Question 376340


Start with the given system of equations:

{{{system(5x+y=-2,3x-2y=4)}}}



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



{{{10x+2y=-4}}} Distribute and multiply.



So we have the new system of equations:

{{{system(10x+2y=-4,3x-2y=4)}}}



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



{{{(10x+2y)+(3x-2y)=(-4)+(4)}}}



{{{(10x+3x)+(2y+-2y)=-4+4}}} Group like terms.



{{{13x+0y=0}}} Combine like terms.



{{{13x=0}}} Simplify.



{{{x=(0)/(13)}}} Divide both sides by {{{13}}} to isolate {{{x}}}.



{{{x=0}}} Reduce.



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{{{10x+2y=-4}}} Now go back to the first equation.



{{{10(0)+2y=-4}}} Plug in {{{x=0}}}.



{{{0+2y=-4}}} Multiply.



{{{2y=-4-0}}} Subtract {{{0}}} from both sides.



{{{2y=-4}}} Combine like terms on the right side.



{{{y=(-4)/(2)}}} Divide both sides by {{{2}}} to isolate {{{y}}}.



{{{y=-2}}} Reduce.



So the solutions are {{{x=0}}} and {{{y=-2}}}.



Which form the ordered pair *[Tex \LARGE \left(0,-2\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(0,-2\right)]. So this visually verifies our answer.



{{{drawing(500,500,-10,10,-12,8,
grid(1),
graph(500,500,-10,10,-12,8,-2-5x,(4-3x)/(-2)),
circle(0,-2,0.05),
circle(0,-2,0.08),
circle(0,-2,0.10)
)}}} Graph of {{{5x+y=-2}}} (red) and {{{3x-2y=4}}} (green) 



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Jim