Question 241745


Start with the given system of equations:

{{{system(2x+y=5,3x-3y=3)}}}



{{{3(2x+y)=3(5)}}} Multiply the both sides of the first equation by 3.



{{{6x+3y=15}}} Distribute and multiply.



So we have the new system of equations:

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



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+3y)+(3x-3y)=(15)+(3)}}}



{{{(6x+3x)+(3y-3y)=15+3}}} Group like terms.



{{{9x+0y=18}}} Combine like terms.



{{{9x=18}}} Simplify.



{{{x=(18)/(9)}}} Divide both sides by {{{9}}} to isolate {{{x}}}.



{{{x=2}}} Reduce.



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{{{6x+3y=15}}} Now go back to the first equation.



{{{6(2)+3y=15}}} Plug in {{{x=2}}}.



{{{12+3y=15}}} Multiply.



{{{3y=15-12}}} Subtract {{{12}}} from both sides.



{{{3y=3}}} Combine like terms on the right side.



{{{y=(3)/(3)}}} Divide both sides by {{{3}}} to isolate {{{y}}}.



{{{y=1}}} Reduce.



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



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



{{{drawing(500,500,-8,12,-9,11,
grid(1),
graph(500,500,-8,12,-9,11,5-2x,(3-3x)/(-3)),
circle(2,1,0.05),
circle(2,1,0.08),
circle(2,1,0.10)
)}}} Graph of {{{2x+y=5}}} (red) and {{{3x-3y=3}}} (green)