Solve by Substitution 2y=x+1 , -2x-y=7

$2 y = x + 1$ , $- 2 x - y = 7$

Step 1

Divide each term in $2 y = x + 1$ by $2$ and simplify.

Divide each term in $2 y = x + 1$ by $2$ .

$\frac{2 y}{2} = \frac{x}{2} + \frac{1}{2}$

$- 2 x - y = 7$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{x}{2} + \frac{1}{2}$

$- 2 x - y = 7$

Divide $y$ by $1$ .

$y = \frac{x}{2} + \frac{1}{2}$

$- 2 x - y = 7$

$y = \frac{x}{2} + \frac{1}{2}$

$- 2 x - y = 7$

$y = \frac{x}{2} + \frac{1}{2}$

$- 2 x - y = 7$

$y = \frac{x}{2} + \frac{1}{2}$

$- 2 x - y = 7$

Step 2

Replace all occurrences of $y$ with $\frac{x}{2} + \frac{1}{2}$ in each equation.

Replace all occurrences of $y$ in $- 2 x - y = 7$ with $\frac{x}{2} + \frac{1}{2}$ .

$- 2 x - (\frac{x}{2} + \frac{1}{2}) = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Simplify the left side.

Simplify $- 2 x - (\frac{x}{2} + \frac{1}{2})$ .

Apply the distributive property.

$- 2 x - \frac{x}{2} - \frac{1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

To write $- 2 x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- 2 x \cdot \frac{2}{2} - \frac{x}{2} - \frac{1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Combine $- 2 x$ and $\frac{2}{2}$ .

$\frac{- 2 x \cdot 2}{2} - \frac{x}{2} - \frac{1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Combine the numerators over the common denominator.

$\frac{- 2 x \cdot 2 - x}{2} - \frac{1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Combine the numerators over the common denominator.

$\frac{- 2 x \cdot 2 - x - 1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Multiply $2$ by $- 2$ .

$\frac{- 4 x - x - 1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Subtract $x$ from $- 4 x$ .

$\frac{- 5 x - 1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Factor $- 1$ out of $- 5 x$ .

$\frac{- (5 x) - 1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Rewrite $- 1$ as $- 1 (1)$ .

$\frac{- (5 x) - 1 \cdot 1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Factor $- 1$ out of $- (5 x) - 1 (1)$ .

$\frac{- (5 x + 1)}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Simplify the expression.

Rewrite $- (5 x + 1)$ as $- 1 (5 x + 1)$ .

$\frac{- 1 (5 x + 1)}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Move the negative in front of the fraction.

$- \frac{5 x + 1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

$- \frac{5 x + 1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

$- \frac{5 x + 1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

$- \frac{5 x + 1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

$- \frac{5 x + 1}{2} = 7$

$y = \frac{x}{2} + \frac{1}{2}$

Step 3

Solve for $x$ in $- \frac{5 x + 1}{2} = 7$ .

Multiply both sides of the equation by $- 2$ .

$- 2 (- \frac{5 x + 1}{2}) = - 2 \cdot 7$

$y = \frac{x}{2} + \frac{1}{2}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- 2 (- \frac{5 x + 1}{2})$ .

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{5 x + 1}{2}$ into the numerator.

$- 2 \frac{- (5 x + 1)}{2} = - 2 \cdot 7$

$y = \frac{x}{2} + \frac{1}{2}$

Factor $2$ out of $- 2$ .

$2 (- 1) (\frac{- (5 x + 1)}{2}) = - 2 \cdot 7$

$y = \frac{x}{2} + \frac{1}{2}$

Cancel the common factor.

$2 \cdot (- 1 \frac{- (5 x + 1)}{2}) = - 2 \cdot 7$

$y = \frac{x}{2} + \frac{1}{2}$

Rewrite the expression.

$5 x + 1 = - 2 \cdot 7$

$y = \frac{x}{2} + \frac{1}{2}$

$5 x + 1 = - 2 \cdot 7$

$y = \frac{x}{2} + \frac{1}{2}$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 (5 x + 1) = - 2 \cdot 7$

$y = \frac{x}{2} + \frac{1}{2}$

Multiply $5 x + 1$ by $1$ .

$5 x + 1 = - 2 \cdot 7$

$y = \frac{x}{2} + \frac{1}{2}$

$5 x + 1 = - 2 \cdot 7$

$y = \frac{x}{2} + \frac{1}{2}$

$5 x + 1 = - 2 \cdot 7$

$y = \frac{x}{2} + \frac{1}{2}$

$5 x + 1 = - 2 \cdot 7$

$y = \frac{x}{2} + \frac{1}{2}$

Simplify the right side.

Multiply $- 2$ by $7$ .

$5 x + 1 = - 14$

$y = \frac{x}{2} + \frac{1}{2}$

$5 x + 1 = - 14$

$y = \frac{x}{2} + \frac{1}{2}$

$5 x + 1 = - 14$

$y = \frac{x}{2} + \frac{1}{2}$

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

Subtract $1$ from both sides of the equation.

$5 x = - 14 - 1$

$y = \frac{x}{2} + \frac{1}{2}$

Subtract $1$ from $- 14$ .

$5 x = - 15$

$y = \frac{x}{2} + \frac{1}{2}$

$5 x = - 15$

$y = \frac{x}{2} + \frac{1}{2}$

Divide each term in $5 x = - 15$ by $5$ and simplify.

Divide each term in $5 x = - 15$ by $5$ .

$\frac{5 x}{5} = \frac{- 15}{5}$

$y = \frac{x}{2} + \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 15}{5}$

$y = \frac{x}{2} + \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 15}{5}$

$y = \frac{x}{2} + \frac{1}{2}$

$x = \frac{- 15}{5}$

$y = \frac{x}{2} + \frac{1}{2}$

$x = \frac{- 15}{5}$

$y = \frac{x}{2} + \frac{1}{2}$

Simplify the right side.

Divide $- 15$ by $5$ .

$x = - 3$

$y = \frac{x}{2} + \frac{1}{2}$

$x = - 3$

$y = \frac{x}{2} + \frac{1}{2}$

$x = - 3$

$y = \frac{x}{2} + \frac{1}{2}$

$x = - 3$

$y = \frac{x}{2} + \frac{1}{2}$

Step 4

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

Replace all occurrences of $x$ in $y = \frac{x}{2} + \frac{1}{2}$ with $- 3$ .

$y = \frac{- 3}{2} + \frac{1}{2}$

$x = - 3$

Simplify the right side.

Simplify $\frac{- 3}{2} + \frac{1}{2}$ .

Combine the numerators over the common denominator.

$y = \frac{- 3 + 1}{2}$

$x = - 3$

Simplify the expression.

Add $- 3$ and $1$ .

$y = \frac{- 2}{2}$

$x = - 3$

Divide $- 2$ by $2$ .

$y = - 1$

$x = - 3$

$y = - 1$

$x = - 3$

$y = - 1$

$x = - 3$

$y = - 1$

$x = - 3$

$y = - 1$

$x = - 3$

Step 5

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

$(- 3, - 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 3, - 1)$

Equation Form:

$x = - 3, y = - 1$

Step 7

Solve by Substitution 2y=x+1 , -2x-y=7

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