Question 183031


{{{2x+(2/3)y =4}}} Start with the first equation.



{{{3(2x)+cross(3)((2/cross(3))y)=3(4)}}} Multiply EVERY term by the LCD {{{3}}} to clear any fractions.



{{{6x+2y=12}}} Distribute and multiply.


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{{{x-(1/2)y=7}}} Move onto the second equation.



{{{2(x)-cross(2)((1/cross(2))y)=2(7)}}} Multiply EVERY term by the LCD {{{2}}} to clear any fractions.



{{{2x-y=14}}} Distribute and multiply.



So we have the system of equations:



{{{system(6x+2y=12,2x-y=14)}}}





{{{2(2x-y)=2(14)}}} Multiply the both sides of the second equation by 2.



{{{4x-2y=28}}} Distribute and multiply.



So we have the new system of equations:

{{{system(6x+2y=12,4x-2y=28)}}}



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



{{{(6x+2y)+(4x-2y)=(12)+(28)}}}



{{{(6x+4x)+(2y+-2y)=12+28}}} Group like terms.



{{{10x+0y=40}}} Combine like terms.



{{{10x=40}}} Simplify.



{{{x=(40)/(10)}}} Divide both sides by {{{10}}} to isolate {{{x}}}.



{{{x=4}}} Reduce.



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{{{6x+2y=12}}} Now go back to the first equation.



{{{6(4)+2y=12}}} Plug in {{{x=4}}}.



{{{24+2y=12}}} Multiply.



{{{2y=12-24}}} Subtract {{{24}}} from both sides.



{{{2y=-12}}} Combine like terms on the right side.



{{{y=(-12)/(2)}}} Divide both sides by {{{2}}} to isolate {{{y}}}.



{{{y=-6}}} Reduce.



So our answer is {{{x=4}}} and {{{y=-6}}}.



Which form the ordered pair *[Tex \LARGE \left(4,-6\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,-6\right)]. So this visually verifies our answer.



{{{drawing(500,500,-6,14,-16,4,
grid(1),
graph(500,500,-6,14,-16,4,(12-6x)/(2),(14-2x)/(-1)),
circle(4,-6,0.05),
circle(4,-6,0.08),
circle(4,-6,0.10)
)}}} Graph of {{{6x+2y=12}}} (red) and {{{2x-y=14}}} (green)