Question 557487


Start with the given system of equations:

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



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



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



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



{{{3x+0y=6}}} Combine like terms.



{{{3x=6}}} Simplify.



{{{x=(6)/(3)}}} Divide both sides by {{{3}}} to isolate {{{x}}}.



{{{x=2}}} Reduce.



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



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



{{{4+y=3}}} Multiply.



{{{y=3-4}}} Subtract {{{4}}} from both sides.



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



So the solutions are {{{x=2}}} and {{{y=-1}}}.



Which form the ordered pair *[Tex \LARGE \left(2,-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(2,-1\right)]. So this visually verifies our answer.



{{{drawing(500,500,-8,12,-11,9,
grid(1),
graph(500,500,-8,12,-11,9,3-2x,(3-x)/(-1)),
circle(2,-1,0.05),
circle(2,-1,0.08),
circle(2,-1,0.10)
)}}} Graph of {{{2x+y=3}}} (red) and {{{x-y=3}}} (green) 


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