Question 452171
{{{2x/(x-3)+1/(x+3)=-6/(x^2-9)}}}...common denominator for left side is {{{(x-3)(x+3)}}}



{{{(2x(x+3)+1(x-3))/((x+3)(x-3))=-6/(x^2-9)}}}....notice that {{{(x-3)(x+3)=x^2-3^2=x^2-9}}}...and it is denominator on the right side, so you can write:


{{{(2x(x+3)+1(x-3))/(x^2-9)=-6/(x^2-9)}}...since you have equal denominators, all you need to find {{{x}}} is nominators

{{{2x(x+3)+1(x-3)=-6}}}

{{{2x^2+6x+1x-3=-6}}}

{{{2x^2+7x-3+6=0}}}

{{{2x^2+7x+3=0}}}.......now, use quadratic formula


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}...notice that {{{a=2}}}, {{{b=7}}} and {{{c=3}}}


{{{x = (-7 +- sqrt( 7^2-4*2*3 ))/(2*2) }}}


{{{x = (-7 +- sqrt( 49-24 ))/4 }}}


{{{x = (-7 +- sqrt( 25 ))/4 }}}


{{{x = (-7 +-5 )/4 }}}

so, solutions are:

{{{x = (-7 +5 )/4 }}}

{{{x = -2 /4 }}}

{{{x = -1 /2 }}}


or

{{{x = (-7 -5 )/4 }}}

{{{x = -12 /4 }}}

{{{x = -3 }}}