Question 234394
If you multiply both sides by {{{x*(x-5)}}}, the
equation will still be true
{{{4x/(x-5) + 2/x  = -4/(x-5)}}}
{{{(x*(x-5))*4x/(x-5) + (x*(x-5))*2/x  = (x*(x-5))*-4/(x-5)}}}
{{{x*4x + 2*(x-5) = -4x}}}
{{{4x^2 + 2x - 10 = -4x}}}
{{{4x^2 + 6x - 10= 0}}}
{{{2x^2 + 3x - 5 = 0}}}
Use the quadratic formula to solve
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 2}}}
{{{b = 3}}}
{{{c = -5}}}
{{{x = (-3 +- sqrt( 3^2-4*2*(-5) ))/(2*2) }}}
{{{x = (-3 +- sqrt( 9 + 40 ))/4 }}}
{{{x = (-3 +- sqrt( 49 ))/4 }}}
{{{x = (-3 + 7)/4}}}
{{{x = 1}}}
and
{{{x = (-3 - 7)/4}}}
{{{x = -5/2}}}
You can check these 2 solutions by plugging back into
the original equation