Question 176233
Start with the given system of equations:

{{{system(x+3y=12,x-3y=30)}}}


Note: I've rearrange the terms in the second equation



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



{{{(x+3y)+(x-3y)=(12)+(30)}}}



{{{(x+x)+(3y-3y)=12+30}}} Group like terms.



{{{2x+0y=42}}} Combine like terms.



{{{2x=42}}} Simplify.



{{{x=(42)/(2)}}} Divide both sides by {{{2}}} to isolate {{{x}}}.



{{{x=21}}} Reduce.



------------------------------------------------------------------



{{{x+3y=12}}} Now go back to the first equation.



{{{21+3y=12}}} Plug in {{{x=21}}}.



{{{21+3y=12}}} Multiply.



{{{3y=12-21}}} Subtract {{{21}}} from both sides.



{{{3y=-9}}} Combine like terms on the right side.



{{{y=(-9)/(3)}}} Divide both sides by {{{3}}} to isolate {{{y}}}.



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



So our answer is {{{x=21}}} and {{{y=-3}}}.



Which form the ordered pair *[Tex \LARGE \left(21,-3\right)].



This means that the system is consistent and independent.