Question 243643


Start with the given system of equations:

{{{system(-x+3y=12,4x+3y=12)}}}



{{{4(-x+3y)=4(12)}}} Multiply the both sides of the first equation by 4.



{{{-4x+12y=48}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-4x+12y=48,4x+3y=12)}}}



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



{{{(-4x+12y)+(4x+3y)=(48)+(12)}}}



{{{(-4x+4x)+(12y+3y)=48+12}}} Group like terms.



{{{0x+15y=60}}} Combine like terms.



{{{15y=60}}} Simplify.



{{{y=(60)/(15)}}} Divide both sides by {{{15}}} to isolate {{{y}}}.



{{{y=4}}} Reduce.



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{{{-4x+12y=48}}} Now go back to the first equation.



{{{-4x+12(4)=48}}} Plug in {{{y=4}}}.



{{{-4x+48=48}}} Multiply.



{{{-4x=48-48}}} Subtract {{{48}}} from both sides.



{{{-4x=0}}} Combine like terms on the right side.



{{{x=(0)/(-4)}}} Divide both sides by {{{-4}}} to isolate {{{x}}}.



{{{x=0}}} Reduce.



So the solutions are {{{x=0}}} and {{{y=4}}}.



Which form the ordered pair *[Tex \LARGE \left(0,4\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(0,4\right)]. So this visually verifies our answer.



{{{drawing(500,500,-10,10,-6,14,
grid(1),
graph(500,500,-10,10,-6,14,(12+x)/(3),(12-4x)/(3)),
circle(0,4,0.05),
circle(0,4,0.08),
circle(0,4,0.10)
)}}} Graph of {{{-x+3y=12}}} (red) and {{{4x+3y=12}}} (green)