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

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

Step 1

Subtract $3 y$ from 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.

$- 1 \cdot 3 - (- 3 y) - 3 y = - 3$

$x = 3 - 3 y$

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 $- 3 + 3 y - 3 y$ .

Subtract $3 y$ from $3 y$ .

$- 3 + 0 = - 3$

$x = 3 - 3 y$

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