Solve the System of Equations 2^(x+y)=16 2^(2x+y)=1/8

$2^{x + y} = 16$ $2^{2 x + y} = \frac{1}{8}$

Solve for $x$ in $2^{x + y} = 16$ .

Create equivalent expressions in the equation that all have equal bases.

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

$2^{2 x + y} = \frac{1}{8}$

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$x + y = 4$

$2^{2 x + y} = \frac{1}{8}$

Subtract $y$ from both sides of the equation.

$x = 4 - y$

$2^{2 x + y} = \frac{1}{8}$

$x = 4 - y$

$2^{2 x + y} = \frac{1}{8}$

Replace all occurrences of $x$ with $4 - y$ in each equation.

Replace all occurrences of $x$ in $2^{2 x + y} = \frac{1}{8}$ with $4 - y$ .

$2^{2 (4 - y) + y} = \frac{1}{8}$

$x = 4 - y$

Simplify the left side.

Simplify $2^{2 (4 - y) + y}$ .

Simplify each term.

Apply the distributive property.

$2^{2 \cdot 4 + 2 (- y) + y} = \frac{1}{8}$

$x = 4 - y$

Multiply $2$ by $4$ .

$2^{8 + 2 (- y) + y} = \frac{1}{8}$

$x = 4 - y$

Multiply $- 1$ by $2$ .

$2^{8 - 2 y + y} = \frac{1}{8}$

$x = 4 - y$

$2^{8 - 2 y + y} = \frac{1}{8}$

$x = 4 - y$

Add $- 2 y$ and $y$ .

$2^{8 - y} = \frac{1}{8}$

$x = 4 - y$

$2^{8 - y} = \frac{1}{8}$

$x = 4 - y$

$2^{8 - y} = \frac{1}{8}$

$x = 4 - y$

$2^{8 - y} = \frac{1}{8}$

$x = 4 - y$

Solve for $y$ in $2^{8 - y} = \frac{1}{8}$ .

Move $8^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$2^{8 - y} = 8^{- 1}$

$x = 4 - y$

Create equivalent expressions in the equation that all have equal bases.

$2^{8 - y} = 2^{- 3}$

$x = 4 - y$

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$8 - y = - 3$

$x = 4 - y$

Solve for $y$ .

Move all terms not containing $y$ to the right side of the equation.

Subtract $8$ from both sides of the equation.

$- y = - 3 - 8$

$x = 4 - y$

Subtract $8$ from $- 3$ .

$- y = - 11$

$x = 4 - y$

$- y = - 11$

$x = 4 - y$

Divide each term in $- y = - 11$ by $- 1$ and simplify.

Divide each term in $- y = - 11$ by $- 1$ .

$\frac{- y}{- 1} = \frac{- 11}{- 1}$

$x = 4 - y$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{- 11}{- 1}$

$x = 4 - y$

Divide $y$ by $1$ .

$y = \frac{- 11}{- 1}$

$x = 4 - y$

$y = \frac{- 11}{- 1}$

$x = 4 - y$

Simplify the right side.

Divide $- 11$ by $- 1$ .

$y = 11$

$x = 4 - y$

$y = 11$

$x = 4 - y$

$y = 11$

$x = 4 - y$

$y = 11$

$x = 4 - y$

$y = 11$

$x = 4 - y$

Replace all occurrences of $y$ with $11$ in each equation.

Replace all occurrences of $y$ in $x = 4 - y$ with $11$ .

$x = 4 - (11)$

$y = 11$

Simplify the right side.

Simplify $4 - (11)$ .

Multiply $- 1$ by $11$ .

$x = 4 - 11$

$y = 11$

Subtract $11$ from $4$ .

$x = - 7$

$y = 11$

$x = - 7$

$y = 11$

$x = - 7$

$y = 11$

$x = - 7$

$y = 11$

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(- 7, 11)$

The result can be shown in multiple forms.

Point Form:

$(- 7, 11)$

Equation Form:

$x = - 7, y = 11$

Solve the System of Equations 2^(x+y)=16 2^(2x+y)=1/8

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