Solve by Substitution 4x-y=20 , -2x-2y=10

$4 x - y = 20$ , $- 2 x - 2 y = 10$

Step 1

Solve for $x$ in $4 x - y = 20$ .

Add $y$ to both sides of the equation.

$4 x = 20 + y$

$- 2 x - 2 y = 10$

Divide each term in $4 x = 20 + y$ by $4$ and simplify.

Divide each term in $4 x = 20 + y$ by $4$ .

$\frac{4 x}{4} = \frac{20}{4} + \frac{y}{4}$

$- 2 x - 2 y = 10$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{20}{4} + \frac{y}{4}$

$- 2 x - 2 y = 10$

Divide $x$ by $1$ .

$x = \frac{20}{4} + \frac{y}{4}$

$- 2 x - 2 y = 10$

$x = \frac{20}{4} + \frac{y}{4}$

$- 2 x - 2 y = 10$

$x = \frac{20}{4} + \frac{y}{4}$

$- 2 x - 2 y = 10$

Simplify the right side.

Divide $20$ by $4$ .

$x = 5 + \frac{y}{4}$

$- 2 x - 2 y = 10$

$x = 5 + \frac{y}{4}$

$- 2 x - 2 y = 10$

$x = 5 + \frac{y}{4}$

$- 2 x - 2 y = 10$

$x = 5 + \frac{y}{4}$

$- 2 x - 2 y = 10$

Step 2

Replace all occurrences of $x$ with $5 + \frac{y}{4}$ in each equation.

Replace all occurrences of $x$ in $- 2 x - 2 y = 10$ with $5 + \frac{y}{4}$ .

$- 2 (5 + \frac{y}{4}) - 2 y = 10$

$x = 5 + \frac{y}{4}$

Simplify the left side.

Simplify $- 2 (5 + \frac{y}{4}) - 2 y$ .

Simplify each term.

Apply the distributive property.

$- 2 \cdot 5 - 2 \frac{y}{4} - 2 y = 10$

$x = 5 + \frac{y}{4}$

Multiply $- 2$ by $5$ .

$- 10 - 2 \frac{y}{4} - 2 y = 10$

$x = 5 + \frac{y}{4}$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$- 10 + 2 (- 1) (\frac{y}{4}) - 2 y = 10$

$x = 5 + \frac{y}{4}$

Factor $2$ out of $4$ .

$- 10 + 2 \cdot (- 1 \frac{y}{2 \cdot 2}) - 2 y = 10$

$x = 5 + \frac{y}{4}$

Cancel the common factor.

$- 10 + 2 \cdot (- 1 \frac{y}{2 \cdot 2}) - 2 y = 10$

$x = 5 + \frac{y}{4}$

Rewrite the expression.

$- 10 - 1 \frac{y}{2} - 2 y = 10$

$x = 5 + \frac{y}{4}$

$- 10 - 1 \frac{y}{2} - 2 y = 10$

$x = 5 + \frac{y}{4}$

Rewrite $- 1 \frac{y}{2}$ as $- \frac{y}{2}$ .

$- 10 - \frac{y}{2} - 2 y = 10$

$x = 5 + \frac{y}{4}$

$- 10 - \frac{y}{2} - 2 y = 10$

$x = 5 + \frac{y}{4}$

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

$- 10 - \frac{y}{2} - 2 y \cdot \frac{2}{2} = 10$

$x = 5 + \frac{y}{4}$

Simplify terms.

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

$- 10 - \frac{y}{2} + \frac{- 2 y \cdot 2}{2} = 10$

$x = 5 + \frac{y}{4}$

Combine the numerators over the common denominator.

$- 10 + \frac{- y - 2 y \cdot 2}{2} = 10$

$x = 5 + \frac{y}{4}$

$- 10 + \frac{- y - 2 y \cdot 2}{2} = 10$

$x = 5 + \frac{y}{4}$

Simplify each term.

Simplify the numerator.

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

Factor $y$ out of $- y$ .

$- 10 + \frac{y \cdot - 1 - 2 y \cdot 2}{2} = 10$

$x = 5 + \frac{y}{4}$

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

$- 10 + \frac{y \cdot - 1 + y (- 2 \cdot 2)}{2} = 10$

$x = 5 + \frac{y}{4}$

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

$- 10 + \frac{y (- 1 - 2 \cdot 2)}{2} = 10$

$x = 5 + \frac{y}{4}$

$- 10 + \frac{y (- 1 - 2 \cdot 2)}{2} = 10$

$x = 5 + \frac{y}{4}$

Multiply $- 2$ by $2$ .

$- 10 + \frac{y (- 1 - 4)}{2} = 10$

$x = 5 + \frac{y}{4}$

Subtract $4$ from $- 1$ .

$- 10 + \frac{y \cdot - 5}{2} = 10$

$x = 5 + \frac{y}{4}$

$- 10 + \frac{y \cdot - 5}{2} = 10$

$x = 5 + \frac{y}{4}$

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

$- 10 + \frac{- 5 \cdot y}{2} = 10$

$x = 5 + \frac{y}{4}$

Move the negative in front of the fraction.

$- 10 - \frac{5 y}{2} = 10$

$x = 5 + \frac{y}{4}$

$- 10 - \frac{5 y}{2} = 10$

$x = 5 + \frac{y}{4}$

$- 10 - \frac{5 y}{2} = 10$

$x = 5 + \frac{y}{4}$

$- 10 - \frac{5 y}{2} = 10$

$x = 5 + \frac{y}{4}$

$- 10 - \frac{5 y}{2} = 10$

$x = 5 + \frac{y}{4}$

Step 3

Solve for $y$ in $- 10 - \frac{5 y}{2} = 10$ .

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

Add $10$ to both sides of the equation.

$- \frac{5 y}{2} = 10 + 10$

$x = 5 + \frac{y}{4}$

Add $10$ and $10$ .

$- \frac{5 y}{2} = 20$

$x = 5 + \frac{y}{4}$

$- \frac{5 y}{2} = 20$

$x = 5 + \frac{y}{4}$

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

$- \frac{2}{5} \cdot (- \frac{5 y}{2}) = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

Simplify both sides of the equation.

Simplify the left side.

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

Cancel the common factor of $2$ .

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

$\frac{- 2}{5} \cdot (- \frac{5 y}{2}) = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

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

$\frac{- 2}{5} \cdot \frac{- 5 y}{2} = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

Factor $2$ out of $- 2$ .

$\frac{2 (- 1)}{5} \cdot \frac{- 5 y}{2} = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

Cancel the common factor.

$\frac{2 \cdot - 1}{5} \cdot \frac{- 5 y}{2} = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

Rewrite the expression.

$\frac{- 1}{5} \cdot (- 5 y) = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

$\frac{- 1}{5} \cdot (- 5 y) = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

Cancel the common factor of $5$ .

Factor $5$ out of $- 5 y$ .

$\frac{- 1}{5} \cdot (5 (- y)) = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

Cancel the common factor.

$\frac{- 1}{5} \cdot (5 (- y)) = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

Rewrite the expression.

$y = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

$y = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 y = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

Multiply $y$ by $1$ .

$y = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

$y = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

$y = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

$y = - \frac{2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

Simplify the right side.

Simplify $- \frac{2}{5} \cdot 20$ .

Cancel the common factor of $5$ .

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

$y = \frac{- 2}{5} \cdot 20$

$x = 5 + \frac{y}{4}$

Factor $5$ out of $20$ .

$y = \frac{- 2}{5} \cdot (5 (4))$

$x = 5 + \frac{y}{4}$

Cancel the common factor.

$y = \frac{- 2}{5} \cdot (5 \cdot 4)$

$x = 5 + \frac{y}{4}$

Rewrite the expression.

$y = - 2 \cdot 4$

$x = 5 + \frac{y}{4}$

$y = - 2 \cdot 4$

$x = 5 + \frac{y}{4}$

Multiply $- 2$ by $4$ .

$y = - 8$

$x = 5 + \frac{y}{4}$

$y = - 8$

$x = 5 + \frac{y}{4}$

$y = - 8$

$x = 5 + \frac{y}{4}$

$y = - 8$

$x = 5 + \frac{y}{4}$

$y = - 8$

$x = 5 + \frac{y}{4}$

Step 4

Replace all occurrences of $y$ with $- 8$ in each equation.

Replace all occurrences of $y$ in $x = 5 + \frac{y}{4}$ with $- 8$ .

$x = 5 + \frac{- 8}{4}$

$y = - 8$

Simplify the right side.

Simplify $5 + \frac{- 8}{4}$ .

Divide $- 8$ by $4$ .

$x = 5 - 2$

$y = - 8$

Subtract $2$ from $5$ .

$x = 3$

$y = - 8$

$x = 3$

$y = - 8$

$x = 3$

$y = - 8$

$x = 3$

$y = - 8$

Step 5

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

$(3, - 8)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, - 8)$

Equation Form:

$x = 3, y = - 8$

Step 7

Solve by Substitution 4x-y=20 , -2x-2y=10

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