Solve by Substitution 15x-20y=-1 , 5y=-1+10x

$15 x - 20 y = - 1$ , $5 y = - 1 + 10 x$

Step 1

Divide each term in $5 y = - 1 + 10 x$ by $5$ and simplify.

Divide each term in $5 y = - 1 + 10 x$ by $5$ .

$\frac{5 y}{5} = \frac{- 1}{5} + \frac{10 x}{5}$

$15 x - 20 y = - 1$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y}{5} = \frac{- 1}{5} + \frac{10 x}{5}$

$15 x - 20 y = - 1$

Divide $y$ by $1$ .

$y = \frac{- 1}{5} + \frac{10 x}{5}$

$15 x - 20 y = - 1$

$y = \frac{- 1}{5} + \frac{10 x}{5}$

$15 x - 20 y = - 1$

$y = \frac{- 1}{5} + \frac{10 x}{5}$

$15 x - 20 y = - 1$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y = - \frac{1}{5} + \frac{10 x}{5}$

$15 x - 20 y = - 1$

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

Factor $5$ out of $10 x$ .

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

$15 x - 20 y = - 1$

Cancel the common factors.

Factor $5$ out of $5$ .

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

$15 x - 20 y = - 1$

Cancel the common factor.

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

$15 x - 20 y = - 1$

Rewrite the expression.

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

$15 x - 20 y = - 1$

Divide $2 x$ by $1$ .

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

$15 x - 20 y = - 1$

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

$15 x - 20 y = - 1$

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

$15 x - 20 y = - 1$

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

$15 x - 20 y = - 1$

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

$15 x - 20 y = - 1$

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

$15 x - 20 y = - 1$

Step 2

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

Replace all occurrences of $y$ in $15 x - 20 y = - 1$ with $- \frac{1}{5} + 2 x$ .

$15 x - 20 (- \frac{1}{5} + 2 x) = - 1$

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

Simplify the left side.

Simplify $15 x - 20 (- \frac{1}{5} + 2 x)$ .

Simplify each term.

Apply the distributive property.

$15 x - 20 (- \frac{1}{5}) - 20 (2 x) = - 1$

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

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{1}{5}$ into the numerator.

$15 x - 20 (\frac{- 1}{5}) - 20 (2 x) = - 1$

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

Factor $5$ out of $- 20$ .

$15 x + 5 (- 4) (\frac{- 1}{5}) - 20 (2 x) = - 1$

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

Cancel the common factor.

$15 x + 5 \cdot (- 4 (\frac{- 1}{5})) - 20 (2 x) = - 1$

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

Rewrite the expression.

$15 x - 4 \cdot - 1 - 20 (2 x) = - 1$

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

$15 x - 4 \cdot - 1 - 20 (2 x) = - 1$

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

Multiply $- 4$ by $- 1$ .

$15 x + 4 - 20 (2 x) = - 1$

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

Multiply $2$ by $- 20$ .

$15 x + 4 - 40 x = - 1$

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

$15 x + 4 - 40 x = - 1$

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

Subtract $40 x$ from $15 x$ .

$- 25 x + 4 = - 1$

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

$- 25 x + 4 = - 1$

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

$- 25 x + 4 = - 1$

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

$- 25 x + 4 = - 1$

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

Step 3

Solve for $x$ in $- 25 x + 4 = - 1$ .

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

Subtract $4$ from both sides of the equation.

$- 25 x = - 1 - 4$

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

Subtract $4$ from $- 1$ .

$- 25 x = - 5$

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

$- 25 x = - 5$

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

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

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

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

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

Simplify the left side.

Cancel the common factor of $- 25$ .

Cancel the common factor.

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

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

Divide $x$ by $1$ .

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

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

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

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

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

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

Simplify the right side.

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

Factor $- 5$ out of $- 5$ .

$x = \frac{- 5 \cdot 1}{- 25}$

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

Cancel the common factors.

Factor $- 5$ out of $- 25$ .

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

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

Cancel the common factor.

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

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

Rewrite the expression.

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

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

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

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

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

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

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

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

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

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

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

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

Step 4

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

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

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

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

Simplify the right side.

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

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

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

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

Combine the numerators over the common denominator.

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

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

Add $- 1$ and $2$ .

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

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

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

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

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

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

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

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

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-20y=-1 , 5y=-1+10x

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