Solve by Substitution 4x+8y=20 , -4x+2y=-30

$4 x + 8 y = 20$ , $- 4 x + 2 y = - 30$

Step 1

Solve for $x$ in $4 x + 8 y = 20$ .

Subtract $8 y$ from both sides of the equation.

$4 x = 20 - 8 y$

$- 4 x + 2 y = - 30$

Divide each term in $4 x = 20 - 8 y$ by $4$ and simplify.

Divide each term in $4 x = 20 - 8 y$ by $4$ .

$\frac{4 x}{4} = \frac{20}{4} + \frac{- 8 y}{4}$

$- 4 x + 2 y = - 30$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{20}{4} + \frac{- 8 y}{4}$

$- 4 x + 2 y = - 30$

Divide $x$ by $1$ .

$x = \frac{20}{4} + \frac{- 8 y}{4}$

$- 4 x + 2 y = - 30$

$x = \frac{20}{4} + \frac{- 8 y}{4}$

$- 4 x + 2 y = - 30$

$x = \frac{20}{4} + \frac{- 8 y}{4}$

$- 4 x + 2 y = - 30$

Simplify the right side.

Simplify each term.

Divide $20$ by $4$ .

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

$- 4 x + 2 y = - 30$

Cancel the common factor of $- 8$ and $4$ .

Factor $4$ out of $- 8 y$ .

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

$- 4 x + 2 y = - 30$

Cancel the common factors.

Factor $4$ out of $4$ .

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

$- 4 x + 2 y = - 30$

Cancel the common factor.

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

$- 4 x + 2 y = - 30$

Rewrite the expression.

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

$- 4 x + 2 y = - 30$

Divide $- 2 y$ by $1$ .

$x = 5 - 2 y$

$- 4 x + 2 y = - 30$

$x = 5 - 2 y$

$- 4 x + 2 y = - 30$

$x = 5 - 2 y$

$- 4 x + 2 y = - 30$

$x = 5 - 2 y$

$- 4 x + 2 y = - 30$

$x = 5 - 2 y$

$- 4 x + 2 y = - 30$

$x = 5 - 2 y$

$- 4 x + 2 y = - 30$

$x = 5 - 2 y$

$- 4 x + 2 y = - 30$

Step 2

Replace all occurrences of $x$ with $5 - 2 y$ in each equation.

Replace all occurrences of $x$ in $- 4 x + 2 y = - 30$ with $5 - 2 y$ .

$- 4 (5 - 2 y) + 2 y = - 30$

$x = 5 - 2 y$

Simplify the left side.

Simplify $- 4 (5 - 2 y) + 2 y$ .

Simplify each term.

Apply the distributive property.

$- 4 \cdot 5 - 4 (- 2 y) + 2 y = - 30$

$x = 5 - 2 y$

Multiply $- 4$ by $5$ .

$- 20 - 4 (- 2 y) + 2 y = - 30$

$x = 5 - 2 y$

Multiply $- 2$ by $- 4$ .

$- 20 + 8 y + 2 y = - 30$

$x = 5 - 2 y$

$- 20 + 8 y + 2 y = - 30$

$x = 5 - 2 y$

Add $8 y$ and $2 y$ .

$- 20 + 10 y = - 30$

$x = 5 - 2 y$

$- 20 + 10 y = - 30$

$x = 5 - 2 y$

$- 20 + 10 y = - 30$

$x = 5 - 2 y$

$- 20 + 10 y = - 30$

$x = 5 - 2 y$

Step 3

Solve for $y$ in $- 20 + 10 y = - 30$ .

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

Add $20$ to both sides of the equation.

$10 y = - 30 + 20$

$x = 5 - 2 y$

Add $- 30$ and $20$ .

$10 y = - 10$

$x = 5 - 2 y$

$10 y = - 10$

$x = 5 - 2 y$

Divide each term in $10 y = - 10$ by $10$ and simplify.

Divide each term in $10 y = - 10$ by $10$ .

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

$x = 5 - 2 y$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

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

$x = 5 - 2 y$

Divide $y$ by $1$ .

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

$x = 5 - 2 y$

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

$x = 5 - 2 y$

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

$x = 5 - 2 y$

Simplify the right side.

Divide $- 10$ by $10$ .

$y = - 1$

$x = 5 - 2 y$

$y = - 1$

$x = 5 - 2 y$

$y = - 1$

$x = 5 - 2 y$

$y = - 1$

$x = 5 - 2 y$

Step 4

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

Replace all occurrences of $y$ in $x = 5 - 2 y$ with $- 1$ .

$x = 5 - 2 \cdot - 1$

$y = - 1$

Simplify the right side.

Simplify $5 - 2 \cdot - 1$ .

Multiply $- 2$ by $- 1$ .

$x = 5 + 2$

$y = - 1$

Add $5$ and $2$ .

$x = 7$

$y = - 1$

$x = 7$

$y = - 1$

$x = 7$

$y = - 1$

$x = 7$

$y = - 1$

Step 5

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

$(7, - 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(7, - 1)$

Equation Form:

$x = 7, y = - 1$

Step 7

Solve by Substitution 4x+8y=20 , -4x+2y=-30

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