Super simultaneous equations #2 Reupload

From $2^{xy} = 64 = 2^6 \implies xy = 6$ . Then:

$\begin{aligned} \left(\frac \blue 6x \right)^2 + 6y - 4 & = 8y - 6 + \frac \blue{xy}6 & \small \blue{\text{Note that }xy=6} \\ \left(\frac \blue{xy}x \right)^2 + 6y - 4 & = 8y - 6 + \frac \blue 66 \\ y^2 + 6y - 4 & = 8y - 5 \\ y^2 - 2y + 1 & = 0 \\ (y-1)^2 & = 0 \\ \implies y & = 1 \end{aligned}$

Therefore $\dfrac {x+y}{x-y} = \dfrac {xy+y^2}{xy-y^2} = \dfrac {6+1^2}{6-1^2} = \dfrac 75 = \boxed{1.4}$ .

$2^{xy}=64=2^6\implies xy=6\implies y=\dfrac 6x$

So, $(\frac 6x)^2+6y-4=8y-6+\frac{xy}{6}$

$\implies y^2-2y+1=0\implies y=1\implies x=6$

Hence $\dfrac {x+y}{x-y}=\dfrac {6+1}{6-1}=\boxed {1.4}$ .