An algebra problem by Rahul Katarki

$\begin{cases} 2^{3x+4}-3^{y+1}=119 \\ 2^{x-1}+3^{y}=4 \end{cases}$ $2^{3x+4}-3(3^{y})=119$ $2^{3x+4}-3(4-2^{x-1})=119$ $2^{3x+4}-12+3(2^{x-1})=119$ $16(2^{x})^{3}+1.5(2^{x})=131$ $32(2^{x})^{3}+3(2^{x})=262$ $2^{x}=2$ $\boxed{x=1}$

Given that $\begin{cases} 2^{3x+4} - 3^{y+1} = 119 & ...(1) \\ 2^{x-1} + 3^y = 4 & ...(2) \end{cases}$

$\implies 3 \times (2): \ 3(2^{x-1}) + 3^{y+1} = 12 \ ...(2a)$ and

$\begin{aligned} (1)+(2a): \quad 2^{3x+4} + 3(2^{x-1}) & = 119 + 12 \\ 2^{3x+4} + (2^2-1)2^{x-1} & = 131 \\ 2^{3x+4} + 2^{x+1} - 2^{x-1} & = 131 \end{aligned}$

Note that the right-hand side $131$ is odd, therefore the left-hand side must also be odd. The only way to make the LHS to be odd is $2^{x-1} = 2^0 = 1$ , $\implies x = 1$ . Then we note that the $LHS = 2^7 + 2^2 - 2^0 = 128 + 4 - 1 = 131 = RHS$ . Therefore $x = \boxed 1$ .