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solve the following system:
\(\displaystyle x^2 +xy +y^2=37\)
\(\displaystyle x^2+xz+z^2=28\)
\(\displaystyle y^2+yz+z^2=19\)
\(\displaystyle x^2 +xy +y^2=37\)
\(\displaystyle x^2+xz+z^2=28\)
\(\displaystyle y^2+yz+z^2=19\)
[sp][sp]
Here's a quick run-down.
1) \(\displaystyle x^2 + xy + y^2 = 37\)
2) \(\displaystyle x^2 + x z + z^2 = 28\)
3) \(\displaystyle y^2 + y z + z^2 = 19\)
We first need to pull a trick. For example, let's take equation 1):
\(\displaystyle x^2 + xy + y^2 = 37\)
\(\displaystyle (x - y)(x^2 + xy + y^2) = 37 (x - y)\)
\(\displaystyle x^3 - y^3 = 37 (x - y)\)
Do this to equations 2) and 3) as well.
Now add equations 1) and 3) and subtract equation 2), giving
\(\displaystyle 0 = x - 2y + z\)
Put this into (the unaltered) equations 1) and 2):
1) \(\displaystyle x^2 + xy + y^2 = 37\)
2) \(\displaystyle x^2 - 2 xy + 4y^2 = 28\)
(We can drop equation 3). It doesn't give us anything new.)
So add 2 times equation 1 and equation 2).
\(\displaystyle x^2 + 2 y^2 = 34\)
Thus
\(\displaystyle y = \pm \sqrt{ \dfrac{1}{2} (34 - x^2) }\)
Now put these y values into equation 1)
\(\displaystyle x^2 \pm x \sqrt{ \dfrac{1}{2} (34 - x^2) } + \dfrac{1}{2} (34 - x^2) = 37\)
To save some time I'll just simply say that we can solve this in the usual way of isolating the square root and squaring the whole thing. etc.
So
\(\displaystyle x = \left \{ \pm 4, ~ \pm \dfrac{10}{\sqrt{3} } \right \}\)
Putting this all together:
\(\displaystyle (x, y, z) = \begin{cases} (4, ~ 3, ~ 2) \\ (-4, ~ -3, ~ -2) \\ \left ( \dfrac{10}{\sqrt{3}}, ~ \dfrac{1}{\sqrt{3}}, ~ - \dfrac{8}{\sqrt{3}} \right ) \\ \left ( - \dfrac{10}{\sqrt{3}}, ~ - \dfrac{1}{\sqrt{3}}, ~ \dfrac{8}{\sqrt{3}} \right ) \end{cases} \)
[/sp]
-Dan
I'm not sure why you posted? I'm guessing that you want a verification of one of the steps. If this not what you are asking, please let me know.[sp]
Subtract the 2nd equation from the 1st and the 3rd from the 2nd and we get:
(y-z)(x+y=z)=9 and (x-y)(x+y+z)=9
Divide those two and we get:
(y-z)/(x-y)=1 => z=2y-x
Substitute the above into the3rd equation and we get \(\displaystyle x^2-2xy+4y^2\) e.t.c e.t.c...[/sp]
Yes. That's on line 11.I just wanted to show that there is another way to get the equation
z=2y-x