Question 183262


Start with the given system of equations:

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



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



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



So we have the new system of equations:

{{{system(-2x-6y=-10,2x+y=5)}}}



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-6y)+(2x+y)=(-10)+(5)}}}



{{{(-2x+2x)+(-6y+1y)=-10+5}}} Group like terms.



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



{{{-5y=-5}}} Simplify.



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



{{{y=1}}} Reduce.



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{{{-2x-6y=-10}}} Now go back to the first equation.



{{{-2x-6(1)=-10}}} Plug in {{{y=1}}}.



{{{-2x-6=-10}}} Multiply.



{{{-2x=-10+6}}} Add {{{6}}} to 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=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-x)/(3),5-2x),
circle(2,1,0.05),
circle(2,1,0.08),
circle(2,1,0.10)
)}}} Graph of {{{x+3y=5}}} (red) and {{{2x+y=5}}} (green)