Question

If 2^x+y = 2^x–y = $\sqrt{8}$, then x = _______ and y = _______.

Answer 1

Given: 2^x+y = 2^x–y = $\sqrt{8}$

Now,
$$\begin{gathered} 2^{x+y}=\sqrt{8} \\ \Rightarrow 2^{x+y}=\sqrt{2^3} \\ \Rightarrow 2^{x+y}=2^{\frac{3}{2}} \\ \Rightarrow x+y=\frac{3}{2}...\left(1\right) \\ {2}^{x-y}=\sqrt{8} \\ \Rightarrow 2^{x-y}=\sqrt{2^3} \\ \Rightarrow 2^{x-y}=2^{\frac{3}{2}} \\ \Rightarrow x-y=\frac{3}{2}...\left(2\right) \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ x=\frac{3}{2}\mathrm{and}y=0 \end{gathered}$$


Hence, x = $\underline{\frac{3}{2}}$ and y = 0.