Solve the System of Equations 2y=x+6 3x=y+7

$2 y = x + 6$ $3 x = y + 7$

Step 1

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

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

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

$3 x = y + 7$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

$3 x = y + 7$

Divide $y$ by $1$ .

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

$3 x = y + 7$

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

$3 x = y + 7$

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

$3 x = y + 7$

Simplify the right side.

Divide $6$ by $2$ .

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

$3 x = y + 7$

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

$3 x = y + 7$

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

$3 x = y + 7$

Step 2

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

Replace all occurrences of $y$ in $3 x = y + 7$ with $\frac{x}{2} + 3$ .

$3 x = (\frac{x}{2} + 3) + 7$

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

Simplify the right side.

Add $3$ and $7$ .

$3 x = \frac{x}{2} + 10$

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

$3 x = \frac{x}{2} + 10$

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

$3 x = \frac{x}{2} + 10$

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

Step 3

Solve for $x$ in $3 x = \frac{x}{2} + 10$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $\frac{x}{2}$ from both sides of the equation.

$3 x - \frac{x}{2} = 10$

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

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

$3 x \cdot \frac{2}{2} - \frac{x}{2} = 10$

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

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

$\frac{3 x \cdot 2}{2} - \frac{x}{2} = 10$

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

Combine the numerators over the common denominator.

$\frac{3 x \cdot 2 - x}{2} = 10$

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

Simplify the numerator.

Factor $x$ out of $3 x \cdot 2 - x$ .

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

$\frac{x (3 \cdot 2) - x}{2} = 10$

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

Factor $x$ out of $- x$ .

$\frac{x (3 \cdot 2) + x \cdot - 1}{2} = 10$

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

Factor $x$ out of $x (3 \cdot 2) + x \cdot - 1$ .

$\frac{x (3 \cdot 2 - 1)}{2} = 10$

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

$\frac{x (3 \cdot 2 - 1)}{2} = 10$

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

Multiply $3$ by $2$ .

$\frac{x (6 - 1)}{2} = 10$

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

Subtract $1$ from $6$ .

$\frac{x \cdot 5}{2} = 10$

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

$\frac{x \cdot 5}{2} = 10$

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

Move $5$ to the left of $x$ .

$\frac{5 x}{2} = 10$

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

$\frac{5 x}{2} = 10$

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

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

$\frac{2}{5} \cdot \frac{5 x}{2} = \frac{2}{5} \cdot 10$

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

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{2}{5} \cdot \frac{5 x}{2}$ .

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2}{5} \cdot \frac{5 x}{2} = \frac{2}{5} \cdot 10$

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

Rewrite the expression.

$\frac{1}{5} \cdot (5 x) = \frac{2}{5} \cdot 10$

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

$\frac{1}{5} \cdot (5 x) = \frac{2}{5} \cdot 10$

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

Cancel the common factor of $5$ .

Factor $5$ out of $5 x$ .

$\frac{1}{5} \cdot (5 (x)) = \frac{2}{5} \cdot 10$

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

Cancel the common factor.

$\frac{1}{5} \cdot (5 x) = \frac{2}{5} \cdot 10$

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

Rewrite the expression.

$x = \frac{2}{5} \cdot 10$

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

$x = \frac{2}{5} \cdot 10$

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

$x = \frac{2}{5} \cdot 10$

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

$x = \frac{2}{5} \cdot 10$

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

Simplify the right side.

Simplify $\frac{2}{5} \cdot 10$ .

Cancel the common factor of $5$ .

Factor $5$ out of $10$ .

$x = \frac{2}{5} \cdot (5 (2))$

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

Cancel the common factor.

$x = \frac{2}{5} \cdot (5 \cdot 2)$

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

Rewrite the expression.

$x = 2 \cdot 2$

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

$x = 2 \cdot 2$

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

Multiply $2$ by $2$ .

$x = 4$

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

$x = 4$

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

$x = 4$

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

$x = 4$

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

$x = 4$

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

Step 4

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

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

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

$x = 4$

Simplify the right side.

Simplify $\frac{4}{2} + 3$ .

Divide $4$ by $2$ .

$y = 2 + 3$

$x = 4$

Add $2$ and $3$ .

$y = 5$

$x = 4$

$y = 5$

$x = 4$

$y = 5$

$x = 4$

$y = 5$

$x = 4$

Step 5

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

$(4, 5)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 5)$

Equation Form:

$x = 4, y = 5$

Step 7

Solve the System of Equations 2y=x+6 3x=y+7

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