# If $\frac{2x-3y}{3z+y}=\frac{z-y}{z-x}=\frac{x+3z}{2y-3x}$, prove that each of these ratios is equal to $\frac{x}{y}$

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.

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

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

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

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

• 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. Commented Jun 8, 2023 at 18:42
• $\frac{a}{b} = \frac{c}{d} = \frac{a \pm b}{c \pm d}$. Then, $\frac{a}{b} = \frac{a(b+d)}{b(b+d)} = \frac{a}{b+d} + \frac{ad}{b(b+d)}$ Since $ad=bc$, $= \frac{a}{b+d} + \frac{bc}{b(b+d)} = \frac{a}{b+d} + \frac{c}{b+d} = \frac{a+c}{b+d}$ Commented Jun 9, 2023 at 13:46
• 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. Commented Jun 9, 2023 at 16:37
• 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. Commented Jun 9, 2023 at 18:29

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.

• 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 $z=x+y$. Commented Jun 8, 2023 at 18:36
• @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? Commented Jun 8, 2023 at 18:39
• Yes I see thank you. Commented Jun 8, 2023 at 18:40

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,

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