$|x+1| = -(x+1)$ and $|2x - 1| = -(2x - 1)$ because $2x - 1 < -3 < 0$. 

Equation: $2^{-(x+1)} - 2^x = -(2x - 1) + 1$ 

Right side: $-2x + 1 + 1 = -2x + 2$ 

Left side: $2^{-(x+1)} = \dfrac{1}{2^{x+1}}$ 

Try $x = -2$: 

LHS: $2^{ -(-2+1) } - 2^{-2} = 2^{1} - \dfrac{1}{4} = 2 - 0.25 = 1.75$ 

RHS: $-2(-2) + 2 = 4 + 2 = 6$ → not equal Try $x = -2$ again carefully: Wait — check systematically. Better approach: substitute $x=-2$ in original equation: Left side: $2^{| -2 + 1 |} - 2^{-2} = 2^{1} - \dfrac{1}{4} = \dfrac{7}{4}$ Right side: $|2(-2) - 1| + 1 = | -5 | + 1 = 6$ Not equal → reject. Try $x = -1.5$ type pattern? Better substitute only options (valid since only one option has $x< -1$): Only option with $x < -1$ is (1) −2. But we tested −2 and it doesn't satisfy? Check original equation carefully: Original: $2^{|x+1|} - 2^x = |2x - 1| + 1$ Try $x=-2$ again: Left: $2^{|-2+1|} - 2^{-2} = 2^{1} - \dfrac{1}{4} = \dfrac{7}{4}$ Right: $|2(-2)-1| + 1 = |-5| +1 = 6$