Question 940173

{{{-4x-y=-21}}}....eq.1
{{{x-2y=12 }}}.....eq.2
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Solve by substitution

{{{x-2y=12 }}}.....eq.2...solve for {{{x}}}

{{{x=2y+12 }}}...substitute in eq.1


{{{-4(2y+12)-y=-21}}}....eq.1...solve for {{{y}}}

{{{-8y-48-y=-21}}}

{{{-9y-48=-21}}}...both sides multiply by {{{-1}}}

{{{9y+48=21}}}

{{{9y=21-48}}}

{{{9y=-27}}}

{{{y=-27/9}}}

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

go to

{{{x=2y+12 }}}...substitute {{{-3}}} for {{{y}}}

{{{x=2(-3)+12 }}}

{{{x=-6+12 }}}

{{{highlight(x=6) }}}

so, intersection point of these two lines is: ({{{6}}},{{{-3}}})


check it on a graph:

{{{drawing( 600, 600, -15, 15, -15, 25,circle(6,-3,.2),locate(6,-3,p(6,-3)), graph( 600, 600, -15, 15, -15, 25, -4x+21,x/2-6)) }}}