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

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

Step 1

Add $3 y$ to both sides of the equation.

$x = - 3 + 3 y$

$- x + 3 y = 3$

Step 2

Replace all occurrences of $x$ with $- 3 + 3 y$ in each equation.

Replace all occurrences of $x$ in $- x + 3 y = 3$ with $- 3 + 3 y$ .

$- (- 3 + 3 y) + 3 y = 3$

$x = - 3 + 3 y$

Simplify the left side.

Simplify $- (- 3 + 3 y) + 3 y$ .

Simplify each term.

Apply the distributive property.

$3 - (3 y) + 3 y = 3$

$x = - 3 + 3 y$

Multiply $- 1$ by $- 3$ .

$3 - (3 y) + 3 y = 3$

$x = - 3 + 3 y$

Multiply $3$ by $- 1$ .

$3 - 3 y + 3 y = 3$

$x = - 3 + 3 y$

$3 - 3 y + 3 y = 3$

$x = - 3 + 3 y$

Combine the opposite terms in $3 - 3 y + 3 y$ .

Add $- 3 y$ and $3 y$ .

$3 + 0 = 3$

$x = - 3 + 3 y$

Add $3$ and $0$ .

$3 = 3$

$x = - 3 + 3 y$

$3 = 3$

$x = - 3 + 3 y$

$3 = 3$

$x = - 3 + 3 y$

$3 = 3$

$x = - 3 + 3 y$

$3 = 3$

$x = - 3 + 3 y$

Step 3

Remove any equations from the system that are always true.

$x = - 3 + 3 y$

Step 4