Question 179036

{{{system(3x-5y=11,x-3y=1)}}}

{{{-3(x-3y)=-3(1)}}} Multiply the both sides of the second equation by -3.

{{{-3x+9y=-3}}} Distribute and multiply.

So we have the new system of equations:

{{{system(3x-5y=11,-3x+9y=-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:

{{{(3x-5y)+(-3x+9y)=(11)+(-3)}}}

{{{(3x+-3x)+(-5y+9y)=11+-3}}} Group like terms.

{{{0x+4y=8}}} Combine like terms.

{{{4y=8}}} Simplify.

{{{y=(8)/(4)}}} Divide both sides by {{{4}}} to isolate {{{y}}}.

{{{y=2}}} Reduce.

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{{{3x-5y=11}}} Now go back to the first equation.

{{{3x-5(2)=11}}} Plug in {{{y=2}}}.

{{{3x-10=11}}} Multiply.

{{{3x=11+10}}} Add {{{10}}} to both sides.

{{{3x=21}}} Combine like terms on the right side.

{{{x=(21)/(3)}}} Divide both sides by {{{3}}} to isolate {{{x}}}.

{{{x=7}}} Reduce.

So our answer is {{{x=7}}} and {{{y=2}}}.

Which form the ordered pair *[Tex \LARGE \left(7,2\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(7,2\right)]. So this visually verifies our answer.

{{{drawing(500,500,-3,17,-8,12,
grid(1),
graph(500,500,-3,17,-8,12,(11-3x)/(-5),(1-x)/(-3)),
circle(7,2,0.05),
circle(7,2,0.08),
circle(7,2,0.10)
)}}} Graph of {{{3x-5y=11}}} (red) and {{{x-3y=1}}} (green)