Question 641351


you have a system to solve:

{{{3x+2y=2}}}......1

{{{-2x+y=8}}}.......2
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{{{3x+2y=2}}}......1...first solve this one for {{{y}}}

{{{2y=2-3x}}}...both sides divide by {{{2}}}

{{{2y/2=2/2-3x/2}}}

{{{y=1-(3/2)x}}}..........now plug in equation 2 and solve for {{{x}}}
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{{{-2x+(1-(3/2)x)=8}}}.......2

{{{-2x+1-(3/2)x=8}}}both sides multiply by {{{2}}}

{{{-2x*2+1*2-(3/2)*2x=8*2}}}

{{{-4x+2-3x=16}}}

{{{-7x+2=16}}}

{{{-16+2=7x}}}

{{{-14=7x}}}

{{{-14/7=x}}}

{{{highlight(-2=x)}}}....now, plug it in {{{y=1-(3/2)x}}} and solve for {{{y}}}
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{{{y=1-(3/2)(-2)}}}

{{{y=1-(3/cross(2))(-cross(2))}}}..you will have.{{{-3/-1=3}}}

{{{y=1+3}}}

{{{highlight(y=4)}}}


check: plug in {{{x=-2}}} and {{{y=4}}}

{{{3x+2y=2}}}......1

{{{3(-2)+2*4=2}}}

{{{-6+8=2}}}

{{{2=2}}}


{{{-2x+y=8}}}.......2

{{{-2(-2)+4=8}}}

{{{4+4=8}}}

{{{8=8}}}