Question 210193

Start with the given system of equations:

{{{system(5x+4y=6,-2x-3y=-1)}}}



{{{2(5x+4y)=2(6)}}} Multiply the both sides of the first equation by 2.



{{{10x+8y=12}}} Distribute and multiply.



{{{5(-2x-3y)=5(-1)}}} Multiply the both sides of the second equation by 5.



{{{-10x-15y=-5}}} Distribute and multiply.



So we have the new system of equations:

{{{system(10x+8y=12,-10x-15y=-5)}}}



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



{{{(10x+8y)+(-10x-15y)=(12)+(-5)}}}



{{{(10x+-10x)+(8y+-15y)=12+-5}}} Group like terms.



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



{{{-7y=7}}} Simplify.



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



{{{y=-1}}} Reduce.



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{{{10x+8y=12}}} Now go back to the first equation.



{{{10x+8(-1)=12}}} Plug in {{{y=-1}}}.



{{{10x-8=12}}} Multiply.



{{{10x=12+8}}} Add {{{8}}} to both sides.



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



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



{{{x=2}}} Reduce.



So the solutions are {{{x=2}}} and {{{y=-1}}}.



Which form the ordered pair *[Tex \LARGE \left(2,-1\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,-1\right)]. So this visually verifies our answer.



{{{drawing(500,500,-8,12,-11,9,
grid(1),
graph(500,500,-8,12,-11,9,(6-5x)/(4),(-1+2x)/(-3)),
circle(2,-1,0.05),
circle(2,-1,0.08),
circle(2,-1,0.10)
)}}} Graph of {{{5x+4y=6}}} (red) and {{{-2x-3y=-1}}} (green)