Find the values of $x,y,z\in\mathbb R$, such that $\frac{xy}{y+2x}=2,\frac{xz}{z+3x}=3,\frac{zy}{3y+2z}=12$

Question

Let $x,y,z\in\mathbb R$, such that $\frac{xy}{y+2x}=2,\frac{xz}{z+3x}=3,\frac{zy}{3y+2z}=12$

Then, find the values of $x,y,z$.

My attempts:

$$xy=2y+4x\implies x=\frac{2y}{y-4}\\ xz=3z+9x\implies z=\frac{9x}{x-3}\\ zy=36y+24z\implies y=\frac{24z}{z-36} $$

But I don't know how to continue from here.

Is there any clever way to do this?

Answer 1

Substituting the variables we obtain $3x=8$, $y=16$ and $z=-72$. More precisely, substituting $y$ and $z$ by $$ y=\frac{4x}{x-2},\quad z=\frac{9x}{x-3}, $$ the last equation becomes $$x(3x-8)=0.$$ Of course, $x=y=z=0$ is not a solution, so that $x=\frac{8}{3}$, and hence $y=16$ and $z=-72$.

The computation is so easy, that one should try this first, before thinking of a more "clever way".

Answer 2

Not clever, but I would like to suggest a possible way:

We see that, the numerator and denominator have the same pair variables: $(x,y);(xz);(yz)$ and unfortunately, we cannot simplify the fractional expressions because the denominator is a polynomial that we cannot factorise.

Therefore, we want to see what we can achieve by reversing fractional expressions. Because in this case the denominator will be a factored polynomial.

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Thus we have,

$$\begin{align}&\begin{cases}\frac{y+2x}{xy}=\frac 1x+\frac 2y=\frac 12\\ \frac{3x+z}{xz}=\frac 3z+\frac 1x=\frac 13\\ \frac{3y+2z}{zy}=\frac 3z+\frac 2y=\frac 1{12}\end{cases}\\ \\ \implies &2\left(\frac 1x+\frac 2y+\frac 3z\right)=\frac 12+\frac 13+\frac 1{12}\\ \implies &\frac 1x+\frac 2y+\frac 3z=\frac{11}{24}\\ \implies &\frac 1x=\frac {11}{24}-\frac{1}{12}= \frac 38\\ \implies &\frac 2y=\frac{11}{24}-\frac 13 =\frac 18\\ \implies &\frac 3z=\frac{11}{24}-\frac 12=-\frac{1}{24}\\ \implies &\left\{x,y,z\right\}=\left\{\frac 83,16, -72\right\}.\end{align}$$

Answer 3

$\frac{xy}{y+2x}=2$, $\frac{y+2x}{xy}=\frac{y}{xy}+\frac{2x}{xy}=\frac{1}{x}+\frac{2}{y}=\frac{1}{2}$

$\frac{xz}{z+3x}=3$, $\frac{z+3x}{xz}=\frac{z}{xz}+\frac{3x}{xz}=\frac{1}{x}+\frac{3}{z}=\frac{1}{3}$

$\frac{zy}{3y+2z}=12$,$\frac{3y+2z}{zy}=\frac{3y}{zy}+\frac{2z}{zy}=\frac{3}{z}+\frac{2}{y}=\frac{1}{12}$

Subtracting first and second equation

$\frac{2}{y}-\frac{3}{z}=\frac{1}{6}$

$\frac{3}{z}+\frac{2}{y}=\frac{1}{12}$

adding with third equation

$\frac{4}{y}=\frac{1}{4}$

$y=16$, $x=\frac{8}{3}$, $z=-72$

Answer 4

Also you can use this simple way:

$(1) xy=2y+4x\implies x=\frac{2y}{y-4}\\ (2) xz=3z+9x\implies xz-3z-9x=0 \\ (3) zy=36y+24z\implies zy-36y-24z=0\\$

We put $x=\frac{2y}{y-4}$ in $(2)$

$(\frac{2y}{y-4})z=3z+9(\frac{2y}{y-4})\implies -zy-18y+12z=0 \\$

Now

\begin{cases}-zy-18y+12z=0\\zy-36y-24z=0\\ \end{cases}

Then we both multiply $\frac{1}{z}$ and $w=\frac{y}{z}$

\begin{align}&\begin{cases}-y-18w+12=0\\y-36w-24=0 \end{cases}\\ \\ \implies &3y-48=0\\ \implies &y=16\\ \end{align}

Then we put $y=16$ in $(3)$ and $(1)$

$zy-36y-24z=0\implies z(16)-36(16)-24z=0\implies z=-72\\$

$xy=2y+4x\implies x(16)=2(16)+4x\implies x=\frac{8}{3}\\$

Answer 5

From $\frac{xy}{y+2x}=2,\frac{xz}{z+3x}=3,\frac{zy}{3y+2z}=12 $, taking reciprocals, $\frac{y+2x}{xy}=\frac12, \frac{z+3x}{xz}=\frac13, \frac{3y+2z}{zy}=\frac1{12} $ or $\frac12=\frac1{x}+\frac{2}{y}, \frac13=\frac1{x}+\frac{3}{z}, \frac1{12}=\frac{3}{z}+\frac{2}{y} $.

Letting $\frac1{x}=u, \frac1{y}=v, \frac1{z}=w $, this becomes $\frac12=u+2v, \frac13=u+3w, \frac1{12}=3w+2v $.

This can be solved in many ways. For example, from the first two, $\frac12-2v=\frac13-3w $ so $v =\frac12(\frac12-\frac13+3w) =\frac1{12}+\frac32 w $.

Substituting in the third, $\frac1{12} =3w+2v =3w+2(\frac1{12}+\frac32 w) =6w+\frac16 $ so $w =\frac16(\frac1{12}-\frac16) =-\frac16(\frac1{12}) =-\frac1{72} $.

Then $v =\frac1{12}-\frac{3}{144} =\frac{3}{48} =\frac1{16} $ and, finally, $u =\frac12-2v =\frac12-\frac18 =\frac38 $.