Question 249039


Start with the given system of equations:

{{{system(x+y=1,4x-3y=11)}}}



{{{3(x+y)=3(1)}}} Multiply the both sides of the first equation by 3.



{{{3x+3y=3}}} Distribute and multiply.



So we have the new system of equations:

{{{system(3x+3y=3,4x-3y=11)}}}



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



{{{(3x+3y)+(4x-3y)=(3)+(11)}}}



{{{(3x+4x)+(3y+-3y)=3+11}}} Group like terms.



{{{7x+0y=14}}} Combine like terms.



{{{7x=14}}} Simplify.



{{{x=(14)/(7)}}} Divide both sides by {{{7}}} to isolate {{{x}}}.



{{{x=2}}} Reduce.



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



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



{{{6+3y=3}}} Multiply.



{{{3y=3-6}}} Subtract {{{6}}} from both sides.



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



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



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



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,-11,9,1-x,(11-4x)/(-3)),
circle(2,-1,0.05),
circle(2,-1,0.08),
circle(2,-1,0.10)
)}}} Graph of {{{x+y=1}}} (red) and {{{4x-3y=11}}} (green)