Question 218289


Start with the given system of equations:

{{{system(x+2y=1,3x-2y=11)}}}



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)+(3x-2y)=(1)+(11)}}}



{{{(1x+3x)+(2y+-2y)=1+11}}} Group like terms.



{{{4x+0y=12}}} Combine like terms. Notice how the y terms cancel out.



{{{4x=12}}} Simplify.



{{{x=(12)/(4)}}} Divide both sides by {{{4}}} to isolate {{{x}}}.



{{{x=3}}} Reduce.



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{{{x+2y=1}}} Now go back to the first equation.



{{{3+2y=1}}} Plug in {{{x=3}}}.



{{{3+2y=1}}} Multiply.



{{{2y=1-3}}} Subtract {{{3}}} from both sides.



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



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



{{{y=-1}}} Reduce.



So our answer is {{{x=3}}} and {{{y=-1}}}.



Which form the ordered pair *[Tex \LARGE \left(3,-1\right)]. Good job. As a check, simply plug in {{{x=3}}} and {{{y=-1}}} and you should get two identities.



This means that the system is consistent and independent.



Notice when we graph the equations, we see that they intersect at *[Tex \LARGE \left(3,-1\right)]. So this visually verifies our answer.



{{{drawing(500,500,-7,13,-11,9,
grid(1),
graph(500,500,-7,13,-11,9,(1-x)/(2),(11-3x)/(-2)),
circle(3,-1,0.05),
circle(3,-1,0.08),
circle(3,-1,0.10)
)}}} Graph of {{{x+2y=1}}} (red) and {{{3x-2y=11}}} (green)