Solve by Substitution (x+3y)/2=5 3x-y=5y

$\frac{x + 3 y}{2} = 5$ $3 x - y = 5 y$

Step 1

Solve for $x$ in $3 x - y = 5 y$ .

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

Add $y$ to both sides of the equation.

$3 x = 5 y + y$

$\frac{x + 3 y}{2} = 5$

Add $5 y$ and $y$ .

$3 x = 6 y$

$\frac{x + 3 y}{2} = 5$

$3 x = 6 y$

$\frac{x + 3 y}{2} = 5$

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

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

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

$\frac{x + 3 y}{2} = 5$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

$\frac{x + 3 y}{2} = 5$

Divide $x$ by $1$ .

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

$\frac{x + 3 y}{2} = 5$

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

$\frac{x + 3 y}{2} = 5$

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

$\frac{x + 3 y}{2} = 5$

Simplify the right side.

Cancel the common factor of $6$ and $3$ .

Factor $3$ out of $6 y$ .

$x = \frac{3 (2 y)}{3}$

$\frac{x + 3 y}{2} = 5$

Cancel the common factors.

Factor $3$ out of $3$ .

$x = \frac{3 (2 y)}{3 (1)}$

$\frac{x + 3 y}{2} = 5$

Cancel the common factor.

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

$\frac{x + 3 y}{2} = 5$

Rewrite the expression.

$x = \frac{2 y}{1}$

$\frac{x + 3 y}{2} = 5$

Divide $2 y$ by $1$ .

$x = 2 y$

$\frac{x + 3 y}{2} = 5$

$x = 2 y$

$\frac{x + 3 y}{2} = 5$

$x = 2 y$

$\frac{x + 3 y}{2} = 5$

$x = 2 y$

$\frac{x + 3 y}{2} = 5$

$x = 2 y$

$\frac{x + 3 y}{2} = 5$

$x = 2 y$

$\frac{x + 3 y}{2} = 5$

Step 2

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

Replace all occurrences of $x$ in $\frac{x + 3 y}{2} = 5$ with $2 y$ .

$\frac{(2 y) + 3 y}{2} = 5$

$x = 2 y$

Simplify the left side.

Add $2 y$ and $3 y$ .

$\frac{5 y}{2} = 5$

$x = 2 y$

$\frac{5 y}{2} = 5$

$x = 2 y$

$\frac{5 y}{2} = 5$

$x = 2 y$

Step 3

Solve for $y$ in $\frac{5 y}{2} = 5$ .

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

$\frac{2}{5} \cdot \frac{5 y}{2} = \frac{2}{5} \cdot 5$

$x = 2 y$

Simplify both sides of the equation.

Simplify the left side.

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

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2}{5} \cdot \frac{5 y}{2} = \frac{2}{5} \cdot 5$

$x = 2 y$

Rewrite the expression.

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

$x = 2 y$

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

$x = 2 y$

Cancel the common factor of $5$ .

Factor $5$ out of $5 y$ .

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

$x = 2 y$

Cancel the common factor.

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

$x = 2 y$

Rewrite the expression.

$y = \frac{2}{5} \cdot 5$

$x = 2 y$

$y = \frac{2}{5} \cdot 5$

$x = 2 y$

$y = \frac{2}{5} \cdot 5$

$x = 2 y$

$y = \frac{2}{5} \cdot 5$

$x = 2 y$

Simplify the right side.

Cancel the common factor of $5$ .

Cancel the common factor.

$y = \frac{2}{5} \cdot 5$

$x = 2 y$

Rewrite the expression.

$y = 2$

$x = 2 y$

$y = 2$

$x = 2 y$

$y = 2$

$x = 2 y$

$y = 2$

$x = 2 y$

$y = 2$

$x = 2 y$

Step 4

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

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

$x = 2 (2)$

$y = 2$

Simplify the right side.

Multiply $2$ by $2$ .

$x = 4$

$y = 2$

$x = 4$

$y = 2$

$x = 4$

$y = 2$

Step 5

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

$(4, 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 2)$

Equation Form:

$x = 4, y = 2$

Step 7

Solve by Substitution (x+3y)/2=5 3x-y=5y

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