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

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

Subtract $x$ from both sides of the equation.

$- 3 y = - 3 - x$

$- x + 3 y = 3$

Divide each term by $- 3$ and simplify.

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

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

$- x + 3 y = 3$

Cancel the common factor of $- 3$ .

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Cancel the common factor.

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

$- x + 3 y = 3$

Divide $y$ by $1$ .

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

$- x + 3 y = 3$

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

$- x + 3 y = 3$

Simplify each term.

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Divide $- 3$ by $- 3$ .

$y = 1 + \frac{- x}{- 3}$

$- x + 3 y = 3$

Dividing two negative values results in a positive value.

$y = 1 + \frac{x}{3}$

$- x + 3 y = 3$

$y = 1 + \frac{x}{3}$

$- x + 3 y = 3$

$y = 1 + \frac{x}{3}$

$- x + 3 y = 3$

Add $x$ to both sides of the equation.

$y = 1 + \frac{x}{3}$

$3 y = 3 + x$

Divide each term by $3$ and simplify.

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Divide each term in $3 y = 3 + x$ by $3$ .

$y = 1 + \frac{x}{3}$

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

Cancel the common factor of $3$ .

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Cancel the common factor.

$y = 1 + \frac{x}{3}$

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

Divide $y$ by $1$ .

$y = 1 + \frac{x}{3}$

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

$y = 1 + \frac{x}{3}$

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

Divide $3$ by $3$ .

$y = 1 + \frac{x}{3}$

The solution is the set of ordered pairs that make $y = 1 + \frac{x}{3}$ true.

$(x, 1 + \frac{x}{3})$

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

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