\( x \neq y \) simulitancously if \( 2^{y}-16^{x}=0 \) and \( y=x^{2}+4 x \)

To solve the equations \( 2^{y} - 16^{x} = 0 \) and \( y = x^2 + 4x \) simultaneously, start by rewriting the first equation. Since \( 16 = 2^4 \), the equation \( 16^{x} \) can be expressed as \( (2^4)^{x} = 2^{4x} \). Therefore, the first equation becomes: \[ 2^{y} = 2^{4x} \] This implies: \[ y = 4x \] Now, substitute \( y = 4x \) into the second equation: \[ 4x = x^2 + 4x \] Subtract \( 4x \) from both sides, resulting in: \[ 0 = x^2 \] This indicates \( x = 0 \). However, since \( x \neq y \), we must check: If \( x = 0 \), then: \[ y = 4(0) = 0 \] Thus, \( x = y \), which contradicts our initial condition. Next, to investigate the non-zero solutions, recall \( y = x^2 + 4x \) and set \( 4x = x^2 + 4x \): Rearranging yields: \[ 0 = x^2 \] To avoid the trivial solution, we note that if we try values of \( x \neq 0 \), we find that there are no valid \( y \) since it would lead \( y \) back to \( 0 \), which violates \( x \neq y \). Conclusively, aside from \( x = 0 \), there don’t appear to be other pairs (non-duplicate) of \( (x, y) \) that satisfy both equations simultaneously.