Solve System of Equalities: x^2+xy+y^2, x^2+xz+z^2, y^2+yz+z^2

[solakis1]

solve the following system:

$$ x^2 +xy +y^2=37$$
$$ x^2+xz+z^2=28$$
$$y^2+yz+z^2=19$$

 

[topsquark]

I presume x, y, and z are real? Are any real and positive?

-Dan

 

[solakis1]

yes, any reals

 

[topsquark]

[sp]
Here's a quick run-down.

1) [math]x^2 + xy + y^2 = 37[/math]

2) [math]x^2 + x z + z^2 = 28[/math]

3) [math]y^2 + y z + z^2 = 19[/math]

We first need to pull a trick. For example, let's take equation 1):
[math]x^2 + xy + y^2 = 37[/math]

[math](x - y)(x^2 + xy + y^2) = 37 (x - y)[/math]

[math]x^3 - y^3 = 37 (x - y)[/math]

Do this to equations 2) and 3) as well.

Now add equations 1) and 3) and subtract equation 2), giving
[math]0 = x - 2y + z[/math]

Put this into (the unaltered) equations 1) and 2):
1) [math]x^2 + xy + y^2 = 37[/math]

2) [math]x^2 - 2 xy + 4y^2 = 28[/math]
(We can drop equation 3). It doesn't give us anything new.)

So add 2 times equation 1 and equation 2).

[math]x^2 + 2 y^2 = 34[/math]

Thus
[math]y = \pm \sqrt{ \dfrac{1}{2} (34 - x^2) }[/math]

Now put these y values into equation 1)
[math]x^2 \pm x \sqrt{ \dfrac{1}{2} (34 - x^2) } + \dfrac{1}{2} (34 - x^2) = 37[/math]

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
[math]x = \left \{ \pm 4, ~ \pm \dfrac{10}{\sqrt{3} } \right \}[/math]

Putting this all together:
[math](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} [/math]
[/sp]
-Dan

 

[solakis1]

[sp]
Here's a quick run-down.

1) [math]x^2 + xy + y^2 = 37[/math]

2) [math]x^2 + x z + z^2 = 28[/math]

3) [math]y^2 + y z + z^2 = 19[/math]

We first need to pull a trick. For example, let's take equation 1):
[math]x^2 + xy + y^2 = 37[/math]

[math](x - y)(x^2 + xy + y^2) = 37 (x - y)[/math]

[math]x^3 - y^3 = 37 (x - y)[/math]

Do this to equations 2) and 3) as well.

Now add equations 1) and 3) and subtract equation 2), giving
[math]0 = x - 2y + z[/math]

Put this into (the unaltered) equations 1) and 2):
1) [math]x^2 + xy + y^2 = 37[/math]

2) [math]x^2 - 2 xy + 4y^2 = 28[/math]
(We can drop equation 3). It doesn't give us anything new.)

So add 2 times equation 1 and equation 2).

[math]x^2 + 2 y^2 = 34[/math]

Thus
[math]y = \pm \sqrt{ \dfrac{1}{2} (34 - x^2) }[/math]

Now put these y values into equation 1)
[math]x^2 \pm x \sqrt{ \dfrac{1}{2} (34 - x^2) } + \dfrac{1}{2} (34 - x^2) = 37[/math]

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
[math]x = \left \{ \pm 4, ~ \pm \dfrac{10}{\sqrt{3} } \right \}[/math]

Putting this all together:
[math](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} [/math]
[/sp]
-Dan

[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 $$ x^2-2xy+4y^2$$ e.t.c e.t.c...[/sp]

 

[topsquark]

[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 $$ x^2-2xy+4y^2$$ e.t.c e.t.c...[/sp]

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]
[math]x^3 - y^3 = 37(x - y)[/math]

[math]x^3 - z^3 = 28(x - z)[/math]

[math]y^3 - z^3 = 19(y - z)[/math]

So
[math](x^3 - y^3) - (x^3 - z^3) + (y^3 - z^3) = (37(x - y) ) - (28(x - z)) + (19(y - z))[/math]

[math]0 = (37 - 28)x + (-37 + 19)y + (28 - 19)z[/math]
[/sp]

-Dan

 

[solakis1]

I just wanted to show that there is another way to get the equation

z=2y-x

 

[topsquark]

I just wanted to show that there is another way to get the equation

z=2y-x

Yes. That's on line 11.

-Dan