Solve by Substitution 3x-7y=33 , -5x+4y=-55

$3 x - 7 y = 33$ , $- 5 x + 4 y = - 55$

Step 1

Solve for $x$ in $3 x - 7 y = 33$ .

Add $7 y$ to both sides of the equation.

$3 x = 33 + 7 y$

$- 5 x + 4 y = - 55$

Divide each term in $3 x = 33 + 7 y$ by $3$ and simplify.

Divide each term in $3 x = 33 + 7 y$ by $3$ .

$\frac{3 x}{3} = \frac{33}{3} + \frac{7 y}{3}$

$- 5 x + 4 y = - 55$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{33}{3} + \frac{7 y}{3}$

$- 5 x + 4 y = - 55$

Divide $x$ by $1$ .

$x = \frac{33}{3} + \frac{7 y}{3}$

$- 5 x + 4 y = - 55$

$x = \frac{33}{3} + \frac{7 y}{3}$

$- 5 x + 4 y = - 55$

$x = \frac{33}{3} + \frac{7 y}{3}$

$- 5 x + 4 y = - 55$

Simplify the right side.

Divide $33$ by $3$ .

$x = 11 + \frac{7 y}{3}$

$- 5 x + 4 y = - 55$

$x = 11 + \frac{7 y}{3}$

$- 5 x + 4 y = - 55$

$x = 11 + \frac{7 y}{3}$

$- 5 x + 4 y = - 55$

$x = 11 + \frac{7 y}{3}$

$- 5 x + 4 y = - 55$

Step 2

Replace all occurrences of $x$ with $11 + \frac{7 y}{3}$ in each equation.

Replace all occurrences of $x$ in $- 5 x + 4 y = - 55$ with $11 + \frac{7 y}{3}$ .

$- 5 (11 + \frac{7 y}{3}) + 4 y = - 55$

$x = 11 + \frac{7 y}{3}$

Simplify the left side.

Simplify $- 5 (11 + \frac{7 y}{3}) + 4 y$ .

Simplify each term.

Apply the distributive property.

$- 5 \cdot 11 - 5 \frac{7 y}{3} + 4 y = - 55$

$x = 11 + \frac{7 y}{3}$

Multiply $- 5$ by $11$ .

$- 55 - 5 \frac{7 y}{3} + 4 y = - 55$

$x = 11 + \frac{7 y}{3}$

Multiply $- 5 \frac{7 y}{3}$ .

Combine $- 5$ and $\frac{7 y}{3}$ .

$- 55 + \frac{- 5 (7 y)}{3} + 4 y = - 55$

$x = 11 + \frac{7 y}{3}$

Multiply $7$ by $- 5$ .

$- 55 + \frac{- 35 y}{3} + 4 y = - 55$

$x = 11 + \frac{7 y}{3}$

$- 55 + \frac{- 35 y}{3} + 4 y = - 55$

$x = 11 + \frac{7 y}{3}$

Move the negative in front of the fraction.

$- 55 - \frac{35 y}{3} + 4 y = - 55$

$x = 11 + \frac{7 y}{3}$

$- 55 - \frac{35 y}{3} + 4 y = - 55$

$x = 11 + \frac{7 y}{3}$

To write $4 y$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- 55 - \frac{35 y}{3} + 4 y \cdot \frac{3}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

Simplify terms.

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

$- 55 - \frac{35 y}{3} + \frac{4 y \cdot 3}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

Combine the numerators over the common denominator.

$- 55 + \frac{- 35 y + 4 y \cdot 3}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

$- 55 + \frac{- 35 y + 4 y \cdot 3}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

Simplify each term.

Simplify the numerator.

Factor $y$ out of $- 35 y + 4 y \cdot 3$ .

Factor $y$ out of $- 35 y$ .

$- 55 + \frac{y \cdot - 35 + 4 y \cdot 3}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

Factor $y$ out of $4 y \cdot 3$ .

$- 55 + \frac{y \cdot - 35 + y (4 \cdot 3)}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

Factor $y$ out of $y \cdot - 35 + y (4 \cdot 3)$ .

$- 55 + \frac{y (- 35 + 4 \cdot 3)}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

$- 55 + \frac{y (- 35 + 4 \cdot 3)}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

Multiply $4$ by $3$ .

$- 55 + \frac{y (- 35 + 12)}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

Add $- 35$ and $12$ .

$- 55 + \frac{y \cdot - 23}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

$- 55 + \frac{y \cdot - 23}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

Move $- 23$ to the left of $y$ .

$- 55 + \frac{- 23 \cdot y}{3} = - 55$

$x = 11 + \frac{7 y}{3}$

Move the negative in front of the fraction.