system of 3 equations

**jmedsy** · January 26th 2010, 04:25 PM

solve for x,y, and z:

x+y = 3z
x^2 + y^2 = 5z
x^3 + y^3 = 9z

**pickslides** · January 26th 2010, 04:30 PM

From

Originally Posted by jmedsy

x+y = 3z

$z = \frac{x}{3}+\frac{y}{3}$

so

$x^2 + y^2 = 5\left(\frac{x}{3}+\frac{y}{3}\right)$

(You might look to complete the sqaure here.)

and

$x^3 + y^3 = 9\left(\frac{x}{3}+\frac{y}{3}\right)$

**jmedsy** · January 26th 2010, 05:28 PM

i still don't really see how its solved without some very lengthy, tedious algebra.

what's the next step

**pickslides** · January 26th 2010, 05:42 PM

Originally Posted by jmedsy

without some very lengthy, tedious algebra.

that's what I also see.

**Soroban** · January 26th 2010, 07:58 PM

Hello, jmedsy!

I think I've solved it . . . I'll get you started.

Solve for $x,y,z\!:\;\;\begin{array}{cccc}x+y &=& 3z & [1] \\ x^2 + y^2 &=& 5z & [2] \\ x^3 + y^3 &=& 9z & [3] \end{array}$

Note that $x=y=z=0$ is a solution.
In the following work, all the variables are nonzero.

Square [1]: . $(x+y) \:=\:(3z)^2 \quad\Rightarrow\quad x^2 + 2xy + y^2 \:=\:9z^2$

. . . $\underbrace{x^2+y^2}_{\text{This is }5z} + 2xy \:=\:9z^2 \quad\Rightarrow\quad 5z + 2xy \:=\:9z^2 \quad\Rightarrow\quad xy \:=\:\frac{9z^2-5z}{2}$ .[1]

Cube [1]: . $(x+y)^3 \:=\:(3z)^3 \quad\Rightarrow\quad x^3 + 3x^2y + 2xy^2 + y^3 \:=\:27z^3$

. . . $\underbrace{x^3+y^3}_{\text{This is }9z} + 3xy\underbrace{(x+y)}_{\text{This is }3x} \:=\:27z^2 \quad\Rightarrow\quad 9z + 9xyz \:=\:27z^3 \quad\Rightarrow\quad xy \:=\:3z^2-1$ .[2]

Equate [1] and [2]: . $\frac{9z^2-5z}{2} \:=\:3z^2-1 \quad\Rightarrow\quad 3z^2 - 5z + 2 \:=\:0$

Factor: . $(z-1)(3z-2) \:=\:0 \quad\Rightarrow\quad z \:=\:1,\:\tfrac{2}{3}$

Your turn . . .

**jmedsy** · January 26th 2010, 09:25 PM

edited, thanks for the help

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