Question 148001
I'll do the first one to get you started



# 1




Start with the given system of equations:



{{{system(-4x + 3y = 10,5x - 7y = -32)}}}



{{{-4x + 3y = 10}}} Start with the first equation.



{{{3y=10+4x}}} Add {{{4x}}} to both sides.



{{{y=(10+4x)/(3)}}} Divide both sides by {{{3}}} to isolate {{{y}}}.



{{{y=(4/3)x+10/3}}} Simplify.



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{{{5x - 7((4/3)x+10/3) = -32}}} Now plug in {{{y=(4/3)x+10/3}}} into the second equation.



{{{5x-(28/3)x-70/3=-32}}} Multiply.



{{{3(5x-(28/cross(3))x-70/cross(3))=3(-32)}}} Multiply both sides by the LCD {{{3}}} to clear any fractions.



{{{15x-28x-70=-96}}} Distribute and multiply.



{{{-13x-70=-96}}} Combine like terms on the left side.



{{{-13x=-96+70}}} Add {{{70}}} to both sides.



{{{-13x=-26}}} Combine like terms on the right side.



{{{x=(-26)/(-13)}}} Divide both sides by {{{-13}}} to isolate {{{x}}}.



{{{x=2}}} Reduce.



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Since we know that {{{x=2}}}, we can use this to find {{{y}}}.



{{{-4x + 3y = 10}}} Go back to the first equation.



{{{-4(2) + 3y = 10}}} Plug in {{{x=2}}}.



{{{-8+3y=10}}} Multiply.



{{{3y=10+8}}} Add {{{8}}} to both sides.



{{{3y=18}}} Combine like terms on the right side.



{{{y=(18)/(3)}}} Divide both sides by {{{3}}} to isolate {{{y}}}.



{{{y=6}}} Reduce.



So the answer is {{{x=2}}} and {{{y=6}}}.



This means that the system is consistent and independent.



Notice when we graph the equations, we see that they intersect at *[Tex \LARGE \left(2,6\right)]. So this visually verifies our answer.



{{{drawing(500,500,-8,12,-4,16,
grid(1),
graph(500,500,-8,12,-4,16,(10+4x)/(3),(-32-5x)/(-7)),
circle(2,6,0.05),
circle(2,6,0.08),
circle(2,6,0.10)
)}}} Graph of {{{-4x+3y=10}}} (red) and {{{5x-7y=-32}}} (green)