Question 202417
Note: I rearranged the second equation {{{8x+62=y}}} to get {{{y=8x+62}}}




Start with the given system of equations:



{{{system(2x-5y=32,y=8x+62)}}}




{{{2x-5y=32}}} Start with the first equation.



{{{2x-5(8x+62)=32}}} Plug in {{{y=8x+62}}}. Take note that the 'y' terms are gone and we can now solve for 'x'



{{{2x-40x-310=32}}} Distribute.



{{{-38x-310=32}}} Combine like terms on the left side.



{{{-38x=32+310}}} Add {{{310}}} to both sides.



{{{-38x=342}}} Combine like terms on the right side.



{{{x=(342)/(-38)}}} Divide both sides by {{{-38}}} to isolate {{{x}}}.



{{{x=-9}}} Reduce.



-------------------------------------------



Since we know that {{{x=-9}}}, we can use this to find {{{y}}}.



{{{2x-5y=32}}} Go back to the first equation.



{{{2(-9)-5y=32}}} Plug in {{{x=-9}}}.



{{{-18-5y=32}}} Multiply.



{{{-5y=32+18}}} Add {{{18}}} to both sides.



{{{-5y=50}}} Combine like terms on the right side.



{{{y=(50)/(-5)}}} Divide both sides by {{{-5}}} to isolate {{{y}}}.



{{{y=-10}}} Reduce.



So the solutions are {{{x=-9}}} and {{{y=-10}}}.



This means that the system is consistent and independent.



Notice when we graph the equations, we see that they intersect at *[Tex \LARGE \left(-9,-10\right)]. So this visually verifies our answer.



{{{drawing(500,500,-19,1,-15,5,
grid(1),
graph(500,500,-19,1,-15,5,(32-2x)/(-5),8x+62),
circle(-9,-10,0.05),
circle(-9,-10,0.08),
circle(-9,-10,0.10)
)}}} Graph of {{{2x-5y=32}}} (red) and {{{y=8x+62}}} (green)