Question 146700
Start with the given system of equations:


{{{system(x=2y-7,3x+2y=-5)}}}



{{{x=2y-7}}} Start with the first equation.



{{{x-2y=-7}}} Subtract {{{2y}}} from both sides.



So we have the new system of equations:


{{{system(x-2y=-7,3x+2y=-5)}}}



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



{{{(x-2y)+(3x+2y)=(-7)+(-5)}}}



{{{(1x+3x)+(-2y+2y)=-7+-5}}} Group like terms.



{{{4x+0y=-12}}} Combine like terms. Notice how the y terms cancel out.



{{{4x=-12}}} Simplify.



{{{x=(-12)/(4)}}} Divide both sides by {{{4}}} to isolate {{{x}}}.



{{{x=-3}}} Reduce.



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{{{x-2y=-7}}} Now go back to the first equation.



{{{-3-2y=-7}}} Plug in {{{x=-3}}}.



{{{-3-2y=-7}}} Multiply.



{{{-2y=-7+3}}} Add {{{3}}} to both sides.



{{{-2y=-4}}} Combine like terms on the right side.



{{{y=(-4)/(-2)}}} Divide both sides by {{{-2}}} to isolate {{{y}}}.



{{{y=2}}} Reduce.



So our answer is {{{x=-3}}} and {{{y=2}}}.



Which form the ordered pair *[Tex \LARGE \left(-3,2\right)].



This means that the two equations are consistent and independent.



Notice when we graph the equations, we see that they intersect at *[Tex \LARGE \left(-3,2\right)]. So this visually verifies our answer.



{{{drawing(500,500,-13,7,-8,12,
grid(1),
graph(500,500,-13,7,-8,12,(-7-x)/(-2),(-5-3x)/(2)),
circle(-3,2,0.05),
circle(-3,2,0.08),
circle(-3,2,0.10)
)}}} Graph of {{{x-2y=-7}}} (red) and {{{3x+2y=-5}}} (green)