Question 202661

Start with the given system of equations:



{{{system(x+2y=6,x-4y=8)}}}



{{{x+2y=6}}} Start with the first equation.



{{{2y=6-x}}} Subtract {{{x}}} from both sides.



{{{y=(6-x)/(2)}}} Divide both sides by {{{2}}} to isolate {{{y}}}.



{{{y=-(1/2)x+3}}} Rearrange the terms and simplify.



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{{{x-4y=8}}} Move onto the second equation.



{{{x-4(-(1/2)x+3)=8}}} Now plug in {{{y=-(1/2)x+3}}}.



{{{x+2x-12=8}}} Distribute.



{{{3x-12=8}}} Combine like terms on the left side.



{{{3x=8+12}}} Add {{{12}}} to both sides.



{{{3x=20}}} Combine like terms on the right side.



{{{x=(20)/(3)}}} Divide both sides by {{{3}}} to isolate {{{x}}}.



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Since we know that {{{x=20/3}}}, we can use this to find {{{y}}}.



{{{x+2y=6}}} Go back to the first equation.



{{{20/3+2y=6}}} Plug in {{{x=20/3}}}.



{{{3(20/cross(3)+2y)=3(6)}}} Multiply both sides by the LCD {{{3}}} to clear any fractions.



{{{20+6y=18}}} Distribute and multiply.



{{{6y=18-20}}} Subtract {{{20}}} from both sides.



{{{6y=-2}}} Combine like terms on the right side.



{{{y=(-2)/(6)}}} Divide both sides by {{{6}}} to isolate {{{y}}}.



{{{y=-1/3}}} Reduce.



So the solutions are {{{x=20/3}}} and {{{y=-1/3}}}.



which form the ordered pair *[Tex \LARGE \left(\frac{20}{3},-\frac{1}{3}\right)]



This means that the system is consistent and independent.