Question 146851

Start with the given system of equations:

{{{system(x+y=2,2x+y=3)}}}



{{{-2(x+y)=-2(2)}}} Multiply the both sides of the first equation by -2.



{{{-2x-2y=-4}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-2x-2y=-4,2x+y=3)}}}



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:



{{{(-2x-2y)+(2x+y)=(-4)+(3)}}}



{{{(-2x+2x)+(-2y+1y)=-4+3}}} Group like terms.



{{{0x-y=-1}}} Combine like terms. Notice how the x terms cancel out.



{{{-y=-1}}} Simplify.



{{{y=(-1)/(-1)}}} Divide both sides by {{{-1}}} to isolate {{{y}}}.



{{{y=1}}} Reduce.



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{{{-2x-2y=-4}}} Now go back to the first equation.



{{{-2x-2(1)=-4}}} Plug in {{{y=1}}}.



{{{-2x-2=-4}}} Multiply.



{{{-2x=-4+2}}} Add {{{2}}} to both sides.



{{{-2x=-2}}} Combine like terms on the right side.



{{{x=(-2)/(-2)}}} Divide both sides by {{{-2}}} to isolate {{{x}}}.



{{{x=1}}} Reduce.



So our answer is {{{x=1}}} and {{{y=1}}}.



Which form the ordered pair *[Tex \LARGE \left(1,1\right)].



This means that the two equations are consistent and independent.



Notice when we graph the equations, we see that they intersect at *[Tex \LARGE \left(1,1\right)]. So this visually verifies our answer.



{{{drawing(500,500,-9,11,-9,11,
grid(1),
graph(500,500,-9,11,-9,11,2-x,3-2x),
circle(1,1,0.05),
circle(1,1,0.08),
circle(1,1,0.10)
)}}} Graph of {{{x+y=2}}} (red) and {{{2x+y=3}}} (green)