Question 147698

Start with the given system of equations:

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



{{{2(4x+y)=2(1)}}} Multiply the both sides of the first equation by 2.



{{{8x+2y=2}}} Distribute and multiply.



So we have the new system of equations:

{{{system(8x+2y=2,x-2y=-2)}}}



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+2y)+(x-2y)=(2)+(-2)}}}



{{{(8x+1x)+(2y+-2y)=2+-2}}} Group like terms.



{{{9x+0y=0}}} Combine like terms. Notice how the y terms cancel out.



{{{9x=0}}} Simplify.



{{{x=(0)/(9)}}} Divide both sides by {{{9}}} to isolate {{{x}}}.



{{{x=0}}} Reduce.



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{{{8x+2y=2}}} Now go back to the first equation.



{{{8(0)+2y=2}}} Plug in {{{x=0}}}.



{{{0+2y=2}}} Multiply.



{{{2y=2}}} Remove any zero terms.



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



{{{y=1}}} Reduce.



So our answer is {{{x=0}}} and {{{y=1}}}.



Which form the ordered pair *[Tex \LARGE \left(0,1\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(0,1\right)]. So this visually verifies our answer.



{{{drawing(500,500,-10,10,-9,11,
grid(1),
graph(500,500,-10,10,-9,11,1-4x,(-2-x)/(-2)),
circle(0,1,0.05),
circle(0,1,0.08),
circle(0,1,0.10)
)}}} Graph of {{{4x+y=1}}} (red) and {{{x-2y=-2}}} (green) 



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Check:



{{{4x+y=1}}} Start with the first equation.



{{{4*(0)+(1)=1}}} Plug in {{{x=0}}} and {{{y=1}}}.



{{{1=1}}} Evaluate and simplify the left side.



Since the equation is <b>true</b>, this means that (0,1) is a solution for the first equation.



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



{{{(0)-2*(1)=-2}}} Plug in {{{x=0}}} and {{{y=1}}}.



{{{-2=-2}}} Evaluate and simplify the left side.



Since the equation is <b>true</b>, this means that (0,1) is a solution for the second equation.



Since <font size="4"><b>all</b></font> of the equations of the system are true, this means that (0,1) is a solution of the system. So this verifies our answer.