Question 184141


Start with the given system of equations:

{{{system(4x-3y=15,x-2y=0)}}}



{{{-4(x-2y)=-4(0)}}} Multiply the both sides of the second equation by -4.



{{{-4x+8y=0}}} Distribute and multiply.



So we have the new system of equations:

{{{system(4x-3y=15,-4x+8y=0)}}}



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



{{{(4x-3y)+(-4x+8y)=(15)+(0)}}}



{{{(4x+-4x)+(-3y+8y)=15+0}}} Group like terms.



{{{0x+5y=15}}} Combine like terms.



{{{5y=15}}} Simplify.



{{{y=(15)/(5)}}} Divide both sides by {{{5}}} to isolate {{{y}}}.



{{{y=3}}} Reduce.



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{{{4x-3y=15}}} Now go back to the first equation.



{{{4x-3(3)=15}}} Plug in {{{y=3}}}.



{{{4x-9=15}}} Multiply.



{{{4x=15+9}}} Add {{{9}}} to both sides.



{{{4x=24}}} Combine like terms on the right side.



{{{x=(24)/(4)}}} Divide both sides by {{{4}}} to isolate {{{x}}}.



{{{x=6}}} Reduce.



So the solutions are {{{x=6}}} and {{{y=3}}}.



Which form the ordered pair *[Tex \LARGE \left(6,3\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(6,3\right)]. So this visually verifies our answer.



{{{drawing(500,500,-4,16,-7,13,
grid(1),
graph(500,500,-4,16,-7,13,(15-4x)/(-3),(0-x)/(-2)),
circle(6,3,0.05),
circle(6,3,0.08),
circle(6,3,0.10)
)}}} Graph of {{{4x-3y=15}}} (red) and {{{x-2y=0}}} (green)