Question 459327
If x^2+y^2=9 and x-3y=3, then what are the values of x and y?
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x^2+y^2=9
x-3y=3
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3y=x-3
y=(x-3)/3
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x^2+y^2=9
x^2+[(x-3)/3]^2=9
x^2+(x^2-6x+9)/9=9
LCD:9
9x^2+x^2-6x+9=81
10x^2-6x-81=0
solve by quadratic formula:
a=10, b=-6, c=-81
x=]-(-6)ħsqrt((-6)^2-4*10*(-81)]/2*10
x=[6ħsqrt(3276)]/20
x=(6ħ57.24)/20
x=3.16
y=(3.16-3)/3=.0533
or
x=-2.56
y=(-2.56-3)/3=-1.85