Finding $x:y:z$ when $\frac{y}{x-z}=\frac{y+x}{z}=\frac{x}{y}$. What is the reason for this peculiar behaviour in the equations?

Question

I am solving this old book and this question is really interesting.

Suppose $$\frac{y}{x-z}=\frac{y+x}{z}=\frac{x}{y}$$ We are asked to find $x:y:z$

Now here is what I've done. Suppose all of this ratios to be equal to k. Then, $$\frac{y}{x-z}=\frac{y+x}{z}=\frac{x}{y}=k$$

Now equating first part and k and then second part and k and so on, we are led to three linear equations, namely: $$kx-y-kz=0 \tag1$$ $$x+y-kz=0 \tag2$$ $$x-ky=0 \tag3$$

Now, using (1) and (2), we could write the ratio of $x:y:z$ , albeit with $k$. Similar is the case with (1) and (3), and (2) and (3). Thus, we get:

$$\frac x{2k}=\frac y{-k+k^2}=\frac z{k+1} \tag4$$ $$\frac x{-k^2}=\frac y{-k}=\frac z{-k-1} \tag5$$ $$\frac x{-k^2}=\frac y{-k}=\frac z{-k^2+1} \tag6$$

Now to find the value $k$, we are again led to three more equations obtained by dividing (5) by (4), (6) by (5) and (4) by (6), that is: $$\frac {2k}{-k^2}=\frac {-k+k^2}{-k}=\frac {k+1}{-k-1} \tag7$$ $$\frac {-k^2}{-k^2}=\frac {-k}{-k}=\frac {-k-1}{-k^2+1} \tag8$$ $$\frac {-k^2}{2k}=\frac {-k}{-k+k^2}=\frac {-k^2+1}{k+1} \tag9$$

From each of these three equations nine equations can be obtained, which gives the values of k, except for one obtained by equating first and the second part of equation $(8)$ which reduces to an identity, namely: $$\frac {-k^2}{-k^2}= \frac{-k}{-k}$$

Now comes the surprising part. The value for k which we obtain by each of those eight equations are same namely $k=0,2,-1$ And even more surprising is the part that each of these three values whether they be substituted in (4) , (5) or (6) gives the same three pair of equations, that are: $$\frac x0=\frac y0=\frac z1 \tag{10}$$ $$\frac x{-1}=\frac y1=\frac z0 \tag{11}$$ $$\frac x4=\frac y2=\frac z3 \tag{12}$$

I want to know the reason for the peculiarity that those eight equations all give the same k. Also when I checked my answer with that given in the book, I found, the equation (10) is not in the answer and only (11) and (12) are given there. What is happening here?

Answer 1

This textbook seems to abide by customs that are a century old. When the original fractions in the problem are written, it is presupposed that they make sense, and hence that $x-z\ne 0$, $y\ne 0$, and $z\ne 0$. This then leads to the answer $x:y:z = 4:2:3$.

Now, if we clear denominators and rewrite the equations without fractions (but still without introducing your $k$), we have $$y^2+xy = xz \qquad\text{and}\qquad y^2=x^2-xz \qquad\text{and, if you insist,} \qquad x(y+x-z)=2yz.$$ Working with the first two, we find that $0=2y^2+xy-x^2=(2y-x)(y+x)$. Thus, we have $x=2y$ or $x=-y$. (The former leads to $x:y:z=2:1:\frac32$. This also checks with the third displayed equation.) The second leads to $z=0$ unless $x=y=0$. This makes perfectly good sense with the displayed equations (rather than the equation of fractions), and we have $1:-1:0$ or $0:0:1$.

Answer 2

$\rm\bf {Task}$

Suppose that,

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

Then, determine the ratio :

$$x:y:z$$

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Let

$$k=\frac{y}{x-z}=\frac{y+x}{z}=\frac{x}{y}$$

We see that, if $x=-y$ then $\dfrac {y}{x-z}=\dfrac xy=0$, which implies $x=y=0$, which means $\dfrac xy$ is undefined. Therefore, we have $0≠x≠-y≠0$ .

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Then, let us recall the following Fraction Arithmetic :

If

$$ \begin{align}\frac {a_1}{b_1}=\frac {a_2}{b_2}=\dots=\frac {a_n}{b_n}=k\end{align} $$

with $\thinspace b_1+b_2+\dots+b_n≠0\thinspace $, then :

$$\frac {a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n}=k\thinspace .$$

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Since $x+y≠0$, then applying the above rule, you have :

$$k=\frac {y+y+x+x}{x-z+z+y}=2$$

This leads to :

$$\frac xy=\frac 21=\frac 42$$

and

$$\begin{align}&\frac{y+x}{z}=\frac {3y}{z}=2\\ \implies &\frac yz=\frac 23\end{align}$$

Thus, the final answer becomes :

$$x:y:z=4:2:3$$