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

$\frac{x(3x+3y)}{y}+2.5x=\frac{(3x+3y)}{y}y$

$\frac{x(3x+3y)}{y}+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+\frac{3yx}{y}+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+3x+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+2.5x=3y$

Here I got stuck. I didn't know how to find the ratio. Can someone help me?

$\frac{y}{x}$

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

$\frac{x(3x+3y)}{y}+2.5x=\frac{(3x+3y)}{y}y$

$\frac{x(3x+3y)}{y}+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+\frac{3yx}{y}+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+3x+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+2.5x=3y$

Here I got stuck. I didn't know how to find the ratio. Can someone help me?

iceniessyoy

Beginner2022-06-23Added 27 answers

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 $\frac{x}{y}$.

$\frac{3{x}^{2}}{{y}^{2}}+\frac{2.5x}{y}=3,$

a quadratic equation for $\frac{x}{y}$.

Hector Petersen

Beginner2022-06-24Added 6 answers

It is given that

$3x+3y=yt=xt+2.5x$

This implies

$t=\frac{3x+3y}{y}=\frac{0.5x+3y}{x}$

$\Rightarrow 3{x}^{2}+2.5xy-3{y}^{2}=0$

$\Rightarrow -3{\left(\frac{y}{x}\right)}^{2}+2.5\frac{y}{x}+3=0$

$\frac{y}{x}=\frac{-2.5\pm \sqrt{42.25}}{-6}$

Hence

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

$3x+3y=yt=xt+2.5x$

This implies

$t=\frac{3x+3y}{y}=\frac{0.5x+3y}{x}$

$\Rightarrow 3{x}^{2}+2.5xy-3{y}^{2}=0$

$\Rightarrow -3{\left(\frac{y}{x}\right)}^{2}+2.5\frac{y}{x}+3=0$

$\frac{y}{x}=\frac{-2.5\pm \sqrt{42.25}}{-6}$

Hence

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