7.287. $\left\{\begin{array}{l}4^{x}-7 \cdot 2^{x-0.5 y}=2^{3-y}, \\ y-x=3 .\end{array}\right.$
## Solution.

From the second equation of the system, we find $y=x+3$. Then

$2^{2 x}-7 \cdot 2^{x-0.5(x+3)}=2^{3-x-3} \Leftrightarrow 2^{2 x}-7 \cdot 2^{0.5 x-1.5}=2^{-x} \Leftrightarrow$

$\Leftrightarrow 2^{3 x}-\frac{7 \cdot 2^{1.5 x}}{2^{1.5}}=1 \Leftrightarrow 2^{1.5} \cdot 2^{3 x}-7 \cdot 2^{1.5 x}-2^{1.5}=0$.

Solving this equation as a quadratic equation in terms of $2^{1.5 x}$, we have $2^{1.5 x}=-2^{-1.5}$, no solutions, or $2^{1.5 x}=2^{1.5} \Rightarrow x=1$. Then $y=4$.

Answer: $(1 ; 4)$.