Question 1152004
{{{x+2y=8}}}........eq.1

{{{4x-2y=12}}}........eq.2
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{{{x+2y=8}}}........eq.1..solve for {{{x}}}

{{{x=8-2y}}}........eq.1a...=> substitute in eq.2


{{{4(8-2y)-2y=12}}}.......solve for {{{y}}}

{{{32-8y-2y=12}}}

{{{32-10y=12}}}

{{{32-12=10y}}}

{{{20=10y}}}

{{{y=2}}}.....=> substitute in eq.1a


{{{x=8-2*2}}}........eq.1a

{{{x=8-4}}}

{{{x=4}}}

solution: 
{{{x=4}}},{{{y=2}}}

=> intersection of these two lines is at  ({{{4}}},{{{2}}})



{{{drawing( 600, 600, -10, 10, -10, 10,
circle(4,2,.12),locate(4,2.3,p(4,2)),
 graph( 600, 600, -10, 10, -10, 10, 4-x/2, 2x-6)) }}}