Question 176474


Start with the given system of equations:

{{{system(3x-2y=0,x+y=-5)}}}



{{{2(x+y)=2(-5)}}} Multiply the both sides of the second equation by 2.



{{{2x+2y=-10}}} Distribute and multiply.



So we have the new system of equations:

{{{system(3x-2y=0,2x+2y=-10)}}}



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



{{{(3x-2y)+(2x+2y)=(0)+(-10)}}}



{{{(3x+2x)+(-2y+2y)=0+-10}}} Group like terms.



{{{5x+0y=-10}}} Combine like terms.



{{{5x=-10}}} Simplify.



{{{x=(-10)/(5)}}} Divide both sides by {{{5}}} to isolate {{{x}}}.



{{{x=-2}}} Reduce.



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{{{3x-2y=0}}} Now go back to the first equation.



{{{3(-2)-2y=0}}} Plug in {{{x=-2}}}.



{{{-6-2y=0}}} Multiply.



{{{-2y=0+6}}} Add {{{6}}} to both sides.



{{{-2y=6}}} Combine like terms on the right side.



{{{y=(6)/(-2)}}} Divide both sides by {{{-2}}} to isolate {{{y}}}.



{{{y=-3}}} Reduce.



So our answer is {{{x=-2}}} and {{{y=-3}}}.



Which form the ordered pair *[Tex \LARGE \left(-2,-3\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(-2,-3\right)]. So this visually verifies our answer.



{{{drawing(500,500,-12,8,-13,7,
grid(1),
graph(500,500,-12,8,-13,7,(0-3x)/(-2),-5-x),
circle(-2,-3,0.05),
circle(-2,-3,0.08),
circle(-2,-3,0.10)
)}}} Graph of {{{3x-2y=0}}} (red) and {{{x+y=-5}}} (green)