Solve by Substitution x+y=4 , y=-3x

$x + y = 4$ , $y = - 3 x$

Step 1

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

Replace all occurrences of $y$ in $x + y = 4$ with $- 3 x$ .

$x - 3 x = 4$

$y = - 3 x$

Simplify the left side.

Simplify $x - 3 x$ .

Remove parentheses.

$x - 3 x = 4$

$y = - 3 x$

Subtract $3 x$ from $x$ .

$- 2 x = 4$

$y = - 3 x$

$- 2 x = 4$

$y = - 3 x$

$- 2 x = 4$

$y = - 3 x$

$- 2 x = 4$

$y = - 3 x$

Step 2

Divide each term in $- 2 x = 4$ by $- 2$ and simplify.

Divide each term in $- 2 x = 4$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{4}{- 2}$

$y = - 3 x$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{4}{- 2}$

$y = - 3 x$

Divide $x$ by $1$ .

$x = \frac{4}{- 2}$

$y = - 3 x$

$x = \frac{4}{- 2}$

$y = - 3 x$

$x = \frac{4}{- 2}$

$y = - 3 x$

Simplify the right side.

Divide $4$ by $- 2$ .

$x = - 2$

$y = - 3 x$

$x = - 2$

$y = - 3 x$

$x = - 2$

$y = - 3 x$

Step 3

Replace all occurrences of $x$ with $- 2$ in each equation.

Replace all occurrences of $x$ in $y = - 3 x$ with $- 2$ .

$y = - 3 \cdot - 2$

$x = - 2$

Simplify the right side.

Multiply $- 3$ by $- 2$ .

$y = 6$

$x = - 2$

$y = 6$

$x = - 2$

$y = 6$

$x = - 2$

Step 4

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

$(- 2, 6)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(- 2, 6)$

Equation Form:

$x = - 2, y = 6$

Step 6

Solve by Substitution x+y=4 , y=-3x

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