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

Question

If for three distinct positive numbers $x$, $y$, and $z$, $$\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}$$

then find the value of $\frac{x}{y}$

I have tried all types of manipulations, even quadratics but can seem to get the answer. Please help!! BTW im a 7th grader so easy solutions will be appreciated.

Answer 1

Solution: $$\frac{x}{y}=\frac{(y)+(x+y)+(x)}{(x-z)+(z)+(y)}=\frac{2(x+y)}{x+y}=2$$ Explaination: We shall prove that if $\frac{a}{b}=\frac{c}{d}$ then, $\frac{a}{b}=\frac{c}{d}=\frac{a+kc}{b+kd}$ (using $k=1$ for the above solution)

Since $\frac{a}{b}=\frac{c}{d}$, then $\frac{d}{b}=\frac{c}{a}$ so... $$\frac{a+kc}{b+kd}=\frac{a(1+k(c/a))}{b(1+k(d/b))}=\frac{a(1+k(c/a))}{b(1+k(c/a))}=\frac{a}{b}$$ (Try with some fractions if your not convinced Ex:$\frac{1}{2}=\frac{2}{4}=\frac{1+2}{2+4}$

Thus, in the problem, we can say x/y is the sum of numerators over the sum of the denominators and we are done.

Answer 2

Hint Let $\alpha=\frac{x}{y}$. Then $$\frac{y}{x-z}=\frac{x+y}{z}=\alpha \\ x=\alpha y$$ giving $$\frac{(\alpha+1)y}{z}=\alpha \Rightarrow z=\frac{(\alpha+1)y}{\alpha} \\ \frac{y}{\alpha y- \frac{(\alpha+1)y}{\alpha}}=\alpha \Rightarrow \\1=\alpha \left( \alpha - \frac{(\alpha+1)}{\alpha}\right)\\$$

Answer 3

Consider (first & third terms) $$\frac{y}{x-z}=\frac{x}{y}$$ $$z=\frac{x^2-y^2}{x}\tag1$$ Similarly, consider (second & third terms) $$\frac{x+y}{z}=\frac{x}{y}$$ $$\frac{x+y}{\frac{x^2-y^2}{x}}=\frac{x}{y}\quad (\text{setting the value of z from (1)})$$ $$\frac{x(x+y)}{(x+y)(x-y)}=\frac{x}{y}$$ $$\frac{x}{x-y}=\frac{x}{y}$$ $$\frac{x/y}{x/y-1}=\frac{x}{y}$$ $$\left(\frac{x}{y}\right)^2-2\frac{x}{y}=0$$ $$\frac xy\left(\frac{x}{y}-2\right)^2=0$$ $$\frac xy=0\ \ \text{or} \ \ \frac xy =2$$ But $x, y, z>0$ hence$$\frac xy =2$$

Answer 4

Given:

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

Note that if

$\frac{a}{b} = \frac{c}{d}$

then each of them is equal to

$\frac{a\pm c}{b \pm d}$ provided $b \pm d \neq 0$

Using this property, from the first two fractions, we obtain

$\frac{x+2y}{x} = \frac{x}{y}$

Simplifying,

$x^2 = xy + 2y^2$

Put $x = ky$ in the above equation to obtain

$k^2 - k - 2 = 0$

Solving $k = 2, -1$

Since $k \neq -1$, we must have $k = 2$

Answer 5

$z=\frac{y(x+y)}{x}$, $x(x-z)=y^2$, $x-z=\frac{x^2-xy-y^2}{x}$, $y^2=x^2-xy-y^2$, let $u=\frac{y}{x}$, $2u.^2+u-1=0$, $u=-1$, or $u=\frac{1}{2}$

Final result $\frac{x}{y}=2$, since $x$ and $y$ have to be positive.