Question 252498



Start with the given system of equations:

{{{system(4x-5y=18,3x-2y=10)}}}



{{{2(4x-5y)=2(18)}}} Multiply the both sides of the first equation by 2.



{{{8x-10y=36}}} Distribute and multiply.



{{{-5(3x-2y)=-5(10)}}} Multiply the both sides of the second equation by -5.



{{{-15x+10y=-50}}} Distribute and multiply.



So we have the new system of equations:

{{{system(8x-10y=36,-15x+10y=-50)}}}



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



{{{(8x-10y)+(-15x+10y)=(36)+(-50)}}}



{{{(8x+-15x)+(-10y+10y)=36+-50}}} Group like terms.



{{{-7x+0y=-14}}} Combine like terms.



{{{-7x=-14}}} Simplify.



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



{{{x=2}}} Reduce.



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{{{8x-10y=36}}} Now go back to the first equation.



{{{8(2)-10y=36}}} Plug in {{{x=2}}}.



{{{16-10y=36}}} Multiply.



{{{-10y=36-16}}} Subtract {{{16}}} from both sides.



{{{-10y=20}}} Combine like terms on the right side.



{{{y=(20)/(-10)}}} Divide both sides by {{{-10}}} to isolate {{{y}}}.



{{{y=-2}}} Reduce.



So the solutions are {{{x=2}}} and {{{y=-2}}}.



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



{{{drawing(500,500,-8,12,-12,8,
grid(1),
graph(500,500,-8,12,-12,8,(18-4x)/(-5),(10-3x)/(-2)),
circle(2,-2,0.05),
circle(2,-2,0.08),
circle(2,-2,0.10)
)}}} Graph of {{{4x-5y=18}}} (red) and {{{3x-2y=10}}} (green)