Question 1189856
{{{5x-2y=46}}}..eq.1
{{{-2x+2y=-16}}}...eq.2
--------------------------add both

{{{5x-2y-2x+2y=46-16}}}

{{{5x-2x=30}}}

{{{3x=30}}}

{{{x=10}}}

go to

{{{-2x+2y=-16}}}...eq.2, substitute {{{x}}}

{{{-2*10+2y=-16}}}...solve for {{{y}}}

{{{-20+2y=-16}}}

{{{2y=20-16}}}

{{{2y=4}}}

{{{y=2}}}

solution: 

{{{x=10}}}
{{{y=2}}}

intersection point:({{{10}}},{{{2}}})