Question 199100

Start with the given system of equations:

{{{system(2x+4y=2,3x-6y=15)}}}



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



{{{6x+12y=6}}} Distribute and multiply.



{{{2(3x-6y)=2(15)}}} Multiply the both sides of the second equation by 2.



{{{6x-12y=30}}} Distribute and multiply.



So we have the new system of equations:

{{{system(6x+12y=6,6x-12y=30)}}}



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)+(6x-12y)=(6)+(30)}}}



{{{(6x+6x)+(12y+-12y)=6+30}}} Group like terms.



{{{12x+0y=36}}} Combine like terms.



{{{12x=36}}} Simplify.



{{{x=(36)/(12)}}} Divide both sides by {{{12}}} to isolate {{{x}}}.



{{{x=3}}} Reduce.



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



{{{6(3)+12y=6}}} Plug in {{{x=3}}}.



{{{18+12y=6}}} Multiply.



{{{12y=6-18}}} Subtract {{{18}}} from both sides.



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



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



{{{y=-1}}} 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,-11,9,
grid(1),
graph(500,500,-7,13,-11,9,(2-2x)/(4),(15-3x)/(-6)),
circle(3,-1,0.05),
circle(3,-1,0.08),
circle(3,-1,0.10)
)}}} Graph of {{{2x+4y=2}}} (red) and {{{3x-6y=15}}} (green) 




I'll let you do the check.