Question 637131


Start with the given system of equations:

{{{system(2x-4y=40,8x-3y=82)}}}



{{{3(2x-4y)=3(40)}}} Multiply the both sides of the first equation by 3.



{{{6x-12y=120}}} Distribute and multiply.



{{{-4(8x-3y)=-4(82)}}} Multiply the both sides of the second equation by -4.



{{{-32x+12y=-328}}} Distribute and multiply.



So we have the new system of equations:

{{{system(6x-12y=120,-32x+12y=-328)}}}



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



{{{(6x-12y)+(-32x+12y)=(120)+(-328)}}}



{{{(6x+-32x)+(-12y+12y)=120+-328}}} Group like terms.



{{{-26x+0y=-208}}} Combine like terms.



{{{-26x=-208}}} Simplify.



{{{x=(-208)/(-26)}}} Divide both sides by {{{-26}}} to isolate {{{x}}}.



{{{x=8}}} Reduce.



------------------------------------------------------------------



{{{6x-12y=120}}} Now go back to the first equation.



{{{6(8)-12y=120}}} Plug in {{{x=8}}}.



{{{48-12y=120}}} Multiply.



{{{-12y=120-48}}} Subtract {{{48}}} from both sides.



{{{-12y=72}}} Combine like terms on the right side.



{{{y=(72)/(-12)}}} Divide both sides by {{{-12}}} to isolate {{{y}}}.



{{{y=-6}}} Reduce.



So the solutions are {{{x=8}}} and {{{y=-6}}}.



Which form the ordered pair *[Tex \LARGE \left(8,-6\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(8,-6\right)]. So this visually verifies our answer.



{{{drawing(500,500,-2,18,-16,4,
grid(1),
graph(500,500,-2,18,-16,4,(40-2x)/(-4),(82-8x)/(-3)),
circle(8,-6,0.05),
circle(8,-6,0.08),
circle(8,-6,0.10)
)}}} Graph of {{{2x-4y=40}}} (red) and {{{8x-3y=82}}} (green)