Question 176030
Note: I've rearranged the terms to get the system


{{{system(-7x+5y=29,x+2y=4)}}}





Start with the given system of equations:

{{{system(-7x+5y=29,x+2y=4)}}}



{{{7(x+2y)=7(4)}}} Multiply the both sides of the second equation by 7.



{{{7x+14y=28}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-7x+5y=29,7x+14y=28)}}}



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:



{{{(-7x+5y)+(7x+14y)=(29)+(28)}}}



{{{(-7x+7x)+(5y+14y)=29+28}}} Group like terms.



{{{0x+19y=57}}} Combine like terms.



{{{19y=57}}} Simplify.



{{{y=(57)/(19)}}} Divide both sides by {{{19}}} to isolate {{{y}}}.



{{{y=3}}} Reduce.



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{{{-7x+5y=29}}} Now go back to the first equation.



{{{-7x+5(3)=29}}} Plug in {{{y=3}}}.



{{{-7x+15=29}}} Multiply.



{{{-7x=29-15}}} Subtract {{{15}}} from both sides.



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



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



{{{x=-2}}} Reduce.



So our answer is {{{x=-2}}} and {{{y=3}}}.



Which form the ordered pair *[Tex \LARGE \left(-2,3\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(-2,3\right)]. So this visually verifies our answer.



{{{drawing(500,500,-12,8,-7,13,
grid(1),
graph(500,500,-12,8,-7,13,(29+7x)/(5),(4-x)/(2)),
circle(-2,3,0.05),
circle(-2,3,0.08),
circle(-2,3,0.10)
)}}} Graph of {{{-7x+5y=29}}} (red) and {{{x+2y=4}}} (green)