Solve by Substitution x+y=16 , y=-5x

$x + y = 16$ , $y = - 5 x$

Step 1

Replace all occurrences of $y$ with $- 5 x$ in each equation.

Replace all occurrences of $y$ in $x + y = 16$ with $- 5 x$ .

$x - 5 x = 16$

$y = - 5 x$

Simplify the left side.

Simplify $x - 5 x$ .

Remove parentheses.

$x - 5 x = 16$

$y = - 5 x$

Subtract $5 x$ from $x$ .

$- 4 x = 16$

$y = - 5 x$

$- 4 x = 16$

$y = - 5 x$

$- 4 x = 16$

$y = - 5 x$

$- 4 x = 16$

$y = - 5 x$

Step 2

Divide each term in $- 4 x = 16$ by $- 4$ and simplify.

Divide each term in $- 4 x = 16$ by $- 4$ .

$\frac{- 4 x}{- 4} = \frac{16}{- 4}$

$y = - 5 x$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 x}{- 4} = \frac{16}{- 4}$

$y = - 5 x$

Divide $x$ by $1$ .

$x = \frac{16}{- 4}$

$y = - 5 x$

$x = \frac{16}{- 4}$

$y = - 5 x$

$x = \frac{16}{- 4}$

$y = - 5 x$

Simplify the right side.

Divide $16$ by $- 4$ .

$x = - 4$

$y = - 5 x$

$x = - 4$

$y = - 5 x$

$x = - 4$

$y = - 5 x$

Step 3

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

Replace all occurrences of $x$ in $y = - 5 x$ with $- 4$ .

$y = - 5 \cdot - 4$

$x = - 4$

Simplify the right side.

Multiply $- 5$ by $- 4$ .

$y = 20$

$x = - 4$

$y = 20$

$x = - 4$

$y = 20$

$x = - 4$

Step 4

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

$(- 4, 20)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(- 4, 20)$

Equation Form:

$x = - 4, y = 20$

Step 6

Solve by Substitution x+y=16 , y=-5x

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