Question 180765


Start with the given system of equations:

{{{system(x+4y=9,3x-2y=13)}}}



{{{2(3x-2y)=2(13)}}} Multiply the both sides of the second equation by 2.



{{{6x-4y=26}}} Distribute and multiply.



So we have the new system of equations:

{{{system(x+4y=9,6x-4y=26)}}}



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+4y)+(6x-4y)=(9)+(26)}}}



{{{(1x+6x)+(4y+-4y)=9+26}}} Group like terms.



{{{7x+0y=35}}} Combine like terms.



{{{7x=35}}} Simplify.



{{{x=(35)/(7)}}} Divide both sides by {{{7}}} to isolate {{{x}}}.



{{{x=5}}} Reduce.



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{{{x+4y=9}}} Now go back to the first equation.



{{{5+4y=9}}} Plug in {{{x=5}}}.



{{{5+4y=9}}} Multiply.



{{{4y=9-5}}} Subtract {{{5}}} from both sides.



{{{4y=4}}} Combine like terms on the right side.



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



{{{y=1}}} Reduce.



So our answer is {{{x=5}}} and {{{y=1}}}.



Which form the ordered pair *[Tex \LARGE \left(5,1\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(5,1\right)]. So this visually verifies our answer.



{{{drawing(500,500,-5,15,-9,11,
grid(1),
graph(500,500,-5,15,-9,11,(9-x)/(4),(13-3x)/(-2)),
circle(5,1,0.05),
circle(5,1,0.08),
circle(5,1,0.10)
)}}} Graph of {{{x+4y=9}}} (red) and {{{3x-2y=13}}} (green) 




So from the graph, we see that the vertex of the angle is the point *[Tex \LARGE \left(5,1\right)]