Question 148095
I'm assuming that the system is: {{{system(x-2y=8,x+2y=0)}}}



Start with the given system of equations:

{{{system(x-2y=8,x+2y=0)}}}



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



{{{(x-2y)+(x+2y)=(8)+(0)}}}



{{{(x+x)+(-2y+2y)=8+0}}} Group like terms.



{{{2x+0y=8}}} Combine like terms. Notice how the y terms cancel out.



{{{2x=8}}} Simplify.



{{{x=(8)/(2)}}} Divide both sides by {{{2}}} to isolate {{{x}}}.



{{{x=4}}} Reduce.



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{{{x-2y=8}}} Now go back to the first equation.



{{{4-2y=8}}} Plug in {{{x=4}}}.



{{{4-2y=8}}} Multiply.



{{{-2y=8-4}}} Subtract {{{4}}} 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 our answer is {{{x=4}}} and {{{y=-2}}}.



Which form the ordered pair *[Tex \LARGE \left(4,-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(4,-2\right)]. So this visually verifies our answer.



{{{drawing(500,500,-6,14,-12,8,
grid(1),
graph(500,500,-6,14,-12,8,(8-x)/(-2),(0-x)/(2)),
circle(4,-2,0.05),
circle(4,-2,0.08),
circle(4,-2,0.10)
)}}} Graph of {{{x-2y=8}}} (red) and {{{x+2y=0}}} (green)