Solve by Substitution 3x=4y+1 , 3y=1-4x

$3 x = 4 y + 1$ , $3 y = 1 - 4 x$

Step 1

Divide each term in $3 x = 4 y + 1$ by $3$ and simplify.

Divide each term in $3 x = 4 y + 1$ by $3$ .

$\frac{3 x}{3} = \frac{4 y}{3} + \frac{1}{3}$

$3 y = 1 - 4 x$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{4 y}{3} + \frac{1}{3}$

$3 y = 1 - 4 x$

Divide $x$ by $1$ .

$x = \frac{4 y}{3} + \frac{1}{3}$

$3 y = 1 - 4 x$

$x = \frac{4 y}{3} + \frac{1}{3}$

$3 y = 1 - 4 x$

$x = \frac{4 y}{3} + \frac{1}{3}$

$3 y = 1 - 4 x$

$x = \frac{4 y}{3} + \frac{1}{3}$

$3 y = 1 - 4 x$

Step 2

Replace all occurrences of $x$ with $\frac{4 y}{3} + \frac{1}{3}$ in each equation.

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

$3 y = 1 - 4 (\frac{4 y}{3} + \frac{1}{3})$

$x = \frac{4 y}{3} + \frac{1}{3}$

Simplify the right side.

Simplify $1 - 4 (\frac{4 y}{3} + \frac{1}{3})$ .

Simplify each term.

Apply the distributive property.

$3 y = 1 - 4 \frac{4 y}{3} - 4 (\frac{1}{3})$

$x = \frac{4 y}{3} + \frac{1}{3}$

Multiply $- 4 \frac{4 y}{3}$ .

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

$3 y = 1 + \frac{- 4 (4 y)}{3} - 4 (\frac{1}{3})$

$x = \frac{4 y}{3} + \frac{1}{3}$

Multiply $4$ by $- 4$ .

$3 y = 1 + \frac{- 16 y}{3} - 4 (\frac{1}{3})$

$x = \frac{4 y}{3} + \frac{1}{3}$

$3 y = 1 + \frac{- 16 y}{3} - 4 (\frac{1}{3})$

$x = \frac{4 y}{3} + \frac{1}{3}$

Combine $- 4$ and $\frac{1}{3}$ .

$3 y = 1 + \frac{- 16 y}{3} + \frac{- 4}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Simplify each term.

Move the negative in front of the fraction.

$3 y = 1 - \frac{(16) y}{3} + \frac{- 4}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Move the negative in front of the fraction.

$3 y = 1 - \frac{16 y}{3} - \frac{4}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$3 y = 1 - \frac{16 y}{3} - \frac{4}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$3 y = 1 - \frac{16 y}{3} - \frac{4}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Simplify the expression.

Write $1$ as a fraction with a common denominator.

$3 y = - \frac{16 y}{3} + \frac{3}{3} - \frac{4}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Combine the numerators over the common denominator.

$3 y = - \frac{16 y}{3} + \frac{3 - 4}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Subtract $4$ from $3$ .

$3 y = - \frac{16 y}{3} + \frac{- 1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Move the negative in front of the fraction.

$3 y = - \frac{16 y}{3} - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$3 y = - \frac{16 y}{3} - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$3 y = - \frac{16 y}{3} - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$3 y = - \frac{16 y}{3} - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$3 y = - \frac{16 y}{3} - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Step 3

Solve for $y$ in $3 y = - \frac{16 y}{3} - \frac{1}{3}$ .

Move all terms containing $y$ to the left side of the equation.

Add $\frac{16 y}{3}$ to both sides of the equation.

$3 y + \frac{16 y}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

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

$3 y \cdot \frac{3}{3} + \frac{16 y}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

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

$\frac{3 y \cdot 3}{3} + \frac{16 y}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Combine the numerators over the common denominator.

$\frac{3 y \cdot 3 + 16 y}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Simplify the numerator.

Factor $y$ out of $3 y \cdot 3 + 16 y$ .

Factor $y$ out of $3 y \cdot 3$ .

$\frac{y (3 \cdot 3) + 16 y}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Factor $y$ out of $16 y$ .

$\frac{y (3 \cdot 3) + y \cdot 16}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Factor $y$ out of $y (3 \cdot 3) + y \cdot 16$ .

$\frac{y (3 \cdot 3 + 16)}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$\frac{y (3 \cdot 3 + 16)}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Multiply $3$ by $3$ .

$\frac{y (9 + 16)}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Add $9$ and $16$ .

$\frac{y \cdot 25}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$\frac{y \cdot 25}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Move $25$ to the left of $y$ .

$\frac{25 y}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$\frac{25 y}{3} = - \frac{1}{3}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$25 y = - 1$

$x = \frac{4 y}{3} + \frac{1}{3}$

Divide each term in $25 y = - 1$ by $25$ and simplify.

Divide each term in $25 y = - 1$ by $25$ .

$\frac{25 y}{25} = \frac{- 1}{25}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $25$ .

Cancel the common factor.

$\frac{25 y}{25} = \frac{- 1}{25}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Divide $y$ by $1$ .

$y = \frac{- 1}{25}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$y = \frac{- 1}{25}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$y = \frac{- 1}{25}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{1}{25}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$y = - \frac{1}{25}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$y = - \frac{1}{25}$

$x = \frac{4 y}{3} + \frac{1}{3}$

$y = - \frac{1}{25}$

$x = \frac{4 y}{3} + \frac{1}{3}$

Step 4

Replace all occurrences of $y$ with $- \frac{1}{25}$ in each equation.

Replace all occurrences of $y$ in $x = \frac{4 y}{3} + \frac{1}{3}$ with $- \frac{1}{25}$ .

$x = \frac{4 (- \frac{1}{25})}{3} + \frac{1}{3}$

$y = - \frac{1}{25}$

Simplify the right side.

Simplify $\frac{4 (- \frac{1}{25})}{3} + \frac{1}{3}$ .

Combine the numerators over the common denominator.

$x = \frac{4 (- \frac{1}{25}) + 1}{3}$

$y = - \frac{1}{25}$

Simplify each term.

Multiply $4 (- \frac{1}{25})$ .

Multiply $- 1$ by $4$ .

$x = \frac{- 4 (\frac{1}{25}) + 1}{3}$

$y = - \frac{1}{25}$

Combine $- 4$ and $\frac{1}{25}$ .

$x = \frac{\frac{- 4}{25} + 1}{3}$

$y = - \frac{1}{25}$

$x = \frac{\frac{- 4}{25} + 1}{3}$

$y = - \frac{1}{25}$

Move the negative in front of the fraction.

$x = \frac{- \frac{4}{25} + 1}{3}$

$y = - \frac{1}{25}$

$x = \frac{- \frac{4}{25} + 1}{3}$

$y = - \frac{1}{25}$

Simplify the expression.

Write $1$ as a fraction with a common denominator.

$x = \frac{- \frac{4}{25} + \frac{25}{25}}{3}$

$y = - \frac{1}{25}$

Combine the numerators over the common denominator.

$x = \frac{\frac{- 4 + 25}{25}}{3}$

$y = - \frac{1}{25}$

Add $- 4$ and $25$ .

$x = \frac{\frac{21}{25}}{3}$

$y = - \frac{1}{25}$

$x = \frac{\frac{21}{25}}{3}$

$y = - \frac{1}{25}$

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{21}{25} \cdot \frac{1}{3}$

$y = - \frac{1}{25}$

Cancel the common factor of $3$ .

Factor $3$ out of $21$ .

$x = \frac{3 (7)}{25} \cdot \frac{1}{3}$

$y = - \frac{1}{25}$

Cancel the common factor.

$x = \frac{3 \cdot 7}{25} \cdot \frac{1}{3}$

$y = - \frac{1}{25}$

Rewrite the expression.

$x = \frac{7}{25}$

$y = - \frac{1}{25}$

$x = \frac{7}{25}$

$y = - \frac{1}{25}$

$x = \frac{7}{25}$

$y = - \frac{1}{25}$

$x = \frac{7}{25}$

$y = - \frac{1}{25}$

$x = \frac{7}{25}$

$y = - \frac{1}{25}$

Step 5

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

$(\frac{7}{25}, - \frac{1}{25})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{7}{25}, - \frac{1}{25})$

Equation Form:

$x = \frac{7}{25}, y = - \frac{1}{25}$

Step 7

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