Simultaneously: \( 3^{x+2}=2-1 \) \( 2^{y}=4^{x+12} \)

Let's address the first equation, \( 3^{x+2} = 2 - 1 \). Here, \( 2 - 1 \) simplifies to \( 1 \). Thus, we have: \[ 3^{x+2} = 1 \] Since any number raised to the power of 0 equals 1, we can equate the exponent to 0: \[ x + 2 = 0 \] \[ x = -2 \] Now, let's turn our attention to the second equation, \( 2^{y} = 4^{x+12} \). Noting that \( 4 \) can be rewritten as \( 2^2 \), we can express \( 4^{x+12} \) in terms of base \( 2 \): \[ 4^{x+12} = (2^2)^{x+12} = 2^{2(x+12)} = 2^{2x + 24} \] Thus, our equation becomes: \[ 2^{y} = 2^{2x + 24} \] Since the bases are the same, we can set the exponents equal to each other: \[ y = 2x + 24 \] Substituting \( x = -2 \) into this equation gives us: \[ y = 2(-2) + 24 = -4 + 24 = 20 \] So the solution to the system of equations is: \[ x = -2, \quad y = 20 \]