Question 183033

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



{{{-x+y=0}}} Subtract "x" from both sides (the goal is to get all the variable terms to the left side).





So we have the system of equations:

{{{system(4x-2y=14,-x+y=0)}}}



{{{2(-x+y)=2(0)}}} Multiply the both sides of the second equation by 2.



{{{-2x+2y=0}}} Distribute and multiply.



So we have the new system of equations:

{{{system(4x-2y=14,-2x+2y=0)}}}



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)+(-2x+2y)=(14)+(0)}}}



{{{(4x+-2x)+(-2y+2y)=14+0}}} Group like terms.



{{{2x+0y=14}}} Combine like terms.



{{{2x=14}}} Simplify.



{{{x=(14)/(2)}}} Divide both sides by {{{2}}} to isolate {{{x}}}.



{{{x=7}}} Reduce.



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{{{4x-2y=14}}} Now go back to the first equation.



{{{4(7)-2y=14}}} Plug in {{{x=7}}}.



{{{28-2y=14}}} Multiply.



{{{-2y=14-28}}} Subtract {{{28}}} from both sides.



{{{-2y=-14}}} Combine like terms on the right side.



{{{y=(-14)/(-2)}}} Divide both sides by {{{-2}}} to isolate {{{y}}}.



{{{y=7}}} Reduce.



So our answer is {{{x=7}}} and {{{y=7}}} (so you are correct).



Which form the ordered pair *[Tex \LARGE \left(7,7\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(7,7\right)]. So this visually verifies our answer.



{{{drawing(500,500,-3,17,-3,17,
grid(1),
graph(500,500,-3,17,-3,17,(14-4x)/(-2),0+x),
circle(7,7,0.05),
circle(7,7,0.08),
circle(7,7,0.10)
)}}} Graph of {{{4x-2y=14}}} (red) and {{{y=x}}} (green)