Question 1182622

assuming you have a system to solve:

{{{25x^2-y^2=36}}}....eq.1
{{{5x+y=2}}}.....eq.2
-------------------------------

{{{5x+y=2}}}.....eq.2.......solve for {{{y}}}


{{{y=2-5x}}}....eq.2a.substitute in eq.1


{{{25x^2-(2-5x)^2=36}}}....eq.1...solve for {{{x}}}

{{{25x^2-(4-20x+25x^2)=36}}}

{{{25x^2-4+20x-25x^2=36}}}

{{{20x=36+4}}}

{{{20x=40}}}

{{{x=2}}}


go to

{{{y=2-5x}}}....eq.2a, substitute {{{x}}}

{{{y=2-5*2}}}

{{{y=2-10}}}

{{{y=-8}}}


solutions:

{{{x = 2}}}, {{{y = -8}}}