Question 153100


Start with the given system of equations:

{{{system(3x-2y=4,2x-5y=21)}}}



{{{-2(3x-2y)=-2(4)}}} Multiply the both sides of the first equation by -2.



{{{-6x+4y=-8}}} Distribute and multiply.



{{{3(2x-5y)=3(21)}}} Multiply the both sides of the second equation by 3.



{{{6x-15y=63}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-6x+4y=-8,6x-15y=63)}}}



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



{{{(-6x+4y)+(6x-15y)=(-8)+(63)}}}



{{{(-6x+6x)+(4y+-15y)=-8+63}}} Group like terms.



{{{0x+-11y=55}}} Combine like terms. Notice how the x terms cancel out.



{{{-11y=55}}} Simplify.



{{{y=(55)/(-11)}}} Divide both sides by {{{-11}}} to isolate {{{y}}}.



{{{y=-5}}} Reduce.



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{{{-6x+4y=-8}}} Now go back to the first equation.



{{{-6x+4(-5)=-8}}} Plug in {{{y=-5}}}.



{{{-6x-20=-8}}} Multiply.



{{{-6x=-8+20}}} Add {{{20}}} to both sides.



{{{-6x=12}}} Combine like terms on the right side.



{{{x=(12)/(-6)}}} Divide both sides by {{{-6}}} to isolate {{{x}}}.



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



So our answer is {{{x=-2}}} and {{{y=-5}}}.



Which form the ordered pair *[Tex \LARGE \left(-2,-5\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,-5\right)]. So this visually verifies our answer.



{{{drawing(500,500,-12,8,-15,5,
grid(1),
graph(500,500,-12,8,-15,5,(4-3x)/(-2),(21-2x)/(-5)),
circle(-2,-5,0.05),
circle(-2,-5,0.08),
circle(-2,-5,0.10)
)}}} Graph of {{{3x-2y=4}}} (red) and {{{2x-5y=21}}} (green)