Solve the simultaneous equations: $16^x \times 2^{y+2} = 8$ and $2^x \div 32 = (\frac{1}{2})^y$.

1. Problem Description

Solve the simultaneous equations:

$16^x \times 2^{y+2} = 8$ and $2^x \div 32 = (\frac{1}{2})^y$.

2. Solution Steps

First, let's rewrite the equations using powers of

2.

Equation 1: $16^x \times 2^{y+2} = 8$

$(2^4)^x \times 2^{y+2} = 2^3$

$2^{4x} \times 2^{y+2} = 2^3$

$2^{4x + y + 2} = 2^3$

Equating the exponents gives us:

$4x + y + 2 = 3$

$4x + y = 1$

Equation 2: $2^x \div 32 = (\frac{1}{2})^y$

$2^x \div 2^5 = (2^{-1})^y$

$2^{x-5} = 2^{-y}$

Equating the exponents gives us:

$x - 5 = -y$

$x + y = 5$

Now we have a system of linear equations:

$4x + y = 1$

$x + y = 5$

Subtract the second equation from the first:

$(4x + y) - (x + y) = 1 - 5$

$3x = -4$

$x = -\frac{4}{3}$

Substitute $x = -\frac{4}{3}$ into the equation $x + y = 5$:

$-\frac{4}{3} + y = 5$

$y = 5 + \frac{4}{3}$

$y = \frac{15}{3} + \frac{4}{3}$

$y = \frac{19}{3}$

3. Final Answer

$x = -\frac{4}{3}, y = \frac{19}{3}$