Solve by Substitution 2x-y=-3 , x+y=0

$2 x - y = - 3$ , $x + y = 0$

Step 1

Subtract $y$ from both sides of the equation.

$x = - y$

$2 x - y = - 3$

Step 2

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

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

$2 (- y) - y = - 3$

$x = - y$

Simplify the left side.

Simplify $2 (- y) - y$ .

Multiply $- 1$ by $2$ .

$- 2 y - y = - 3$

$x = - y$

Subtract $y$ from $- 2 y$ .

$- 3 y = - 3$

$x = - y$

$- 3 y = - 3$

$x = - y$

$- 3 y = - 3$

$x = - y$

$- 3 y = - 3$

$x = - y$

Step 3

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

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

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

$x = - y$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

$x = - y$

Divide $y$ by $1$ .

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

$x = - y$

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

$x = - y$

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

$x = - y$

Simplify the right side.

Divide $- 3$ by $- 3$ .

$y = 1$

$x = - y$

$y = 1$

$x = - y$

$y = 1$

$x = - y$

Step 4

Replace all occurrences of $y$ with $1$ in each equation.

Replace all occurrences of $y$ in $x = - y$ with $1$ .

$x = - (1)$

$y = 1$

Simplify the right side.

Multiply $- 1$ by $1$ .

$x = - 1$

$y = 1$

$x = - 1$

$y = 1$

$x = - 1$

$y = 1$

Step 5

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

$(- 1, 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 1, 1)$

Equation Form:

$x = - 1, y = 1$

Step 7

Solve by Substitution 2x-y=-3 , x+y=0

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