Solve by Substitution 15x-10y=-5 , 15y=4+5x

$15 x - 10 y = - 5$ , $15 y = 4 + 5 x$

Step 1

Divide each term in $15 y = 4 + 5 x$ by $15$ and simplify.

Divide each term in $15 y = 4 + 5 x$ by $15$ .

$\frac{15 y}{15} = \frac{4}{15} + \frac{5 x}{15}$

$15 x - 10 y = - 5$

Simplify the left side.

Cancel the common factor of $15$ .

Cancel the common factor.

$\frac{15 y}{15} = \frac{4}{15} + \frac{5 x}{15}$

$15 x - 10 y = - 5$

Divide $y$ by $1$ .

$y = \frac{4}{15} + \frac{5 x}{15}$

$15 x - 10 y = - 5$

$y = \frac{4}{15} + \frac{5 x}{15}$

$15 x - 10 y = - 5$

$y = \frac{4}{15} + \frac{5 x}{15}$

$15 x - 10 y = - 5$

Simplify the right side.

Cancel the common factor of $5$ and $15$ .

Factor $5$ out of $5 x$ .

$y = \frac{4}{15} + \frac{5 (x)}{15}$

$15 x - 10 y = - 5$

Cancel the common factors.

Factor $5$ out of $15$ .

$y = \frac{4}{15} + \frac{5 x}{5 \cdot 3}$

$15 x - 10 y = - 5$

Cancel the common factor.

$y = \frac{4}{15} + \frac{5 x}{5 \cdot 3}$

$15 x - 10 y = - 5$

Rewrite the expression.

$y = \frac{4}{15} + \frac{x}{3}$

$15 x - 10 y = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

$15 x - 10 y = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

$15 x - 10 y = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

$15 x - 10 y = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

$15 x - 10 y = - 5$

Step 2

Replace all occurrences of $y$ with $\frac{4}{15} + \frac{x}{3}$ in each equation.

Replace all occurrences of $y$ in $15 x - 10 y = - 5$ with $\frac{4}{15} + \frac{x}{3}$ .

$15 x - 10 (\frac{4}{15} + \frac{x}{3}) = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Simplify the left side.

Simplify $15 x - 10 (\frac{4}{15} + \frac{x}{3})$ .

Simplify each term.

Apply the distributive property.

$15 x - 10 (\frac{4}{15}) - 10 \frac{x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Cancel the common factor of $5$ .

Factor $5$ out of $- 10$ .

$15 x + 5 (- 2) (\frac{4}{15}) - 10 \frac{x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Factor $5$ out of $15$ .

$15 x + 5 \cdot (- 2 \frac{4}{5 \cdot 3}) - 10 \frac{x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Cancel the common factor.

$15 x + 5 \cdot (- 2 \frac{4}{5 \cdot 3}) - 10 \frac{x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Rewrite the expression.

$15 x - 2 (\frac{4}{3}) - 10 \frac{x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

$15 x - 2 (\frac{4}{3}) - 10 \frac{x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Combine $- 2$ and $\frac{4}{3}$ .

$15 x + \frac{- 2 \cdot 4}{3} - 10 \frac{x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Multiply $- 2$ by $4$ .

$15 x + \frac{- 8}{3} - 10 \frac{x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Combine $- 10$ and $\frac{x}{3}$ .

$15 x + \frac{- 8}{3} + \frac{- 10 x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Simplify each term.

Move the negative in front of the fraction.

$15 x - \frac{8}{3} + \frac{- 10 x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Move the negative in front of the fraction.

$15 x - \frac{8}{3} - \frac{10 x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

$15 x - \frac{8}{3} - \frac{10 x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

$15 x - \frac{8}{3} - \frac{10 x}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

To write $15 x$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$15 x \cdot \frac{3}{3} - \frac{10 x}{3} - \frac{8}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Combine $15 x$ and $\frac{3}{3}$ .

$\frac{15 x \cdot 3}{3} - \frac{10 x}{3} - \frac{8}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Combine the numerators over the common denominator.

$\frac{15 x \cdot 3 - 10 x}{3} - \frac{8}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Combine the numerators over the common denominator.

$\frac{15 x \cdot 3 - 10 x - 8}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Multiply $3$ by $15$ .

$\frac{45 x - 10 x - 8}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Subtract $10 x$ from $45 x$ .

$\frac{35 x - 8}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

$\frac{35 x - 8}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

$\frac{35 x - 8}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

$\frac{35 x - 8}{3} = - 5$

$y = \frac{4}{15} + \frac{x}{3}$

Step 3

Solve for $x$ in $\frac{35 x - 8}{3} = - 5$ .

Multiply both sides by $3$ .

$\frac{35 x - 8}{3} \cdot 3 = - 5 \cdot 3$

$y = \frac{4}{15} + \frac{x}{3}$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{35 x - 8}{3} \cdot 3 = - 5 \cdot 3$

$y = \frac{4}{15} + \frac{x}{3}$

Rewrite the expression.

$35 x - 8 = - 5 \cdot 3$

$y = \frac{4}{15} + \frac{x}{3}$

$35 x - 8 = - 5 \cdot 3$

$y = \frac{4}{15} + \frac{x}{3}$

$35 x - 8 = - 5 \cdot 3$

$y = \frac{4}{15} + \frac{x}{3}$

Simplify the right side.

Multiply $- 5$ by $3$ .

$35 x - 8 = - 15$

$y = \frac{4}{15} + \frac{x}{3}$

$35 x - 8 = - 15$

$y = \frac{4}{15} + \frac{x}{3}$

$35 x - 8 = - 15$

$y = \frac{4}{15} + \frac{x}{3}$

Solve for $x$ .

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

Add $8$ to both sides of the equation.

$35 x = - 15 + 8$

$y = \frac{4}{15} + \frac{x}{3}$

Add $- 15$ and $8$ .

$35 x = - 7$

$y = \frac{4}{15} + \frac{x}{3}$

$35 x = - 7$

$y = \frac{4}{15} + \frac{x}{3}$

Divide each term in $35 x = - 7$ by $35$ and simplify.

Divide each term in $35 x = - 7$ by $35$ .

$\frac{35 x}{35} = \frac{- 7}{35}$

$y = \frac{4}{15} + \frac{x}{3}$

Simplify the left side.

Cancel the common factor of $35$ .

Cancel the common factor.

$\frac{35 x}{35} = \frac{- 7}{35}$

$y = \frac{4}{15} + \frac{x}{3}$

Divide $x$ by $1$ .

$x = \frac{- 7}{35}$

$y = \frac{4}{15} + \frac{x}{3}$

$x = \frac{- 7}{35}$

$y = \frac{4}{15} + \frac{x}{3}$

$x = \frac{- 7}{35}$

$y = \frac{4}{15} + \frac{x}{3}$

Simplify the right side.

Cancel the common factor of $- 7$ and $35$ .

Factor $7$ out of $- 7$ .

$x = \frac{7 (- 1)}{35}$

$y = \frac{4}{15} + \frac{x}{3}$

Cancel the common factors.

Factor $7$ out of $35$ .

$x = \frac{7 \cdot - 1}{7 \cdot 5}$

$y = \frac{4}{15} + \frac{x}{3}$

Cancel the common factor.

$x = \frac{7 \cdot - 1}{7 \cdot 5}$

$y = \frac{4}{15} + \frac{x}{3}$

Rewrite the expression.

$x = \frac{- 1}{5}$

$y = \frac{4}{15} + \frac{x}{3}$

$x = \frac{- 1}{5}$

$y = \frac{4}{15} + \frac{x}{3}$

$x = \frac{- 1}{5}$

$y = \frac{4}{15} + \frac{x}{3}$

Move the negative in front of the fraction.

$x = - \frac{1}{5}$

$y = \frac{4}{15} + \frac{x}{3}$

$x = - \frac{1}{5}$

$y = \frac{4}{15} + \frac{x}{3}$

$x = - \frac{1}{5}$

$y = \frac{4}{15} + \frac{x}{3}$

$x = - \frac{1}{5}$

$y = \frac{4}{15} + \frac{x}{3}$

$x = - \frac{1}{5}$

$y = \frac{4}{15} + \frac{x}{3}$

Step 4

Replace all occurrences of $x$ with $- \frac{1}{5}$ in each equation.

Replace all occurrences of $x$ in $y = \frac{4}{15} + \frac{x}{3}$ with $- \frac{1}{5}$ .

$y = \frac{4}{15} + \frac{- \frac{1}{5}}{3}$

$x = - \frac{1}{5}$

Simplify the right side.

Simplify $\frac{4}{15} + \frac{- \frac{1}{5}}{3}$ .

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$y = \frac{4}{15} - \frac{1}{5} \cdot \frac{1}{3}$

$x = - \frac{1}{5}$

Multiply $- \frac{1}{5} \cdot \frac{1}{3}$ .

Multiply $\frac{1}{3}$ by $\frac{1}{5}$ .

$y = \frac{4}{15} - \frac{1}{3 \cdot 5}$

$x = - \frac{1}{5}$

Multiply $3$ by $5$ .

$y = \frac{4}{15} - \frac{1}{15}$

$x = - \frac{1}{5}$

$y = \frac{4}{15} - \frac{1}{15}$

$x = - \frac{1}{5}$

$y = \frac{4}{15} - \frac{1}{15}$

$x = - \frac{1}{5}$

Simplify terms.

Combine the numerators over the common denominator.

$y = \frac{4 - 1}{15}$

$x = - \frac{1}{5}$

Subtract $1$ from $4$ .

$y = \frac{3}{15}$

$x = - \frac{1}{5}$

Cancel the common factor of $3$ and $15$ .

Factor $3$ out of $3$ .

$y = \frac{3 (1)}{15}$

$x = - \frac{1}{5}$

Cancel the common factors.

Factor $3$ out of $15$ .

$y = \frac{3 \cdot 1}{3 \cdot 5}$

$x = - \frac{1}{5}$

Cancel the common factor.

$y = \frac{3 \cdot 1}{3 \cdot 5}$

$x = - \frac{1}{5}$

Rewrite the expression.

$y = \frac{1}{5}$

$x = - \frac{1}{5}$

$y = \frac{1}{5}$

$x = - \frac{1}{5}$

$y = \frac{1}{5}$

$x = - \frac{1}{5}$

$y = \frac{1}{5}$

$x = - \frac{1}{5}$

$y = \frac{1}{5}$

$x = - \frac{1}{5}$

$y = \frac{1}{5}$

$x = - \frac{1}{5}$

$y = \frac{1}{5}$

$x = - \frac{1}{5}$

Step 5

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

$(- \frac{1}{5}, \frac{1}{5})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- \frac{1}{5}, \frac{1}{5})$

Equation Form:

$x = - \frac{1}{5}, y = \frac{1}{5}$

Step 7

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