seems like this one went by yesterday. But, it's not in the related questions, so, ignoring the hint, we have
x^2+y^2+z^2 = 9
3x^2-y^2-z^2 = 7
0x^2+y^2+2z^2 = 6
adding #1 to #2, we have
x^2+y^2+z^2 = 9
4x^2+0y^2+0z^2 = 16
so, 4x^2=16 --> x^2 = 4 --> x = 2
Since all these goodies are squared, negative answers also apply.
Now we have, using x^2=4,
y^2+z^2 = 5
y^2+2z^2 = 6
subtracting #1 from #2, we get
z^1 = 1 --> z=1
That means y^2 = 4 --> y=2
So, the solutions are (±2,±1,±2)
Solve the system of equations to find x,y,z (hint let d=x2 e=y2 and d=z2 . Find d e and f first) !!where ever you see 2 it's supposed to be a squared sign!!
X2+y2+z2=9
3x2-y2-z2=7
Y2+2z2 =6
1 answer
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