Question 148146


Start with the given system of equations:

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



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



{{{-4x-2y=-6}}} Distribute and multiply.



So we have the new system of equations:

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



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-2y)+(4x+3y)=(-6)+(1)}}}



{{{(-4x+4x)+(-2y+3y)=-6+1}}} Group like terms.



{{{0x+y=-5}}} Combine like terms. Notice how the x terms cancel out.



{{{y=-5}}} Simplify.



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{{{-4x-2y=-6}}} Now go back to the first equation.



{{{-4x-2(-5)=-6}}} Plug in {{{y=-5}}}.



{{{-4x+10=-6}}} Multiply.



{{{-4x=-6-10}}} Subtract {{{10}}} from both sides.



{{{-4x=-16}}} Combine like terms on the right side.



{{{x=(-16)/(-4)}}} Divide both sides by {{{-4}}} to isolate {{{x}}}.



{{{x=4}}} Reduce.



So our answer is {{{x=4}}} and {{{y=-5}}}.



Which form the ordered pair *[Tex \LARGE \left(4,-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(4,-5\right)]. So this visually verifies our answer.



{{{drawing(500,500,-6,14,-15,5,
grid(1),
graph(500,500,-6,14,-15,5,3-2x,(1-4x)/(3)),
circle(4,-5,0.05),
circle(4,-5,0.08),
circle(4,-5,0.10)
)}}} Graph of {{{2x+y=3}}} (red) and {{{4x+3y=1}}} (green)