Question 154283


Start with the given system of equations:

{{{system(-4x+2y=10,6x+2y=10)}}}



{{{-1(6x+2y)=-1(10)}}} Multiply the both sides of the second equation by -1.



{{{-6x-2y=-10}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-4x+2y=10,-6x-2y=-10)}}}



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+2y)+(-6x-2y)=(10)+(-10)}}}



{{{(-4x+-6x)+(2y+-2y)=10+-10}}} Group like terms.



{{{-10x+0y=0}}} Combine like terms. Notice how the y terms cancel out.



{{{-10x=0}}} Simplify.



{{{x=(0)/(-10)}}} Divide both sides by {{{-10}}} to isolate {{{x}}}.



{{{x=0}}} Reduce.



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{{{-4x+2y=10}}} Now go back to the first equation.



{{{-4(0)+2y=10}}} Plug in {{{x=0}}}.



{{{0+2y=10}}} Multiply.



{{{2y=10}}} Remove any zero terms.



{{{y=(10)/(2)}}} Divide both sides by {{{2}}} to isolate {{{y}}}.



{{{y=5}}} Reduce.



So our answer is {{{x=0}}} and {{{y=5}}}.



Which form the ordered pair *[Tex \LARGE \left(0,5\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(0,5\right)]. So this visually verifies our answer.



{{{drawing(500,500,-10,10,-5,15,
grid(1),
graph(500,500,-10,10,-5,15,(10+4x)/(2),(10-6x)/(2)),
circle(0,5,0.05),
circle(0,5,0.08),
circle(0,5,0.10)
)}}} Graph of {{{-4x+2y=10}}} (red) and {{{6x+2y=10}}} (green)