Question 202943
Start with the given system of equations:

{{{system(2x+7y=12,5x-4y=-13)}}}

{{{2x+7y=12}}} Start with the first equation.

{{{7y=12-2x}}} Subtract {{{2x}}} from both sides.

{{{y=(12-2x)/(7)}}} Divide both sides by {{{7}}} to isolate {{{y}}}.

{{{y=-(2/7)x+12/7}}} Rearrange the terms and simplify.

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{{{5x-4y=-13}}} Move onto the second equation.

{{{5x-4(-(2/7)x+12/7)=-13}}} Now plug in {{{y=-(2/7)x+12/7}}}.

{{{5x+(8/7)x-48/7=-13}}} Distribute.

{{{7(5x+(8/cross(7))x-48/cross(7))=7(-13)}}} Multiply both sides by the LCD {{{7}}} to clear any fractions.

{{{35x+8x-48=-91}}} Distribute and multiply.

{{{43x-48=-91}}} Combine like terms on the left side.

{{{43x=-91+48}}} Add {{{48}}} to both sides.

{{{43x=-43}}} Combine like terms on the right side.

{{{x=(-43)/(43)}}} Divide both sides by {{{43}}} to isolate {{{x}}}.

{{{x=-1}}} Reduce.

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Since we know that {{{x=-1}}}, we can use this to find {{{y}}}.

{{{2x+7y=12}}} Go back to the first equation.

{{{2(-1)+7y=12}}} Plug in {{{x=-1}}}.

{{{-2+7y=12}}} Multiply.

{{{7y=12+2}}} Add {{{2}}} to both sides.

{{{7y=14}}} Combine like terms on the right side.

{{{y=(14)/(7)}}} Divide both sides by {{{7}}} to isolate {{{y}}}.

{{{y=2}}} Reduce.

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

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,-11,9,-8,12,
grid(1),
graph(500,500,-11,9,-8,12,(12-2x)/(7),(-13-5x)/(-4)),
circle(-1,2,0.05),
circle(-1,2,0.08),
circle(-1,2,0.10)
)}}} Graph of {{{2x+7y=12}}} (red) and {{{5x-4y=-13}}} (green)