Solve by Graphing x+y=-1 , x-3y=15

$x + y = - 1$ , $x - 3 y = 15$

Step 1

Subtract $x$ from both sides of the equation.

$y = - 1 - x$

$x - 3 y = 15$

Step 2

Subtract $x$ from both sides of the equation.

$- 3 y = 15 - x$

$y = - 1 - x$

Step 3

Divide each term in $- 3 y = 15 - x$ by $- 3$ and simplify.

Divide each term in $- 3 y = 15 - x$ by $- 3$ .

$\frac{- 3 y}{- 3} = \frac{15}{- 3} + \frac{- x}{- 3}$

$y = - 1 - x$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 y}{- 3} = \frac{15}{- 3} + \frac{- x}{- 3}$

$y = - 1 - x$

Divide $y$ by $1$ .

$y = \frac{15}{- 3} + \frac{- x}{- 3}$

$y = - 1 - x$

$y = \frac{15}{- 3} + \frac{- x}{- 3}$

$y = - 1 - x$

$y = \frac{15}{- 3} + \frac{- x}{- 3}$

$y = - 1 - x$

Simplify the right side.

Simplify each term.

Divide $15$ by $- 3$ .

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

$y = - 1 - x$

Dividing two negative values results in a positive value.

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

$y = - 1 - x$

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

$y = - 1 - x$

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

$y = - 1 - x$

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

$y = - 1 - x$

Step 4

Create a graph to locate the intersection of the equations. The intersection of the system of equations is the solution.

$(3, - 4)$

Step 5