Solve by Substitution 15x-10y=-1 , 15y=-1+10x

$15 x - 10 y = - 1$ , $15 y = - 1 + 10 x$

Step 1

Divide each term in $15 y = - 1 + 10 x$ by $15$ and simplify.

Divide each term in $15 y = - 1 + 10 x$ by $15$ .

$\frac{15 y}{15} = \frac{- 1}{15} + \frac{10 x}{15}$

$15 x - 10 y = - 1$

Simplify the left side.

Cancel the common factor of $15$ .

Cancel the common factor.

$\frac{15 y}{15} = \frac{- 1}{15} + \frac{10 x}{15}$

$15 x - 10 y = - 1$

Divide $y$ by $1$ .

$y = \frac{- 1}{15} + \frac{10 x}{15}$

$15 x - 10 y = - 1$

$y = \frac{- 1}{15} + \frac{10 x}{15}$

$15 x - 10 y = - 1$

$y = \frac{- 1}{15} + \frac{10 x}{15}$

$15 x - 10 y = - 1$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y = - \frac{1}{15} + \frac{10 x}{15}$

$15 x - 10 y = - 1$

Cancel the common factor of $10$ and $15$ .

Factor $5$ out of $10 x$ .

$y = - \frac{1}{15} + \frac{5 (2 x)}{15}$

$15 x - 10 y = - 1$

Cancel the common factors.

Factor $5$ out of $15$ .

$y = - \frac{1}{15} + \frac{5 (2 x)}{5 (3)}$

$15 x - 10 y = - 1$

Cancel the common factor.

$y = - \frac{1}{15} + \frac{5 (2 x)}{5 \cdot 3}$

$15 x - 10 y = - 1$

Rewrite the expression.

$y = - \frac{1}{15} + \frac{2 x}{3}$

$15 x - 10 y = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$15 x - 10 y = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$15 x - 10 y = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$15 x - 10 y = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$15 x - 10 y = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$15 x - 10 y = - 1$

Step 2

Replace all occurrences of $y$ with $- \frac{1}{15} + \frac{2 x}{3}$ in each equation.

Replace all occurrences of $y$ in $15 x - 10 y = - 1$ with $- \frac{1}{15} + \frac{2 x}{3}$ .

$15 x - 10 (- \frac{1}{15} + \frac{2 x}{3}) = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Simplify the left side.

Simplify $15 x - 10 (- \frac{1}{15} + \frac{2 x}{3})$ .

Simplify each term.

Apply the distributive property.

$15 x - 10 (- \frac{1}{15}) - 10 \frac{2 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{1}{15}$ into the numerator.

$15 x - 10 (\frac{- 1}{15}) - 10 \frac{2 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Factor $5$ out of $- 10$ .

$15 x + 5 (- 2) (\frac{- 1}{15}) - 10 \frac{2 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Factor $5$ out of $15$ .

$15 x + 5 \cdot (- 2 \frac{- 1}{5 \cdot 3}) - 10 \frac{2 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Cancel the common factor.

$15 x + 5 \cdot (- 2 \frac{- 1}{5 \cdot 3}) - 10 \frac{2 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Rewrite the expression.

$15 x - 2 (\frac{- 1}{3}) - 10 \frac{2 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$15 x - 2 (\frac{- 1}{3}) - 10 \frac{2 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Combine $- 2$ and $\frac{- 1}{3}$ .

$15 x + \frac{- 2 \cdot - 1}{3} - 10 \frac{2 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Multiply $- 2$ by $- 1$ .

$15 x + \frac{2}{3} - 10 \frac{2 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Multiply $- 10 \frac{2 x}{3}$ .

Combine $- 10$ and $\frac{2 x}{3}$ .

$15 x + \frac{2}{3} + \frac{- 10 (2 x)}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Multiply $2$ by $- 10$ .

$15 x + \frac{2}{3} + \frac{- 20 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$15 x + \frac{2}{3} + \frac{- 20 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Move the negative in front of the fraction.

$15 x + \frac{2}{3} - \frac{20 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$15 x + \frac{2}{3} - \frac{20 x}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

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

$15 x \cdot \frac{3}{3} - \frac{20 x}{3} + \frac{2}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

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

$\frac{15 x \cdot 3}{3} - \frac{20 x}{3} + \frac{2}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Combine the numerators over the common denominator.

$\frac{15 x \cdot 3 - 20 x}{3} + \frac{2}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Combine the numerators over the common denominator.

$\frac{15 x \cdot 3 - 20 x + 2}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Multiply $3$ by $15$ .

$\frac{45 x - 20 x + 2}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Subtract $20 x$ from $45 x$ .

$\frac{25 x + 2}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$\frac{25 x + 2}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$\frac{25 x + 2}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$\frac{25 x + 2}{3} = - 1$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Step 3

Solve for $x$ in $\frac{25 x + 2}{3} = - 1$ .

Multiply both sides by $3$ .

$\frac{25 x + 2}{3} \cdot 3 = - 1 \cdot 3$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{25 x + 2}{3} \cdot 3 = - 1 \cdot 3$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Rewrite the expression.

$25 x + 2 = - 1 \cdot 3$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$25 x + 2 = - 1 \cdot 3$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$25 x + 2 = - 1 \cdot 3$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Simplify the right side.

Multiply $- 1$ by $3$ .

$25 x + 2 = - 3$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$25 x + 2 = - 3$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$25 x + 2 = - 3$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Solve for $x$ .

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

Subtract $2$ from both sides of the equation.

$25 x = - 3 - 2$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Subtract $2$ from $- 3$ .

$25 x = - 5$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$25 x = - 5$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Divide each term in $25 x = - 5$ by $25$ and simplify.

Divide each term in $25 x = - 5$ by $25$ .

$\frac{25 x}{25} = \frac{- 5}{25}$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Simplify the left side.

Cancel the common factor of $25$ .

Cancel the common factor.

$\frac{25 x}{25} = \frac{- 5}{25}$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Divide $x$ by $1$ .

$x = \frac{- 5}{25}$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$x = \frac{- 5}{25}$

$y = - \frac{1}{15} + \frac{2 x}{3}$

$x = \frac{- 5}{25}$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Simplify the right side.

Cancel the common factor of $- 5$ and $25$ .

Factor $5$ out of $- 5$ .

$x = \frac{5 (- 1)}{25}$

$y = - \frac{1}{15} + \frac{2 x}{3}$

Cancel the common factors.

Factor $5$ out of $25$ .

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

$y = - \frac{1}{15} + \frac{2 x}{3}$

Cancel the common factor.

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

$y = - \frac{1}{15} + \frac{2 x}{3}$

Rewrite the expression.

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

$y = - \frac{1}{15} + \frac{2 x}{3}$

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

$y = - \frac{1}{15} + \frac{2 x}{3}$

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

$y = - \frac{1}{15} + \frac{2 x}{3}$

Move the negative in front of the fraction.

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

$y = - \frac{1}{15} + \frac{2 x}{3}$

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

$y = - \frac{1}{15} + \frac{2 x}{3}$

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

$y = - \frac{1}{15} + \frac{2 x}{3}$

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

$y = - \frac{1}{15} + \frac{2 x}{3}$

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

$y = - \frac{1}{15} + \frac{2 x}{3}$

Step 4

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

Replace all occurrences of $x$ in $y = - \frac{1}{15} + \frac{2 x}{3}$ with $- \frac{1}{5}$ .

$y = - \frac{1}{15} + \frac{2 (- \frac{1}{5})}{3}$

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

Simplify the right side.

Simplify $- \frac{1}{15} + \frac{2 (- \frac{1}{5})}{3}$ .

Simplify each term.

Simplify the numerator.

Multiply $- 1$ by $2$ .

$y = - \frac{1}{15} + \frac{- 2 (\frac{1}{5})}{3}$

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

Combine $- 2$ and $\frac{1}{5}$ .

$y = - \frac{1}{15} + \frac{\frac{- 2}{5}}{3}$

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

$y = - \frac{1}{15} + \frac{\frac{- 2}{5}}{3}$

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

Move the negative in front of the fraction.

$y = - \frac{1}{15} + \frac{- \frac{2}{5}}{3}$

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

Multiply the numerator by the reciprocal of the denominator.

$y = - \frac{1}{15} - \frac{2}{5} \cdot \frac{1}{3}$

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

Multiply $- \frac{2}{5} \cdot \frac{1}{3}$ .

Multiply $\frac{1}{3}$ by $\frac{2}{5}$ .

$y = - \frac{1}{15} - \frac{2}{3 \cdot 5}$

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

Multiply $3$ by $5$ .

$y = - \frac{1}{15} - \frac{2}{15}$

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

$y = - \frac{1}{15} - \frac{2}{15}$

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

$y = - \frac{1}{15} - \frac{2}{15}$

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

Simplify terms.

Combine the numerators over the common denominator.

$y = \frac{- 1 - 2}{15}$

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

Subtract $2$ from $- 1$ .

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

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

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.

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

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

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

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

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

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

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

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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