Solve by Substitution x-3y=-15 and x=-3

$x - 3 y = - 15$ and $x = - 3$

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

Replace all occurrences of $x$ in $x - 3 y = - 15$ with $- 3$ .

$(- 3) - 3 y = - 15$

$x = - 3$

Simplify the left side.

Remove parentheses.

$- 3 - 3 y = - 15$

$x = - 3$

$- 3 - 3 y = - 15$

$x = - 3$

$- 3 - 3 y = - 15$

$x = - 3$

Solve for $y$ in $- 3 - 3 y = - 15$ .

Move all terms not containing $y$ to the right side of the equation.

Add $3$ to both sides of the equation.

$- 3 y = - 15 + 3$

$x = - 3$

Add $- 15$ and $3$ .

$- 3 y = - 12$

$x = - 3$

$- 3 y = - 12$

$x = - 3$

Divide each term in $- 3 y = - 12$ by $- 3$ and simplify.

Divide each term in $- 3 y = - 12$ by $- 3$ .

$\frac{- 3 y}{- 3} = \frac{- 12}{- 3}$

$x = - 3$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 y}{- 3} = \frac{- 12}{- 3}$

$x = - 3$

Divide $y$ by $1$ .

$y = \frac{- 12}{- 3}$

$x = - 3$

$y = \frac{- 12}{- 3}$

$x = - 3$

$y = \frac{- 12}{- 3}$

$x = - 3$

Simplify the right side.

Divide $- 12$ by $- 3$ .

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

Solve the system of equations $y = 4 x = - 3$ .

$y = 4 x = - 3$

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

$(- 3, 4)$

The result can be shown in multiple forms.

Point Form:

$(- 3, 4)$

Equation Form:

$x = - 3, y = 4$

Solve by Substitution x-3y=-15 and x=-3

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