Question 133758
{{{5x/(x^2-25)=2/(x-5)}}}


Note that {{{x^2-25}}} is the difference of two squares and can be factored to {{{(x+5)(x-5)}}}.  Looking at the other denominator, we see that {{{x^2-25}}} is the common denominator and the equation can be re-written:


{{{5x/(x^2-25)=(2(x+5))/(x^2-25)}}}


Before we continue, note that the factors of the denominator are {{{(x+5)(x-5)}}}.  This means that the domain of the original function excludes the values 5 and -5 because either of those values would cause the denominator on the left to go to zero.  Keep this in mind as we develop the solution and remember to exclude either of these values if they come up as possible roots of the equation.


Now move the expression on the right side of the equation to the left by adding the additive inverse of the right to both sides:


{{{5x/(x^2-25)-(2(x+5))/(x^2-25)=0}}}


Next add the numerators:
{{{(5x-2x-10)/(x^2-25)=0}}}
{{{(3x-10)/(x^2-25)=0}}}


Since we know that {{{a/b=0}}} if and only if {{{a=0}}} and {{{b<>0}}}, we can simply set the numerator of our equation equal to zero and solve, then exclude any root that would make the denominator go to zero.


{{{3x-10=0}}}
{{{3x=10}}}
{{{x=10/3}}}


So the solution set is the value {{{10/3}}} and we have no roots that need to be excluded.


Check:

{{{(5(10/3))/((10/3)^2-25)=2/((10/3)-5)}}}
{{{(50/3)/(100/9-225/9)=2/(10/3-15/3)}}}
{{{(50/3)/(-125/9)=2/(-5/3)}}}
{{{(50/3)(-9/125)=2(-3/5)}}}
{{{(2/1)(-3/5)=-6/5}}}
{{{-6/5=-6/5}}}, Answer checks.


For a visual check, make the original equation into a function {{{f(x)=5x/(x^2-25)-2/(x-5)}}} and sketch a graph of the function:


{{{drawing(600,600,-10,10,-10,10,
grid(1),
graph(600,600,-10,10,-10,10,(3x-10)/(x^2-25)),
circle(10/3,0,.2)
)}}} 


Note the asymptotes at x = 5 and x = -5 and the x-intercept at {{{10/3}}}