Question 207121

Start with the given system of equations:



{{{system(3y-x=5,-x+2y=-4)}}}



{{{3y-x=5}}} Start with the first equation.



{{{3y=5+x}}} Add {{{x}}} to both sides.



{{{y=(5+x)/(3)}}} Divide both sides by {{{3}}} to isolate {{{y}}}.



{{{y=(1/3)x+5/3}}} Rearrange the terms and simplify.



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{{{-x+2y=-4}}} Move onto the second equation.



{{{-x+2((1/3)x+5/3)=-4}}} Now plug in {{{y=(1/3)x+5/3}}}.



{{{-x+(2/3)x+10/3=-4}}} Distribute.



{{{3(-x+(2/cross(3))x+10/cross(3))=3(-4)}}} Multiply both sides by the LCD {{{3}}} to clear any fractions.



{{{-3x+2x+10=-12}}} Distribute and multiply.



{{{-x+10=-12}}} Combine like terms on the left side.



{{{-x=-12-10}}} Subtract {{{10}}} from both sides.



{{{-x=-22}}} Combine like terms on the right side.



{{{x=(-22)/(-1)}}} Divide both sides by {{{-1}}} to isolate {{{x}}}.



{{{x=22}}} Reduce.



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Since we know that {{{x=22}}}, we can use this to find {{{y}}}.



{{{3y-x=5}}} Go back to the first equation.



{{{3y-22=5}}} Plug in {{{x=22}}}.



{{{3y=5+22}}} Add {{{22}}} to both sides.



{{{3y=27}}} Combine like terms on the right side.



{{{y=(27)/(3)}}} Divide both sides by {{{3}}} to isolate {{{y}}}.



{{{y=9}}} Reduce.



So the solutions are {{{x=22}}} and {{{y=9}}}.



which form the ordered pair (22,9)



This means that the system is consistent and independent.