Question 197921
Ignore the other solution, I don't know what that is...





Start with the given system of equations:

{{{system(2x+3y=8,3x+2y=7)}}}



{{{3(2x+3y)=3(8)}}} Multiply the both sides of the first equation by 3.



{{{6x+9y=24}}} Distribute and multiply.



{{{-2(3x+2y)=-2(7)}}} Multiply the both sides of the second equation by -2.



{{{-6x-4y=-14}}} Distribute and multiply.



So we have the new system of equations:

{{{system(6x+9y=24,-6x-4y=-14)}}}



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+9y)+(-6x-4y)=(24)+(-14)}}}



{{{(6x+-6x)+(9y+-4y)=24+-14}}} Group like terms.



{{{0x+5y=10}}} Combine like terms.



{{{5y=10}}} Simplify.



{{{y=(10)/(5)}}} Divide both sides by {{{5}}} to isolate {{{y}}}.



{{{y=2}}} Reduce.



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{{{6x+9y=24}}} Now go back to the first equation.



{{{6x+9(2)=24}}} Plug in {{{y=2}}}.



{{{6x+18=24}}} Multiply.



{{{6x=24-18}}} Subtract {{{18}}} from both sides.



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



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



{{{x=1}}} Reduce.



So the solutions are {{{x=1}}} and {{{y=2}}}.



Which form the ordered pair *[Tex \LARGE \left(1,2\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(1,2\right)]. So this visually verifies our answer.



{{{drawing(500,500,-9,11,-8,12,
grid(1),
graph(500,500,-9,11,-8,12,(8-2x)/(3),(7-3x)/(2)),
circle(1,2,0.05),
circle(1,2,0.08),
circle(1,2,0.10)
)}}} Graph of {{{2x+3y=8}}} (red) and {{{3x+2y=7}}} (green)