- Thread starter
- #1

- Thread starter solakis
- Start date

- Thread starter
- #1

- Aug 30, 2012

- 1,120

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

-Dan

-Dan

- Thread starter
- #3

- Aug 30, 2012

- 1,120

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:

\(\displaystyle (x, y, z) = \begin{cases} (4, ~ 3, ~ 2) \\ (-4, ~ -3, ~ -2) \\ \left ( \dfrac{10}{\sqrt{3}}, ~ \dfrac{1}{\sqrt{3}}, ~ - \dfrac{8}{\sqrt{3}} \right ) \\ \left ( - \dfrac{10}{\sqrt{3}}, ~ - \dfrac{1}{\sqrt{3}}, ~ \dfrac{8}{\sqrt{3}} \right ) \end{cases} \)

-Dan

- Thread starter
- #5

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:

\(\displaystyle (x, y, z) = \begin{cases} (4, ~ 3, ~ 2) \\ (-4, ~ -3, ~ -2) \\ \left ( \dfrac{10}{\sqrt{3}}, ~ \dfrac{1}{\sqrt{3}}, ~ - \dfrac{8}{\sqrt{3}} \right ) \\ \left ( - \dfrac{10}{\sqrt{3}}, ~ - \dfrac{1}{\sqrt{3}}, ~ \dfrac{8}{\sqrt{3}} \right ) \end{cases} \)

-Dan

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.......

- Aug 30, 2012

- 1,120

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.

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.......

\(\displaystyle x^3 - y^3 = 37(x - y)\)

\(\displaystyle x^3 - z^3 = 28(x - z)\)

\(\displaystyle y^3 - z^3 = 19(y - z)\)

So

\(\displaystyle (x^3 - y^3) - (x^3 - z^3) + (y^3 - z^3) = (37(x - y) ) - (28(x - z)) + (19(y - z))\)

\(\displaystyle 0 = (37 - 28)x + (-37 + 19)y + (28 - 19)z\)

-Dan

- Thread starter
- #7

- Aug 30, 2012

- 1,120

Yes. That's on line 11.I just wanted to show that there is another way to get the equation

z=2y-x

-Dan