Solve by Substitution 15x-10y=1 , 5y=-2+15x

$15 x - 10 y = 1$ , $5 y = - 2 + 15 x$

Step 1

Divide each term in $5 y = - 2 + 15 x$ by $5$ and simplify.

Divide each term in $5 y = - 2 + 15 x$ by $5$ .

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

$15 x - 10 y = 1$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

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

$15 x - 10 y = 1$

Divide $y$ by $1$ .

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

$15 x - 10 y = 1$

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

$15 x - 10 y = 1$

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

$15 x - 10 y = 1$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

$15 x - 10 y = 1$

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

Factor $5$ out of $15 x$ .

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

$15 x - 10 y = 1$

Cancel the common factors.

Factor $5$ out of $5$ .

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

$15 x - 10 y = 1$

Cancel the common factor.

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

$15 x - 10 y = 1$

Rewrite the expression.

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

$15 x - 10 y = 1$

Divide $3 x$ by $1$ .

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

$15 x - 10 y = 1$

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

$15 x - 10 y = 1$

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

$15 x - 10 y = 1$

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

$15 x - 10 y = 1$

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

$15 x - 10 y = 1$

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

$15 x - 10 y = 1$

Step 2

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

Replace all occurrences of $y$ in $15 x - 10 y = 1$ with $- \frac{2}{5} + 3 x$ .

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

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

Simplify the left side.

Simplify $15 x - 10 (- \frac{2}{5} + 3 x)$ .

Simplify each term.

Apply the distributive property.

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

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

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{2}{5}$ into the numerator.

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

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

Factor $5$ out of $- 10$ .

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

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

Cancel the common factor.

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

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

Rewrite the expression.

$15 x - 2 \cdot - 2 - 10 (3 x) = 1$

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

$15 x - 2 \cdot - 2 - 10 (3 x) = 1$

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

Multiply $- 2$ by $- 2$ .

$15 x + 4 - 10 (3 x) = 1$

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

Multiply $3$ by $- 10$ .

$15 x + 4 - 30 x = 1$

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

$15 x + 4 - 30 x = 1$

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

Subtract $30 x$ from $15 x$ .

$- 15 x + 4 = 1$

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

$- 15 x + 4 = 1$

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

$- 15 x + 4 = 1$

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

$- 15 x + 4 = 1$

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

Step 3

Solve for $x$ in $- 15 x + 4 = 1$ .

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

Subtract $4$ from both sides of the equation.

$- 15 x = 1 - 4$

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

Subtract $4$ from $1$ .

$- 15 x = - 3$

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

$- 15 x = - 3$

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

Divide each term in $- 15 x = - 3$ by $- 15$ and simplify.

Divide each term in $- 15 x = - 3$ by $- 15$ .

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

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

Simplify the left side.

Cancel the common factor of $- 15$ .

Cancel the common factor.

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

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

Divide $x$ by $1$ .

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

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

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

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

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

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

Simplify the right side.

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

Factor $- 3$ out of $- 3$ .

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

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

Cancel the common factors.

Factor $- 3$ out of $- 15$ .

$x = \frac{- 3 \cdot 1}{- 3 \cdot 5}$

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

Cancel the common factor.

$x = \frac{- 3 \cdot 1}{- 3 \cdot 5}$

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

Rewrite the expression.

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

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

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

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

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

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

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

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

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

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

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

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

Step 4

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

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

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

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

Simplify the right side.

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

Combine $3$ and $\frac{1}{5}$ .

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

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

Combine the numerators over the common denominator.

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

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

Add $- 2$ and $3$ .

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

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

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

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

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

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

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

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

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

Solve by Substitution 15x-10y=1 , 5y=-2+15x

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