Solve the System of Equations 3x-2y=6 (2x+2y)=(4)

$3 x - 2 y = 6$ $(2 x + 2 y) = (4)$

Solve for $x$ in $3 x - 2 y = 6$ .

Add $2 y$ to both sides of the equation.

$3 x = 6 + 2 y$

$2 x + 2 y = 4$

Divide each term in $3 x = 6 + 2 y$ by $3$ and simplify.

Divide each term in $3 x = 6 + 2 y$ by $3$ .

$\frac{3 x}{3} = \frac{6}{3} + \frac{2 y}{3}$

$2 x + 2 y = 4$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{6}{3} + \frac{2 y}{3}$

$2 x + 2 y = 4$

Divide $x$ by $1$ .

$x = \frac{6}{3} + \frac{2 y}{3}$

$2 x + 2 y = 4$

$x = \frac{6}{3} + \frac{2 y}{3}$

$2 x + 2 y = 4$

$x = \frac{6}{3} + \frac{2 y}{3}$

$2 x + 2 y = 4$

Simplify the right side.

Divide $6$ by $3$ .

$x = 2 + \frac{2 y}{3}$

$2 x + 2 y = 4$

$x = 2 + \frac{2 y}{3}$

$2 x + 2 y = 4$

$x = 2 + \frac{2 y}{3}$

$2 x + 2 y = 4$

$x = 2 + \frac{2 y}{3}$

$2 x + 2 y = 4$

Replace all occurrences of $x$ with $2 + \frac{2 y}{3}$ in each equation.

Replace all occurrences of $x$ in $2 x + 2 y = 4$ with $2 + \frac{2 y}{3}$ .

$2 (2 + \frac{2 y}{3}) + 2 y = 4$

$x = 2 + \frac{2 y}{3}$

Simplify the left side.

Simplify $2 (2 + \frac{2 y}{3}) + 2 y$ .

Simplify each term.

Apply the distributive property.

$2 \cdot 2 + 2 (\frac{2 y}{3}) + 2 y = 4$

$x = 2 + \frac{2 y}{3}$

Multiply $2$ by $2$ .

$4 + 2 (\frac{2 y}{3}) + 2 y = 4$

$x = 2 + \frac{2 y}{3}$

Multiply $2 \frac{2 y}{3}$ .

Combine $2$ and $\frac{2 y}{3}$ .

$4 + \frac{2 (2 y)}{3} + 2 y = 4$

$x = 2 + \frac{2 y}{3}$

Multiply $2$ by $2$ .

$4 + \frac{4 y}{3} + 2 y = 4$

$x = 2 + \frac{2 y}{3}$

$4 + \frac{4 y}{3} + 2 y = 4$

$x = 2 + \frac{2 y}{3}$

$4 + \frac{4 y}{3} + 2 y = 4$

$x = 2 + \frac{2 y}{3}$

To write $2 y$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$4 + \frac{4 y}{3} + 2 y \cdot \frac{3}{3} = 4$

$x = 2 + \frac{2 y}{3}$

Simplify terms.

Combine $2 y$ and $\frac{3}{3}$ .

$4 + \frac{4 y}{3} + \frac{2 y \cdot 3}{3} = 4$

$x = 2 + \frac{2 y}{3}$

Combine the numerators over the common denominator.

$4 + \frac{4 y + 2 y \cdot 3}{3} = 4$

$x = 2 + \frac{2 y}{3}$

$4 + \frac{4 y + 2 y \cdot 3}{3} = 4$

$x = 2 + \frac{2 y}{3}$

Simplify the numerator.

Factor $2 y$ out of $4 y + 2 y \cdot 3$ .

Factor $2 y$ out of $4 y$ .

$4 + \frac{2 y (2) + 2 y \cdot 3}{3} = 4$

$x = 2 + \frac{2 y}{3}$

Factor $2 y$ out of $2 y \cdot 3$ .

$4 + \frac{2 y (2) + 2 y (3)}{3} = 4$

$x = 2 + \frac{2 y}{3}$

Factor $2 y$ out of $2 y (2) + 2 y (3)$ .

$4 + \frac{2 y (2 + 3)}{3} = 4$

$x = 2 + \frac{2 y}{3}$

$4 + \frac{2 y (2 + 3)}{3} = 4$

$x = 2 + \frac{2 y}{3}$

Add $2$ and $3$ .

$4 + \frac{2 y \cdot 5}{3} = 4$

$x = 2 + \frac{2 y}{3}$

Multiply $5$ by $2$ .

$4 + \frac{10 y}{3} = 4$

$x = 2 + \frac{2 y}{3}$

$4 + \frac{10 y}{3} = 4$

$x = 2 + \frac{2 y}{3}$

$4 + \frac{10 y}{3} = 4$

$x = 2 + \frac{2 y}{3}$

$4 + \frac{10 y}{3} = 4$

$x = 2 + \frac{2 y}{3}$

$4 + \frac{10 y}{3} = 4$

$x = 2 + \frac{2 y}{3}$

Solve for $y$ in $4 + \frac{10 y}{3} = 4$ .

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

Subtract $4$ from both sides of the equation.

$\frac{10 y}{3} = 4 - 4$

$x = 2 + \frac{2 y}{3}$

Subtract $4$ from $4$ .

$\frac{10 y}{3} = 0$

$x = 2 + \frac{2 y}{3}$

$\frac{10 y}{3} = 0$

$x = 2 + \frac{2 y}{3}$

Set the numerator equal to zero.

$10 y = 0$

$x = 2 + \frac{2 y}{3}$

Divide each term in $10 y = 0$ by $10$ and simplify.

Divide each term in $10 y = 0$ by $10$ .

$\frac{10 y}{10} = \frac{0}{10}$

$x = 2 + \frac{2 y}{3}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 y}{10} = \frac{0}{10}$

$x = 2 + \frac{2 y}{3}$

Divide $y$ by $1$ .

$y = \frac{0}{10}$

$x = 2 + \frac{2 y}{3}$

$y = \frac{0}{10}$

$x = 2 + \frac{2 y}{3}$

$y = \frac{0}{10}$

$x = 2 + \frac{2 y}{3}$

Simplify the right side.

Divide $0$ by $10$ .

$y = 0$

$x = 2 + \frac{2 y}{3}$

$y = 0$

$x = 2 + \frac{2 y}{3}$

$y = 0$

$x = 2 + \frac{2 y}{3}$

$y = 0$

$x = 2 + \frac{2 y}{3}$

Replace all occurrences of $y$ with $0$ in each equation.

Replace all occurrences of $y$ in $x = 2 + \frac{2 y}{3}$ with $0$ .

$x = 2 + \frac{2 (0)}{3}$

$y = 0$

Simplify the right side.

Simplify $2 + \frac{2 (0)}{3}$ .

Multiply $2$ by $0$ .

$x = 2 + \frac{0}{3}$

$y = 0$

Divide $0$ by $3$ .

$x = 2 + 0$

$y = 0$

Add $2$ and $0$ .

$x = 2$

$y = 0$

$x = 2$

$y = 0$

$x = 2$

$y = 0$

$x = 2$

$y = 0$

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

$(2, 0)$

The result can be shown in multiple forms.

Point Form:

$(2, 0)$

Equation Form:

$x = 2, y = 0$

Solve the System of Equations 3x-2y=6 (2x+2y)=(4)

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