Question 360764

Start with the given system of equations:

{{{system(x-y=1,2x-y=5)}}}



{{{-1(2x-y)=-1(5)}}} Multiply the both sides of the second equation by -1.



{{{-2x+y=-5}}} Distribute and multiply.



So we have the new system of equations:

{{{system(x-y=1,-2x+y=-5)}}}



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



{{{(x-y)+(-2x+y)=(1)+(-5)}}}



{{{(x-2x)+(-y+y)=1-5}}} Group like terms.



{{{-x+0y=-4}}} Combine like terms.



{{{-x=-4}}} Simplify.



{{{x=(-4)/(-1)}}} Divide both sides by {{{-1}}} to isolate {{{x}}}.



{{{x=4}}} Reduce.



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



{{{4-y=1}}} Plug in {{{x=4}}}.



{{{-y=1-4}}} Subtract {{{4}}} from both sides.



{{{-y=-3}}} Combine like terms on the right side.



{{{y=(-3)/(-1)}}} Divide both sides by {{{-1}}} to isolate {{{y}}}.



{{{y=3}}} Reduce.



So the solutions are {{{x=4}}} and {{{y=3}}}.



Which form the ordered pair *[Tex \LARGE \left(4,3\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(4,3\right)]. So this visually verifies our answer.



{{{drawing(500,500,-6,14,-7,13,
grid(1),
graph(500,500,-6,14,-7,13,(1-x)/(-1),(5-2x)/(-1)),
circle(4,3,0.05),
circle(4,3,0.08),
circle(4,3,0.10)
)}}} Graph of {{{x-y=1}}} (red) and {{{2x-y=5}}} (green) 



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