Question 31115
{{{ 4/(x-2) = 1 + (6/(x+2)) }}}


Multiply every term by (x-2)(x+2)
{{{ (4/(x-2))*((x-2)(x+2)) = ((x-2)(x+2)) + (6/(x+2))*((x-2)(x+2)) }}}
{{{ 4*(x+2) = ((x-2)(x+2)) + 6*(x-2) }}}
{{{ 4x+8 = ((x-2)(x+2)) + 6x-12 }}}
{{{ 4x+8 = x^2-4 + 6x-12 }}}
{{{ 4x+8 = x^2+6x-16 }}}
{{{ 8 = x^2+2x-16 }}}
{{{ 0 = x^2+2x-24 }}}
{{{ x^2+2x-24 = 0 }}}


Now factorise into (x+6)(x-4) = 0
which means that either x+6 = 0 or x-4 = 0


so that either x=-6 or x=4


You try the second question


jon.