Approach to solving this set of equations with constant p?

Question

Does anybody have an approach to solve this by hand for variables $x,y,z,a$? Of course I could plug it into wolfram, but I wanted to know the fastest / smartest way to get it done by hand and would appreciate every help:

$yz+2ay+2az=0$

$xz+2xa+2za=0$

$xy+2ax+2ay=0$

$2xy+2yz+2xz-p=0$

with $p$ being a constant.

So far I have tried solving for and then replacing but it didn't get me anywhere.

Answer 1

Playing around.

Multiply first by $x$, second by $y$.

$xyz+2axy+2axz=0$

$xyz+2xya+2yza=0$

Subtract.

$2axz = 2yza$ so, if $az \ne 0$, $x = y$.

Multiply third by $z$.

$xyz+2axz+2ayz=0$.

Subtract $xyz+2xya+2yza=0$ to get $2axz = 2xya$. Therefore, if $ax \ne 0$, $y = z$.

Therefore if $axy \ne 0$, $x = y = z$.

The equations then become

$0 =x^2+4ax$

$0 =x^2+4ax$

$6x^2-p=0$

We have both $a = -x/4$ and $p = 6x^2 $.

Answer 2

Factoring the first three equations... $$yz+2ay+2az=(y+2a)(z+2a)-4a^2$$ $$xz+2ax+2az=(x+2a)(z+2a)-4a^2$$ $$yx+2ay+2ax=(y+2a)(x+2a)-4a^2$$ Now we can say $(y+2a)(z+2a)=4a^2$ and so on. Because all three of the equations equal $4a^2$ we can say $x=y=z$.

Thus we can write... $$(x+2a)^2=(y+2a)^2=(z+2a)^2=4a^2$$

Assuming $x=y=z\neq0$ $$(x+2a)=-2a$$

So, $x=y=z=-4a$

Answer 3

Here is a way of solving the system which produces also solutions which are missing (while writing this answer) in the other answers.

This way is rather systematic and may be transferred to similar problems.

First $2$ observations:

• $3$ equations contain plenty of $a$ while the last one does not contain $a \Rightarrow$ It might be helpful to consider $a=0$ and $a\neq 0$ separately.
• All equations contain cyclicly replaced products of two variables. $\Rightarrow$ It might be helpful to consider differences of the equations to factor out $(x-y)$, $(y-z)$ etc.

Let's number the equations by $(1)$ through $(4)$ from top to bottom.

$\pmb{a=0}:$

We get $xy=yz=xz= 0 \stackrel{eq.\; (4)}{\Rightarrow}$ solutions can only exist for $\pmb{p=0}$ and are not only $\pmb{x=y=z=0}$, but also for $\pmb{t \in \mathbb{R}}$:

• $\pmb{x=t,y=0,z=0}$
• $\pmb{x=0,y=t,z=0}$
• $\pmb{x=0,y=0,z=t}$

$\pmb{a\neq0}:$

Let's consider the difference of the equations $(2)$ and $(1)$: $$\stackrel{(2)-(1)}{\Longrightarrow}(z+2a)(x-y)=0$$

• Case $\pmb{z =-2a} \stackrel{into \;(2)}{\Longrightarrow}a=0$ Contradiction to $a \neq 0$. So, this is not an option.
• Case $\pmb{x=y} \stackrel{into \;(3)}{\Longrightarrow}x^2+4ax=0\Rightarrow x=0$ or $x=-4a$
• Subcase $\pmb{x=0}\stackrel{into \;(2)}{\Longrightarrow}2az=0 \stackrel{a \neq 0}{\Rightarrow}z =0 \Rightarrow$ solution only for $\pmb{p=0}$ is $\pmb{x=y=z=0}$.
• Subcase $\pmb{x=-4a}\stackrel{into \;(2)}{\Longrightarrow}-2az-8a^2=0 \stackrel{a \neq 0}{\Rightarrow}z =-4a \Rightarrow$ solution only for $\pmb{p=96a^2}$ is $\pmb{x=y=z=-4a}$.