Question 407336


Start with the given system of equations:

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



{{{-1(2x+y)=-1(4)}}} Multiply the both sides of the second equation by -1.



{{{-2x-1y=-4}}} Distribute and multiply.



So we have the new system of equations:

{{{system(x+y=3,-2x-1y=-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:



{{{(x+y)+(-2x-1y)=(3)+(-4)}}}



{{{(1x+-2x)+(1y+-1y)=3+-4}}} Group like terms.



{{{-x+0y=-1}}} Combine like terms. Notice how the y terms cancel out.



{{{-x=-1}}} Simplify.



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



{{{x=1}}} Reduce.



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



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



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



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



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



So our answer is {{{x=1}}} and {{{y=2}}}.



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



{{{drawing(500,500,-9,11,-8,12,
grid(1),
graph(500,500,-9,11,-8,12,3-x,4-2x),
circle(1,2,0.05),
circle(1,2,0.08),
circle(1,2,0.10)
)}}} Graph of {{{x+y=3}}} (red) and {{{2x+y=4}}} (green) 



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