Question 768477
{{{y=5x+1}}}....eq.1
{{{y=8x+2}}}.....eq.2......substitute {{{y}}} in eq.1
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{{{8x+2=5x+1}}}....eq.1.....solve for {{{x}}}

{{{8x-5x=1-2}}}

{{{3x=-1}}}

{{{highlight(x=-1/3)}}}........substitute {{{x}}} in eq.2

{{{y=8(-1/3)+2}}}.....eq.2.....solve for {{{y}}}

{{{y=-8/3+2}}}

{{{y=-8/3+6/3}}}

{{{highlight(y=-2/3)}}}


{{{ graph( 600, 600, -5, 5, -5, 5, 5x+1, 8x+2) }}}