Solve the System of Equations -3x+9y=42 and 2x-2y=4

$- 3 x + 9 y = 42$ and $2 x - 2 y = 4$

Step 1

Solve for $x$ in $- 3 x + 9 y = 42$ .

Subtract $9 y$ from both sides of the equation.

$- 3 x = 42 - 9 y$

$2 x - 2 y = 4$

Divide each term in $- 3 x = 42 - 9 y$ by $- 3$ and simplify.

Divide each term in $- 3 x = 42 - 9 y$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{42}{- 3} + \frac{- 9 y}{- 3}$

$2 x - 2 y = 4$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{42}{- 3} + \frac{- 9 y}{- 3}$

$2 x - 2 y = 4$

Divide $x$ by $1$ .

$x = \frac{42}{- 3} + \frac{- 9 y}{- 3}$

$2 x - 2 y = 4$

$x = \frac{42}{- 3} + \frac{- 9 y}{- 3}$

$2 x - 2 y = 4$

$x = \frac{42}{- 3} + \frac{- 9 y}{- 3}$

$2 x - 2 y = 4$

Simplify the right side.

Simplify each term.

Divide $42$ by $- 3$ .

$x = - 14 + \frac{- 9 y}{- 3}$

$2 x - 2 y = 4$

Cancel the common factor of $- 9$ and $- 3$ .

Factor $- 3$ out of $- 9 y$ .

$x = - 14 + \frac{- 3 (3 y)}{- 3}$

$2 x - 2 y = 4$

Cancel the common factors.

Factor $- 3$ out of $- 3$ .

$x = - 14 + \frac{- 3 (3 y)}{- 3 \cdot 1}$

$2 x - 2 y = 4$

Cancel the common factor.

$x = - 14 + \frac{- 3 (3 y)}{- 3 \cdot 1}$

$2 x - 2 y = 4$

Rewrite the expression.

$x = - 14 + \frac{3 y}{1}$

$2 x - 2 y = 4$

Divide $3 y$ by $1$ .

$x = - 14 + 3 y$

$2 x - 2 y = 4$

$x = - 14 + 3 y$

$2 x - 2 y = 4$

$x = - 14 + 3 y$

$2 x - 2 y = 4$

$x = - 14 + 3 y$

$2 x - 2 y = 4$

$x = - 14 + 3 y$

$2 x - 2 y = 4$

$x = - 14 + 3 y$

$2 x - 2 y = 4$

$x = - 14 + 3 y$

$2 x - 2 y = 4$

Step 2

Replace all occurrences of $x$ with $- 14 + 3 y$ in each equation.

Replace all occurrences of $x$ in $2 x - 2 y = 4$ with $- 14 + 3 y$ .

$2 (- 14 + 3 y) - 2 y = 4$

$x = - 14 + 3 y$

Simplify the left side.

Simplify $2 (- 14 + 3 y) - 2 y$ .

Simplify each term.

Apply the distributive property.

$2 \cdot - 14 + 2 (3 y) - 2 y = 4$

$x = - 14 + 3 y$

Multiply $2$ by $- 14$ .

$- 28 + 2 (3 y) - 2 y = 4$

$x = - 14 + 3 y$

Multiply $3$ by $2$ .

$- 28 + 6 y - 2 y = 4$

$x = - 14 + 3 y$

$- 28 + 6 y - 2 y = 4$

$x = - 14 + 3 y$

Subtract $2 y$ from $6 y$ .

$- 28 + 4 y = 4$

$x = - 14 + 3 y$

$- 28 + 4 y = 4$

$x = - 14 + 3 y$

$- 28 + 4 y = 4$

$x = - 14 + 3 y$

$- 28 + 4 y = 4$

$x = - 14 + 3 y$

Step 3

Solve for $y$ in $- 28 + 4 y = 4$ .

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

Add $28$ to both sides of the equation.

$4 y = 4 + 28$

$x = - 14 + 3 y$

Add $4$ and $28$ .

$4 y = 32$

$x = - 14 + 3 y$

$4 y = 32$

$x = - 14 + 3 y$

Divide each term in $4 y = 32$ by $4$ and simplify.

Divide each term in $4 y = 32$ by $4$ .

$\frac{4 y}{4} = \frac{32}{4}$

$x = - 14 + 3 y$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{32}{4}$

$x = - 14 + 3 y$

Divide $y$ by $1$ .

$y = \frac{32}{4}$

$x = - 14 + 3 y$

$y = \frac{32}{4}$

$x = - 14 + 3 y$

$y = \frac{32}{4}$

$x = - 14 + 3 y$

Simplify the right side.

Divide $32$ by $4$ .

$y = 8$

$x = - 14 + 3 y$

$y = 8$

$x = - 14 + 3 y$

$y = 8$

$x = - 14 + 3 y$

$y = 8$

$x = - 14 + 3 y$

Step 4

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

Replace all occurrences of $y$ in $x = - 14 + 3 y$ with $8$ .

$x = - 14 + 3 (8)$

$y = 8$

Simplify the right side.

Simplify $- 14 + 3 (8)$ .

Multiply $3$ by $8$ .

$x = - 14 + 24$

$y = 8$

Add $- 14$ and $24$ .

$x = 10$

$y = 8$

$x = 10$

$y = 8$

$x = 10$

$y = 8$

$x = 10$

$y = 8$

Step 5

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

$(10, 8)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(10, 8)$

Equation Form:

$x = 10, y = 8$

Step 7

Solve the System of Equations -3x+9y=42 and 2x-2y=4

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