Question 201631
I'm assuming that the second "equation" should be {{{2x+6y=12}}}. You need to be more careful...




Start with the given system of equations:

{{{system(-x+3y=0,2x+6y=12)}}}



{{{2(-x+3y)=2(0)}}} Multiply the both sides of the first equation by 2.



{{{-2x+6y=0}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-2x+6y=0,2x+6y=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:



{{{(-2x+6y)+(2x+6y)=(0)+(12)}}}



{{{(-2x+2x)+(6y+6y)=0+12}}} Group like terms.



{{{0x+12y=12}}} Combine like terms.



{{{12y=12}}} Simplify.



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



{{{y=1}}} Reduce.



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



{{{-2x+6y=0}}} Now go back to the first equation.



{{{-2x+6(1)=0}}} Plug in {{{y=1}}}.



{{{-2x+6=0}}} Multiply.



{{{-2x=0-6}}} Subtract {{{6}}} from both sides.



{{{-2x=-6}}} Combine like terms on the right side.



{{{x=(-6)/(-2)}}} Divide both sides by {{{-2}}} to isolate {{{x}}}.



{{{x=3}}} Reduce.



So the solutions are {{{x=3}}} and {{{y=1}}}.



Which form the ordered pair *[Tex \LARGE \left(3,1\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(3,1\right)]. So this visually verifies our answer.



{{{drawing(500,500,-7,13,-9,11,
grid(1),
graph(500,500,-7,13,-9,11,(0+x)/(3),(12-2x)/(6)),
circle(3,1,0.05),
circle(3,1,0.08),
circle(3,1,0.10)
)}}} Graph of {{{-x+3y=0}}} (red) and {{{2x+6y=12}}} (green)