Find $\frac{y}{x}$ from $3x + 3y = yt = xt + 2.5x$ I need to find the ratio of
 $$\frac{y}{x}$$
If given that
$$3x + 3y = yt = xt + 2.5x$$
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?

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 $$\frac{y}{x}=\frac{3}{2},\frac{-2}{3}$$
