Solve by Substitution x+y=1 , 2x-y=-2

$x + y = 1$ , $2 x - y = - 2$

Step 1

Subtract $y$ from both sides of the equation.

$x = 1 - y$

$2 x - y = - 2$

Step 2

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

Replace all occurrences of $x$ in $2 x - y = - 2$ with $1 - y$ .

$2 (1 - y) - y = - 2$

$x = 1 - y$

Simplify the left side.

Simplify $2 (1 - y) - y$ .

Simplify each term.

Apply the distributive property.

$2 \cdot 1 + 2 (- y) - y = - 2$

$x = 1 - y$

Multiply $2$ by $1$ .

$2 + 2 (- y) - y = - 2$

$x = 1 - y$

Multiply $- 1$ by $2$ .

$2 - 2 y - y = - 2$

$x = 1 - y$

$2 - 2 y - y = - 2$

$x = 1 - y$

Subtract $y$ from $- 2 y$ .

$2 - 3 y = - 2$

$x = 1 - y$

$2 - 3 y = - 2$

$x = 1 - y$

$2 - 3 y = - 2$

$x = 1 - y$

$2 - 3 y = - 2$

$x = 1 - y$

Step 3

Solve for $y$ in $2 - 3 y = - 2$ .

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

Subtract $2$ from both sides of the equation.

$- 3 y = - 2 - 2$

$x = 1 - y$

Subtract $2$ from $- 2$ .

$- 3 y = - 4$

$x = 1 - y$

$- 3 y = - 4$

$x = 1 - y$

Divide each term in $- 3 y = - 4$ by $- 3$ and simplify.

Divide each term in $- 3 y = - 4$ by $- 3$ .

$\frac{- 3 y}{- 3} = \frac{- 4}{- 3}$

$x = 1 - y$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 y}{- 3} = \frac{- 4}{- 3}$

$x = 1 - y$

Divide $y$ by $1$ .

$y = \frac{- 4}{- 3}$

$x = 1 - y$

$y = \frac{- 4}{- 3}$

$x = 1 - y$

$y = \frac{- 4}{- 3}$

$x = 1 - y$

Simplify the right side.

Dividing two negative values results in a positive value.

$y = \frac{4}{3}$

$x = 1 - y$

$y = \frac{4}{3}$

$x = 1 - y$

$y = \frac{4}{3}$

$x = 1 - y$

$y = \frac{4}{3}$

$x = 1 - y$

Step 4

Replace all occurrences of $y$ with $\frac{4}{3}$ in each equation.

Replace all occurrences of $y$ in $x = 1 - y$ with $\frac{4}{3}$ .

$x = 1 - (\frac{4}{3})$

$y = \frac{4}{3}$

Simplify the right side.

Simplify $1 - (\frac{4}{3})$ .

Write $1$ as a fraction with a common denominator.

$x = \frac{3}{3} - \frac{4}{3}$

$y = \frac{4}{3}$

Combine the numerators over the common denominator.

$x = \frac{3 - 4}{3}$

$y = \frac{4}{3}$

Subtract $4$ from $3$ .

$x = \frac{- 1}{3}$

$y = \frac{4}{3}$

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

$y = \frac{4}{3}$

$x = - \frac{1}{3}$

$y = \frac{4}{3}$

$x = - \frac{1}{3}$

$y = \frac{4}{3}$

$x = - \frac{1}{3}$

$y = \frac{4}{3}$

Step 5

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

$(- \frac{1}{3}, \frac{4}{3})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- \frac{1}{3}, \frac{4}{3})$

Equation Form:

$x = - \frac{1}{3}, y = \frac{4}{3}$

Step 7

Solve by Substitution x+y=1 , 2x-y=-2

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