Solve the System of Equations y+2x=-1 y=1/2x+4

$y + 2 x = - 1$ $y = \frac{1}{2} x + 4$

Step 1

Subtract $2 x$ from both sides of the equation.

$y = - 1 - 2 x$

$y = \frac{1}{2} x + 4$

Step 2

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

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

$- 1 - 2 x = - 1 - 2 x$

$y = \frac{1}{2} x + 4$

Simplify the left side.

Remove parentheses.

$- 1 - 2 x = - 1 - 2 x$

$y = \frac{1}{2} x + 4$

$- 1 - 2 x = - 1 - 2 x$

$y = \frac{1}{2} x + 4$

Replace all occurrences of $y$ in $y = \frac{1}{2} x + 4$ with $- 1 - 2 x$ .

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

$- 1 - 2 x = - 1 - 2 x$

Simplify $- 1 - 2 x = \frac{1}{2} x + 4$ .

Simplify the left side.

Remove parentheses.

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

$- 1 - 2 x = - 1 - 2 x$

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

$- 1 - 2 x = - 1 - 2 x$

Simplify the right side.

Combine $\frac{1}{2}$ and $x$ .

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

$- 1 - 2 x = - 1 - 2 x$

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

$- 1 - 2 x = - 1 - 2 x$

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

$- 1 - 2 x = - 1 - 2 x$

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

$- 1 - 2 x = - 1 - 2 x$

Step 3

Remove any equations from the system that are always true.

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

Step 4

Solve for $x$ in $- 1 - 2 x = \frac{x}{2} + 4$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $\frac{x}{2}$ from both sides of the equation.

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

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

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

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

$- 1 + \frac{- 2 x \cdot 2}{2} - \frac{x}{2} = 4$

Combine the numerators over the common denominator.

$- 1 + \frac{- 2 x \cdot 2 - x}{2} = 4$

Simplify each term.

Simplify the numerator.

Factor $x$ out of $- 2 x \cdot 2 - x$ .

Factor $x$ out of $- 2 x \cdot 2$ .

$- 1 + \frac{x (- 2 \cdot 2) - x}{2} = 4$

Factor $x$ out of $- x$ .

$- 1 + \frac{x (- 2 \cdot 2) + x \cdot - 1}{2} = 4$

Factor $x$ out of $x (- 2 \cdot 2) + x \cdot - 1$ .

$- 1 + \frac{x (- 2 \cdot 2 - 1)}{2} = 4$

Multiply $- 2$ by $2$ .

$- 1 + \frac{x (- 4 - 1)}{2} = 4$

Subtract $1$ from $- 4$ .

$- 1 + \frac{x \cdot - 5}{2} = 4$

Move $- 5$ to the left of $x$ .

$- 1 + \frac{- 5 \cdot x}{2} = 4$

Move the negative in front of the fraction.

$- 1 - \frac{5 x}{2} = 4$

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

Add $1$ to both sides of the equation.

$- \frac{5 x}{2} = 4 + 1$

Add $4$ and $1$ .

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

Multiply both sides of the equation by $- \frac{2}{5}$ .

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

Simplify both sides of the equation.

Simplify the left side.

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

Cancel the common factor of $2$ .

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

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

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

$\frac{- 2}{5} \cdot \frac{- 5 x}{2} = - \frac{2}{5} \cdot 5$

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $5$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

$- - x = - \frac{2}{5} \cdot 5$

Multiply.

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

$x = - \frac{2}{5} \cdot 5$

Simplify the right side.

Cancel the common factor of $5$ .

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

$x = \frac{- 2}{5} \cdot 5$

Cancel the common factor.

$x = \frac{- 2}{5} \cdot 5$

Rewrite the expression.

$x = - 2$

Step 5

Solve the System of Equations y+2x=-1 y=1/2x+4

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