Solve the System of Equations -3x+3y=12 and 5x+5y=10

$- 3 x + 3 y = 12$ and $5 x + 5 y = 10$

Step 1

Solve for $x$ in $- 3 x + 3 y = 12$ .

Subtract $3 y$ from both sides of the equation.

$- 3 x = 12 - 3 y$

$5 x + 5 y = 10$

Divide each term in $- 3 x = 12 - 3 y$ by $- 3$ and simplify.

Divide each term in $- 3 x = 12 - 3 y$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{12}{- 3} + \frac{- 3 y}{- 3}$

$5 x + 5 y = 10$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{12}{- 3} + \frac{- 3 y}{- 3}$

$5 x + 5 y = 10$

Divide $x$ by $1$ .

$x = \frac{12}{- 3} + \frac{- 3 y}{- 3}$

$5 x + 5 y = 10$

$x = \frac{12}{- 3} + \frac{- 3 y}{- 3}$

$5 x + 5 y = 10$

$x = \frac{12}{- 3} + \frac{- 3 y}{- 3}$

$5 x + 5 y = 10$

Simplify the right side.

Simplify each term.

Divide $12$ by $- 3$ .

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

$5 x + 5 y = 10$

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

$5 x + 5 y = 10$

Divide $y$ by $1$ .

$x = - 4 + y$

$5 x + 5 y = 10$

$x = - 4 + y$

$5 x + 5 y = 10$

$x = - 4 + y$

$5 x + 5 y = 10$

$x = - 4 + y$

$5 x + 5 y = 10$

$x = - 4 + y$

$5 x + 5 y = 10$

$x = - 4 + y$

$5 x + 5 y = 10$

Step 2

Replace all occurrences of $x$ with $- 4 + y$ in each equation.

Replace all occurrences of $x$ in $5 x + 5 y = 10$ with $- 4 + y$ .

$5 (- 4 + y) + 5 y = 10$

$x = - 4 + y$

Simplify the left side.

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

Simplify each term.

Apply the distributive property.

$5 \cdot - 4 + 5 y + 5 y = 10$

$x = - 4 + y$

Multiply $5$ by $- 4$ .

$- 20 + 5 y + 5 y = 10$

$x = - 4 + y$

$- 20 + 5 y + 5 y = 10$

$x = - 4 + y$

Add $5 y$ and $5 y$ .

$- 20 + 10 y = 10$

$x = - 4 + y$

$- 20 + 10 y = 10$

$x = - 4 + y$

$- 20 + 10 y = 10$

$x = - 4 + y$

$- 20 + 10 y = 10$

$x = - 4 + y$

Step 3

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

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

Add $20$ to both sides of the equation.

$10 y = 10 + 20$

$x = - 4 + y$

Add $10$ and $20$ .

$10 y = 30$

$x = - 4 + y$

$10 y = 30$

$x = - 4 + y$

Divide each term in $10 y = 30$ by $10$ and simplify.

Divide each term in $10 y = 30$ by $10$ .

$\frac{10 y}{10} = \frac{30}{10}$

$x = - 4 + y$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 y}{10} = \frac{30}{10}$

$x = - 4 + y$

Divide $y$ by $1$ .

$y = \frac{30}{10}$

$x = - 4 + y$

$y = \frac{30}{10}$

$x = - 4 + y$

$y = \frac{30}{10}$

$x = - 4 + y$

Simplify the right side.

Divide $30$ by $10$ .

$y = 3$

$x = - 4 + y$

$y = 3$

$x = - 4 + y$

$y = 3$

$x = - 4 + y$

$y = 3$

$x = - 4 + y$

Step 4

Replace all occurrences of $y$ with $3$ in each equation.

Replace all occurrences of $y$ in $x = - 4 + y$ with $3$ .

$x = - 4 + 3$

$y = 3$

Simplify $x = - 4 + 3$ .

Simplify the left side.

Remove parentheses.

$x = - 4 + 3$

$y = 3$

$x = - 4 + 3$

$y = 3$

Simplify the right side.

Add $- 4$ and $3$ .

$x = - 1$

$y = 3$

$x = - 1$

$y = 3$

$x = - 1$

$y = 3$

$x = - 1$

$y = 3$

Step 5

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

$(- 1, 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 1, 3)$

Equation Form:

$x = - 1, y = 3$

Step 7

Solve the System of Equations -3x+3y=12 and 5x+5y=10

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