Solving equation in three variables

Question

please help me understand how the following equation with 3 variables and power of 2 is solved and what solution approach is the quickest.

$$3y^2 - 3 = 0$$ $$4x - 3z^2 = 0$$ $$-6xz+ 6z = 0 $$

score 1 · Accepted Answer · answered Oct 04 '16 at 17:38

1

$$ \begin{cases} 3y^2-3=0\\ 4x-3z^2=0\\ 6z-6xz=0 \end{cases}\Longleftrightarrow \begin{cases} 3y^2-3=0\\ x=\frac{3z^2}{4}\\ 6z(1-x)=0 \end{cases}\Longleftrightarrow \begin{cases} 3y^2-3=0\\ x=\frac{3z^2}{4}\\ 6z\left(1-\frac{3z^2}{4}\right)=0 \end{cases} $$

So, when you solve:

$$6z\left(1-\frac{3z^2}{4}\right)=0$$

We get three solutions for $z$, $z=0$ or $z=\pm\frac{2}{\sqrt{3}}$

answered Oct 04 '16 at 17:38

Jan Eerland

28,671

is it that simple ? I then plug the three z values into x to get three x-s ? y is +/-1 ? where do I get my third y ? – HGO HGO Oct 04 '16 at 17:43
@HGOHGO I see no relation between y and the other two varibables – Jan Eerland Oct 04 '16 at 17:45
So, for now $y=\pm1$ and $x=0$ or $x=1$ – Jan Eerland Oct 04 '16 at 17:46
should'nt x be 0 or 4/3. The above is the gradient of a function in three variables : the function is
$$f(xyz)= 2x^2-3xz^2+y^3+3z^2-3y+3$$
– HGO HGO Oct 04 '16 at 18:00

score 1 · Answer 2 · 2016-10-05T06:48:51.043

1

A quick solution:

By simplification, rewrite

$$\begin{cases}y^2=1\\(1-x)z=0\\3z^2=4x.\end{cases}$$

Then mentally,

$$y=\pm1,\\x=z=0\lor x=1,z=\pm\frac2{\sqrt3}.$$

edited Oct 05 '16 at 06:48

answered Oct 05 '16 at 06:36

score 0 · Answer 3 · answered Oct 05 '16 at 06:23

$3y^2-3=0.............(1)$

$4x-3z^2=0...........(2)$

$6z-6xz=0...........(3)$

First, solving the equation(1),

$3y^2-3=0\;\;\;\implies y^2=1\;\;\;\implies y=\pm1$

Now, solving the equation(3),

$6z-6xz=0\;\;\;\implies 6z(1-x)=0$

$z=0,\;1-x=0$

$z=0,\;x=1$

Solving the equation(2),

$4x-3z^2=0$

Put the value of $x$ in equation(2)

$4(1)-3z^2=0\;\;\implies-3z^2=-4\;\;\implies z=\sqrt{\dfrac{4}{3}}\;\implies z=\pm \dfrac{2}{\sqrt{3}}$

Thus, $x=1,\;y=\pm1,\;z=0\;or\;z=\pm \dfrac{2}{\sqrt{3}}$

Solving equation in three variables

3 Answers3