Finley Mckinney

2022-06-22

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
$\frac{x\left(3x+3y\right)}{y}+2.5x=\frac{\left(3x+3y\right)}{y}y$
$\frac{x\left(3x+3y\right)}{y}+2.5x=3x+3y$
$\frac{3{x}^{2}}{y}+\frac{3yx}{y}+2.5x=3x+3y$
$\frac{3{x}^{2}}{y}+3x+2.5x=3x+3y$
$\frac{3{x}^{2}}{y}+2.5x=3y$
Here I got stuck. I didn't know how to find the ratio. Can someone help me?

iceniessyoy

Assuming your calculations so far are correct (I didn't check), you are almost there. Divide both sides by $y$, you will get
$\frac{3{x}^{2}}{{y}^{2}}+\frac{2.5x}{y}=3,$
a quadratic equation for $\frac{x}{y}$.

Hector Petersen

It is given that
$3x+3y=yt=xt+2.5x$
This implies
$t=\frac{3x+3y}{y}=\frac{0.5x+3y}{x}$
$⇒3{x}^{2}+2.5xy-3{y}^{2}=0$
$⇒-3{\left(\frac{y}{x}\right)}^{2}+2.5\frac{y}{x}+3=0$
$\frac{y}{x}=\frac{-2.5±\sqrt{42.25}}{-6}$
Hence
$\frac{y}{x}=\frac{3}{2},\frac{-2}{3}$

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