x-x+4-4+6sqrt(x-5)}}}
{{{x-5=6sqrt(x-5)}}}
{{{(x-5)^2=(6sqrt(x-5))^2}}}
{{{(x-5)(x-5)=(6)^2*(sqrt(x-5))^2}}}
{{{x^2-5x-5x+25=36(x-5)}}}
{{{x^2-10x+25=36x-180}}}
{{{x^2-10x-36x+25+180=36x-36x-180+180}}}
{{{x^2-46x+205=0}}}
{{{(x-41)(x-5)=0}}}
x-41=0
x-41+41=0+41
x=41
x-5=0
x-5+5=0+5
x=5
It appear that the solutions are x={5,41}.  However, you have to be especially careful to check problems that involve even roots for extraneous (false) solutions.  
Substitute 5 back into the original equation:
{{{sqrt(2(5)-1)-sqrt((5)-5)=3}}}
{{{sqrt(10-1)-sqrt(5-5)=3}}}
{{{sqrt(9)-sqrt(0)=3}}}
3-0=3
3=3 (x=5 checks out)
Substitute 41 back into the original equation:
{{{sqrt(2(41)-1)-sqrt((41)-5)=3}}}
{{{sqrt(82-1)-sqrt(41-5)=3}}}
{{{sqrt(81)-sqrt(36)=3}}}
9-6=3
3=3 (x=41 checks out)
We can now safely say that x={5,41}
Happy Calculating!!!!