Question 557490


Start with the given system of equations:

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



{{{-1(x-3y)=-1(2)}}} Multiply the both sides of the first equation by -1.



{{{-1x+3y=-2}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-1x+3y=-2,x+2y=-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:



{{{(-1x+3y)+(x+2y)=(-2)+(-3)}}}



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



{{{0x+5y=-5}}} Combine like terms.



{{{5y=-5}}} Simplify.



{{{y=(-5)/(5)}}} Divide both sides by {{{5}}} to isolate {{{y}}}.



{{{y=-1}}} Reduce.



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{{{-1x+3y=-2}}} Now go back to the first equation.



{{{-1x+3(-1)=-2}}} Plug in {{{y=-1}}}.



{{{-x-3=-2}}} Multiply.



{{{-x=-2+3}}} Add {{{3}}} to both sides.



{{{-x=1}}} Combine like terms on the right side.



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



{{{x=-1}}} Reduce.



So the solutions are {{{x=-1}}} and {{{y=-1}}}.



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



This means that the system is 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,-11,9,-11,9,
grid(1),
graph(500,500,-11,9,-11,9,(2-x)/(-3),(-3-x)/(2)),
circle(-1,-1,0.05),
circle(-1,-1,0.08),
circle(-1,-1,0.10)
)}}} Graph of {{{x-3y=2}}} (red) and {{{x+2y=-3}}} (green) 


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