$x^2 - y^2=9$, $2y-x=3$, Find maximum value of $\dfrac{y}{x}$

Question

I came across this question and have no idea what to do. Given, $$ x^2-y^2=9 \\ 2y-x=3 $$

find maximum value of $\dfrac{y}{x}$

Answer 1

Because $x = 2y - 3$, we have:

$$(2y-3)^{2} - y^{2} = 9$$

$$4y^{2} - 12y + 9 - y^{2} = 9$$

$$3y^{2} - 12y = 0$$

$$3y(y - 4) = 0$$

$$y = 0, 4$$

$$(x, y) = (-3, 0), (5, 4)$$

$$\frac{y}{x} = 0, \frac{4}{5}$$

$$\boxed{\max(\frac{y}{x}) = \frac{4}{5}}$$

Answer 2

Solve the system of equations, for instance, with $2y-x=3\iff x=2y-3$; insert this value in the first equation to obtain $(2y-3)^2-y^2=9\iff ...\iff y\cdot(y-4)=0$. Thus the solutions $(x,y)$ are $(3, 0)$ and $(5, 4)$. Which one of them yields the maximum ratio $\frac yx$?