Solve by Substitution 5x=15+10y 3x+7y=31

$5 x = 15 + 10 y$ $3 x + 7 y = 31$

Step 1

Divide each term in $5 x = 15 + 10 y$ by $5$ and simplify.

Divide each term in $5 x = 15 + 10 y$ by $5$ .

$\frac{5 x}{5} = \frac{15}{5} + \frac{10 y}{5}$

$3 x + 7 y = 31$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{15}{5} + \frac{10 y}{5}$

$3 x + 7 y = 31$

Divide $x$ by $1$ .

$x = \frac{15}{5} + \frac{10 y}{5}$

$3 x + 7 y = 31$

$x = \frac{15}{5} + \frac{10 y}{5}$

$3 x + 7 y = 31$

$x = \frac{15}{5} + \frac{10 y}{5}$

$3 x + 7 y = 31$

Simplify the right side.

Simplify each term.

Divide $15$ by $5$ .

$x = 3 + \frac{10 y}{5}$

$3 x + 7 y = 31$

Cancel the common factor of $10$ and $5$ .

Factor $5$ out of $10 y$ .

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

$3 x + 7 y = 31$

Cancel the common factors.

Factor $5$ out of $5$ .

$x = 3 + \frac{5 (2 y)}{5 (1)}$

$3 x + 7 y = 31$

Cancel the common factor.

$x = 3 + \frac{5 (2 y)}{5 \cdot 1}$

$3 x + 7 y = 31$

Rewrite the expression.

$x = 3 + \frac{2 y}{1}$

$3 x + 7 y = 31$

Divide $2 y$ by $1$ .

$x = 3 + 2 y$

$3 x + 7 y = 31$

$x = 3 + 2 y$

$3 x + 7 y = 31$

$x = 3 + 2 y$

$3 x + 7 y = 31$

$x = 3 + 2 y$

$3 x + 7 y = 31$

$x = 3 + 2 y$

$3 x + 7 y = 31$

$x = 3 + 2 y$

$3 x + 7 y = 31$

Step 2

Replace all occurrences of $x$ with $3 + 2 y$ in each equation.

Replace all occurrences of $x$ in $3 x + 7 y = 31$ with $3 + 2 y$ .

$3 (3 + 2 y) + 7 y = 31$

$x = 3 + 2 y$

Simplify the left side.

Simplify $3 (3 + 2 y) + 7 y$ .

Simplify each term.

Apply the distributive property.

$3 \cdot 3 + 3 (2 y) + 7 y = 31$

$x = 3 + 2 y$

Multiply $3$ by $3$ .

$9 + 3 (2 y) + 7 y = 31$

$x = 3 + 2 y$

Multiply $2$ by $3$ .

$9 + 6 y + 7 y = 31$

$x = 3 + 2 y$

$9 + 6 y + 7 y = 31$

$x = 3 + 2 y$

Add $6 y$ and $7 y$ .

$9 + 13 y = 31$

$x = 3 + 2 y$

$9 + 13 y = 31$

$x = 3 + 2 y$

$9 + 13 y = 31$

$x = 3 + 2 y$

$9 + 13 y = 31$

$x = 3 + 2 y$

Step 3

Solve for $y$ in $9 + 13 y = 31$ .

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

Subtract $9$ from both sides of the equation.

$13 y = 31 - 9$

$x = 3 + 2 y$

Subtract $9$ from $31$ .

$13 y = 22$

$x = 3 + 2 y$

$13 y = 22$

$x = 3 + 2 y$

Divide each term in $13 y = 22$ by $13$ and simplify.

Divide each term in $13 y = 22$ by $13$ .

$\frac{13 y}{13} = \frac{22}{13}$

$x = 3 + 2 y$

Simplify the left side.

Cancel the common factor of $13$ .

Cancel the common factor.

$\frac{13 y}{13} = \frac{22}{13}$

$x = 3 + 2 y$

Divide $y$ by $1$ .

$y = \frac{22}{13}$

$x = 3 + 2 y$

$y = \frac{22}{13}$

$x = 3 + 2 y$

$y = \frac{22}{13}$

$x = 3 + 2 y$

$y = \frac{22}{13}$

$x = 3 + 2 y$

$y = \frac{22}{13}$

$x = 3 + 2 y$

Step 4

Replace all occurrences of $y$ with $\frac{22}{13}$ in each equation.

Replace all occurrences of $y$ in $x = 3 + 2 y$ with $\frac{22}{13}$ .

$x = 3 + 2 (\frac{22}{13})$

$y = \frac{22}{13}$

Simplify the right side.

Simplify $3 + 2 (\frac{22}{13})$ .

Multiply $2 (\frac{22}{13})$ .

Combine $2$ and $\frac{22}{13}$ .

$x = 3 + \frac{2 \cdot 22}{13}$

$y = \frac{22}{13}$

Multiply $2$ by $22$ .

$x = 3 + \frac{44}{13}$

$y = \frac{22}{13}$

$x = 3 + \frac{44}{13}$

$y = \frac{22}{13}$

To write $3$ as a fraction with a common denominator, multiply by $\frac{13}{13}$ .

$x = 3 \cdot \frac{13}{13} + \frac{44}{13}$

$y = \frac{22}{13}$

Combine $3$ and $\frac{13}{13}$ .

$x = \frac{3 \cdot 13}{13} + \frac{44}{13}$

$y = \frac{22}{13}$

Combine the numerators over the common denominator.

$x = \frac{3 \cdot 13 + 44}{13}$

$y = \frac{22}{13}$

Simplify the numerator.

Multiply $3$ by $13$ .

$x = \frac{39 + 44}{13}$

$y = \frac{22}{13}$

Add $39$ and $44$ .

$x = \frac{83}{13}$

$y = \frac{22}{13}$

$x = \frac{83}{13}$

$y = \frac{22}{13}$

$x = \frac{83}{13}$

$y = \frac{22}{13}$

$x = \frac{83}{13}$

$y = \frac{22}{13}$

$x = \frac{83}{13}$

$y = \frac{22}{13}$

Step 5

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

$(\frac{83}{13}, \frac{22}{13})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{83}{13}, \frac{22}{13})$

Equation Form:

$x = \frac{83}{13}, y = \frac{22}{13}$

Step 7

Solve by Substitution 5x=15+10y 3x+7y=31

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