Question 92298
I assume it looks like this right?

{{{x^2+y^2=10}}} 

{{{-3x+y=0}}} 



{{{y=3x}}} Solve the 2nd equation for y


{{{x^2+(3x)^2=10}}} Plug in y=3x


{{{x^2+9x^2=10}}} Square 3x to get {{{9x^2}}}


{{{x^2+9x^2-10=0}}} Subtract 10 from both sides



{{{10x^2-10=0}}} Combine like terms


{{{10(x^2-1)=0}}} Factor out 10


{{{x^2-1=0}}} Set each factor equal to zero


{{{x^2=1}}} Add one to both sides


*[Tex \LARGE x=\pm1] Take the square root of both sides



So the solution breaks down to:


{{{x=1}}} or {{{x=-1}}}








Now plug in x=1 to find y


{{{y=3(1)=3}}}


So when x=1, y=3



Now plug in x=-1 to find y


{{{y=3(-1)=-3}}}


So when x=-1, y=-3




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Answer:

So our solutions are


{{{x=1}}} and {{{y=3}}} which makes the point (1,3)



and



{{{x=-1}}} and {{{y=-3}}} which makes the point (-1, -3)