Question 186755


Start with the given system of equations:

{{{system(2x+4y=12,x+y=4)}}}



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



{{{-2x-2y=-8}}} Distribute and multiply.



So we have the new system of equations:

{{{system(2x+4y=12,-2x-2y=-8)}}}



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+4y)+(-2x-2y)=(12)+(-8)}}}



{{{(2x+-2x)+(4y+-2y)=12+-8}}} Group like terms.



{{{0x+2y=4}}} Combine like terms.



{{{2y=4}}} Simplify.



{{{y=(4)/(2)}}} Divide both sides by {{{2}}} to isolate {{{y}}}.



{{{y=2}}} Reduce.



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{{{2x+4y=12}}} Now go back to the first equation.



{{{2x+4(2)=12}}} Plug in {{{y=2}}}.



{{{2x+8=12}}} Multiply.



{{{2x=12-8}}} Subtract {{{8}}} from both sides.



{{{2x=4}}} Combine like terms on the right side.



{{{x=(4)/(2)}}} Divide both sides by {{{2}}} to isolate {{{x}}}.



{{{x=2}}} Reduce.



So the solutions are {{{x=2}}} and {{{y=2}}}.



Which form the ordered pair *[Tex \LARGE \left(2,2\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(2,2\right)]. So this visually verifies our answer.



{{{drawing(500,500,-8,12,-8,12,
grid(1),
graph(500,500,-8,12,-8,12,(12-2x)/(4),4-x),
circle(2,2,0.05),
circle(2,2,0.08),
circle(2,2,0.10)
)}}} Graph of {{{2x+4y=12}}} (red) and {{{x+y=4}}} (green)