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If $\dfrac{2x-3y}{3z+y}=\dfrac{z-y}{z-x}=\dfrac{x+3z}{2y-3x}$, prove that each of these ratios is equal to $\dfrac{x}{y}$; hence show that either $x=y$, or $z=x+y$

$\Rightarrow \dfrac{2x-3y}{3z+y}=\dfrac{z-y}{z-x}=\dfrac{x+3z}{2y-3x}=k$

$$2x-3y=(3z+y)k\tag{1}$$ $$z-y=(z-x)k\tag{2}$$ $$x+3z=(2y-3x)k\tag{3}$$

Then $3 \cdot (2)-(3)-(1)$ to get $\dfrac{x}{y}=k$, proving all the ratios are equal to it. My question is how do I show $x=y$ or $z=x+y$? I could for example just plug in $z=x+y$ into $(2)$ to get $\dfrac{x}{y}=k$, but I'm not sure I'm meant to just literally plug it back in. I think I'm suppose to prove it somehow, but I don't know how. Thanks for the help.

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3 Answers 3

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This is another method: $$\frac{2x-3y}{3z+y} = \frac{z-y}{z-x} = \frac {x+3z}{2y-3x}$$

$$\frac{2x-3y}{3z+y} = \frac{3(z-y)}{3(z-x)} = \frac {x+3z}{2y-3x}$$

$$= \frac{(2x-3y)-3(z-y)+x+3z}{(3z+y)-3(z-x)+(2y-3x)}$$

$$= \frac{2x-3y-3z+3y+x+3z}{3z+y-3z+3x+2y-3x} = \frac{3x}{3y} = \frac{x}{y}$$

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  • $\begingroup$ How did you go from $\frac{2x-3y}{3z+y} = \frac{3(z-y)}{3(z-x)} = \frac {x+3z}{2y-3x}$ to $\frac{(2x-3y)-3(z-y)+x+3z}{(3z+y)-3(z-x)+(2y-3x)}$? Thanks. $\endgroup$
    – ronald christenkkson
    Commented Jun 8, 2023 at 18:42
  • $\begingroup$ $\frac{a}{b} = \frac{c}{d} = \frac{a \pm b}{c \pm d}$. Then, $\frac{a}{b} = \frac{a(b+d)}{b(b+d)} = \frac{a}{b+d} + \frac{ad}{b(b+d)}$ Since $ad=bc$, $= \frac{a}{b+d} + \frac{bc}{b(b+d)} = \frac{a}{b+d} + \frac{c}{b+d} = \frac{a+c}{b+d}$ $\endgroup$
    – Johnny Lenin
    Commented Jun 9, 2023 at 13:46
  • $\begingroup$ In $\frac{a}{b}=\frac{a(b+d)}{b(b+d)}$, where did the $(b+d)$ come from? Otherwise I understand what you are saying here, but I can't see how the question fits this pattern. For example if I set $a=2x-3y$ it doesn't work. Thank you. $\endgroup$
    – ronald christenkkson
    Commented Jun 9, 2023 at 16:37
  • $\begingroup$ I used the theorem "If $\frac{a}{b}=\frac{c}{d}$, then they are also equal to $\frac{a \pm b}{c \pm d}$. The other part of that comment was proving the theorem. $\endgroup$
    – Johnny Lenin
    Commented Jun 9, 2023 at 18:29
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Since you've already shown that each ratio is equal to $\frac x y$, we can use $$\frac{z-y}{z-x} = \frac{x}{y}$$ Clearing out the denominators, we get $$yz - y^2 = xz - x^2$$ i.e., $$x^2 - y^2 = xz - yz$$ Factoring $x-y$, we have $$(x-y)(x+y) = (x-y)z$$ i.e., $$(x-y)(x+y-z) = 0$$ implying $x = y$ or $x + y = z$ as required.

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  • $\begingroup$ Thanks for the answer. How did you go from $(x-y)(x+y) = (x-y)z$ to $(x-y)(x+y-z) = 0$? Because if I see $(x-y)(x+y) = (x-y)z$ I would instinctively take out the $(x-y)$ and get $z=x+y$. $\endgroup$
    – ronald christenkkson
    Commented Jun 8, 2023 at 18:36
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    $\begingroup$ @ronaldchristenkkson You can "take out" $(x-y)$ only if $x - y \ne 0$, because "taking out" is really dividing, and dividing by zero is illegal. The best way to avoid falling into such traps is to take all terms to the left-hand side of the equality and factorize. That's precisely what I did. I went from $$(x-y)(x+y) = (x-y)z$$ to $$(x-y)(x+y) - (x-y)z = 0$$ and finally $$(x-y)(x+y-z) = 0$$ Do you see it now? $\endgroup$
    – stoic-santiago
    Commented Jun 8, 2023 at 18:39
  • $\begingroup$ Yes I see thank you. $\endgroup$
    – ronald christenkkson
    Commented Jun 8, 2023 at 18:40
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If $\;z=x+y\;,\;$ from $\;\dfrac{2x-3y}{3z+y}=\dfrac{z-y}{z-x}=\dfrac{x+3z}{2y-3x}\;,\quad\color{blue}{(*)}$

it follows that

$\dfrac{2x-3y}{3x+4y}=\dfrac xy=\dfrac{4x+3y}{2y-3x}\;,$

hence,

$3x^2+2xy+3y^2=0\;,$

therefore,

$\begin{cases}x\neq0\\[3pt]y=\dfrac{-1+2i\sqrt2}3x\\[3pt]z=\dfrac{2+2i\sqrt2}3x\end{cases}$

Consequently, if $\;x,y,z\;$ were real numbers, then it would be impossible that $\;z=x+y\;$ and in this case, from $\,(*)\,,\,$ it would only follow that $\;x=y\;,\;$ more precisely,

$\begin{cases}x\neq0\\[3pt]y=x\\[3pt]z=-\dfrac23x\end{cases}$

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