Question 175694

Start with the given system of equations:

{{{system(2x+3y=40,-2x+2y=20)}}}



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



{{{(2x+3y)+(-2x+2y)=(40)+(20)}}}



{{{(2x+-2x)+(3y+2y)=40+20}}} Group like terms.



{{{0x+5y=60}}} Combine like terms. Notice how the x terms cancel out.



{{{5y=60}}} Simplify.



{{{y=(60)/(5)}}} Divide both sides by {{{5}}} to isolate {{{y}}}.



{{{y=12}}} Reduce.



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{{{2x+3y=40}}} Now go back to the first equation.



{{{2x+3(12)=40}}} Plug in {{{y=12}}}.



{{{2x+36=40}}} Multiply.



{{{2x=40-36}}} Subtract {{{36}}} 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 our answer is {{{x=2}}} and {{{y=12}}}.



Which form the ordered pair *[Tex \LARGE \left(2,12\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(2,12\right)]. So this visually verifies our answer.



{{{drawing(500,500,-8,12,-3,15,
grid(1),
graph(500,500,-8,12,-3,15,(40-2x)/(3),(20+2x)/(2)),
circle(2,12,0.05),
circle(2,12,0.08),
circle(2,12,0.10)
)}}} Graph of {{{2x+3y=40}}} (red) and {{{-2x+2y=20}}} (green)