Question 182817


Start with the given system of equations:

{{{system(x+3y=19,x-y=-1)}}}



{{{3(x-y)=3(-1)}}} Multiply the both sides of the second equation by 3. The goal here is to get the "y" coefficients equal but opposite.



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



So we have the new system of equations:

{{{system(x+3y=19,3x-3y=-3)}}}



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+3y)+(3x-3y)=(19)+(-3)}}}



{{{(x+3x)+(3y+-3y)=19+-3}}} Group like terms.



{{{4x+0y=16}}} Combine like terms.



{{{4x=16}}} Simplify.



{{{x=(16)/(4)}}} Divide both sides by {{{4}}} to isolate {{{x}}}.



{{{x=4}}} Reduce.



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{{{x+3y=19}}} Now go back to the first equation.



{{{4+3y=19}}} Plug in {{{x=4}}}.



{{{4+3y=19}}} Multiply.



{{{3y=19-4}}} Subtract {{{4}}} from both sides.



{{{3y=15}}} Combine like terms on the right side.



{{{y=(15)/(3)}}} Divide both sides by {{{3}}} to isolate {{{y}}}.



{{{y=5}}} Reduce.



So our answer is {{{x=4}}} and {{{y=5}}}.



Which form the ordered pair *[Tex \LARGE \left(4,5\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(4,5\right)]. So this visually verifies our answer.



{{{drawing(500,500,-6,14,-5,15,
grid(1),
graph(500,500,-6,14,-5,15,(19-x)/(3),(-1-x)/(-1)),
circle(4,5,0.05),
circle(4,5,0.08),
circle(4,5,0.10)
)}}} Graph of {{{x+3y=19}}} (red) and {{{x-y=-1}}} (green)