I need to find the ratio of
If given that
So what I tried is:
$$t = \frac{3x + 3y}{y}$$ And then put it in the equation $$\frac{x(3x + 3y)}{y} + 2.5x = \frac{(3x + 3y)}{y}y$$ $$\frac{x(3x + 3y)}{y} + 2.5x = 3x + 3y$$
$$\frac{3x^2}{y} + \frac{3yx}{y} + 2.5x = 3x + 3y$$ $$\frac{3x^2}{y} + 3x + 2.5x = 3x + 3y$$ $$\frac{3x^2}{y} + 2.5x = 3y$$
Here I got stuck. I didn't know how to find the ratio. Can someone help me?
On
It is given that $$3x+3y=yt=xt+2.5x$$ This implies $$t=\frac{3x+3y}{y}=\frac{0.5x+3y}{x}$$ $$\Rightarrow 3x^2+2.5xy-3y^2=0$$ $$\Rightarrow -3\left(\frac{y}{x}\right)^2 +2.5\frac{y}{x}+3=0$$ $$\frac{y}{x}= \frac{-2.5 \pm \sqrt{42.25}}{-6}$$
Hence $$\frac{y}{x}=\frac{3}{2},\frac{-2}{3}$$
On
Solve to get
$$t = \frac{5x}{2y-2x} = \frac{3(x+y)}{y}$$
This gives the quadratic equation:
$$ 6x^2 - 5xy - 6y^2 = 0 $$
Factorize to get:
$$(3x + 2y) (2x - 3y) = 0$$
yielding
$$\frac{y}{x} = \frac{-3}{2} \,\,\,\text{or} \,\,\,\frac{2}{3}$$
Thus unless some sort of restriction is placed on $x$ or $y$
(for e.g both > $0$), $\frac yx$ can assume $2$ distinct values.
Assuming your calculations so far are correct (I didn't check), you are almost there. Divide both sides by $y$, you will get $$\frac {3x^2} {y^2} + \frac {2.5x} y = 3,$$ a quadratic equation for $\frac xy$.