Solve the System of Equations 3x+4y=3 and 2x-4y=12

$3 x + 4 y = 3$ and $2 x - 4 y = 12$

Solve for $x$ in $3 x + 4 y = 3$ .

Subtract $4 y$ from both sides of the equation.

$3 x = 3 - 4 y$

$2 x - 4 y = 12$

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

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

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

$2 x - 4 y = 12$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

$2 x - 4 y = 12$

Divide $x$ by $1$ .

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

$2 x - 4 y = 12$

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

$2 x - 4 y = 12$

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

$2 x - 4 y = 12$

Simplify the right side.

Simplify each term.

Divide $3$ by $3$ .

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

$2 x - 4 y = 12$

Move the negative in front of the fraction.

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

$2 x - 4 y = 12$

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

$2 x - 4 y = 12$

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

$2 x - 4 y = 12$

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

$2 x - 4 y = 12$

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

$2 x - 4 y = 12$

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

Replace all occurrences of $x$ in $2 x - 4 y = 12$ with $1 - \frac{4 y}{3}$ .

$2 (1 - \frac{4 y}{3}) - 4 y = 12$

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

Simplify the left side.

Simplify $2 (1 - \frac{4 y}{3}) - 4 y$ .

Simplify each term.

Apply the distributive property.

$2 \cdot 1 + 2 (- \frac{4 y}{3}) - 4 y = 12$

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

Multiply $2$ by $1$ .

$2 + 2 (- \frac{4 y}{3}) - 4 y = 12$

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

Multiply $2 (- \frac{4 y}{3})$ .

Multiply $- 1$ by $2$ .

$2 - 2 \frac{4 y}{3} - 4 y = 12$

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

Combine $- 2$ and $\frac{4 y}{3}$ .

$2 + \frac{- 2 (4 y)}{3} - 4 y = 12$

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

Multiply $4$ by $- 2$ .

$2 + \frac{- 8 y}{3} - 4 y = 12$

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

$2 + \frac{- 8 y}{3} - 4 y = 12$

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

Move the negative in front of the fraction.

$2 - \frac{8 y}{3} - 4 y = 12$

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

$2 - \frac{8 y}{3} - 4 y = 12$

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

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

$2 - \frac{8 y}{3} - 4 y \cdot \frac{3}{3} = 12$

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

Simplify terms.

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

$2 - \frac{8 y}{3} + \frac{- 4 y \cdot 3}{3} = 12$

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

Combine the numerators over the common denominator.

$2 + \frac{- 8 y - 4 y \cdot 3}{3} = 12$

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

$2 + \frac{- 8 y - 4 y \cdot 3}{3} = 12$

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

Simplify each term.

Simplify the numerator.

Factor $4 y$ out of $- 8 y - 4 y \cdot 3$ .

Factor $4 y$ out of $- 8 y$ .

$2 + \frac{4 y (- 2) - 4 y \cdot 3}{3} = 12$

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

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

$2 + \frac{4 y (- 2) + 4 y (- 1 \cdot 3)}{3} = 12$

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

Factor $4 y$ out of $4 y (- 2) + 4 y (- 1 \cdot 3)$ .

$2 + \frac{4 y (- 2 - 1 \cdot 3)}{3} = 12$

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

$2 + \frac{4 y (- 2 - 1 \cdot 3)}{3} = 12$

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

Multiply $- 1$ by $3$ .

$2 + \frac{4 y (- 2 - 3)}{3} = 12$

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

Subtract $3$ from $- 2$ .

$2 + \frac{4 y \cdot - 5}{3} = 12$

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

Multiply $- 5$ by $4$ .

$2 + \frac{- 20 y}{3} = 12$

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

$2 + \frac{- 20 y}{3} = 12$

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

Move the negative in front of the fraction.

$2 - \frac{20 y}{3} = 12$

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

$2 - \frac{20 y}{3} = 12$

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

$2 - \frac{20 y}{3} = 12$

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

$2 - \frac{20 y}{3} = 12$

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

$2 - \frac{20 y}{3} = 12$

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

Solve for $y$ in $2 - \frac{20 y}{3} = 12$ .

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

Subtract $2$ from both sides of the equation.

$- \frac{20 y}{3} = 12 - 2$

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

Subtract $2$ from $12$ .

$- \frac{20 y}{3} = 10$

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

$- \frac{20 y}{3} = 10$

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

Multiply both sides of the equation by $- \frac{3}{20}$ .

$- \frac{3}{20} \cdot (- \frac{20 y}{3}) = - \frac{3}{20} \cdot 10$

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

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{3}{20} (- \frac{20 y}{3})$ .

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{20}$ into the numerator.

$\frac{- 3}{20} \cdot (- \frac{20 y}{3}) = - \frac{3}{20} \cdot 10$

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

Move the leading negative in $- \frac{20 y}{3}$ into the numerator.

$\frac{- 3}{20} \cdot \frac{- 20 y}{3} = - \frac{3}{20} \cdot 10$

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

Factor $3$ out of $- 3$ .

$\frac{3 (- 1)}{20} \cdot \frac{- 20 y}{3} = - \frac{3}{20} \cdot 10$

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

Cancel the common factor.

$\frac{3 \cdot - 1}{20} \cdot \frac{- 20 y}{3} = - \frac{3}{20} \cdot 10$

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

Rewrite the expression.

$\frac{- 1}{20} \cdot (- 20 y) = - \frac{3}{20} \cdot 10$

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

$\frac{- 1}{20} \cdot (- 20 y) = - \frac{3}{20} \cdot 10$

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

Cancel the common factor of $20$ .

Factor $20$ out of $- 20 y$ .

$\frac{- 1}{20} \cdot (20 (- y)) = - \frac{3}{20} \cdot 10$

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

Cancel the common factor.

$\frac{- 1}{20} \cdot (20 (- y)) = - \frac{3}{20} \cdot 10$

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

Rewrite the expression.

$y = - \frac{3}{20} \cdot 10$

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

$y = - \frac{3}{20} \cdot 10$

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

Multiply.

Multiply $- 1$ by $- 1$ .

$1 y = - \frac{3}{20} \cdot 10$

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

Multiply $y$ by $1$ .

$y = - \frac{3}{20} \cdot 10$

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

$y = - \frac{3}{20} \cdot 10$

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

$y = - \frac{3}{20} \cdot 10$

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

$y = - \frac{3}{20} \cdot 10$

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

Simplify the right side.

Simplify $- \frac{3}{20} \cdot 10$ .

Cancel the common factor of $10$ .

Move the leading negative in $- \frac{3}{20}$ into the numerator.

$y = \frac{- 3}{20} \cdot 10$

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

Factor $10$ out of $20$ .

$y = \frac{- 3}{10 (2)} \cdot 10$

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

Cancel the common factor.

$y = \frac{- 3}{10 \cdot 2} \cdot 10$

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

Rewrite the expression.

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

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

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

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

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

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

Replace all occurrences of $y$ in $x = 1 - \frac{4 y}{3}$ with $- \frac{3}{2}$ .

$x = 1 - \frac{4 (- \frac{3}{2})}{3}$

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

Simplify the right side.

Simplify $1 - \frac{4 (- \frac{3}{2})}{3}$ .

Simplify each term.

Simplify the numerator.

Multiply $- 1$ by $4$ .

$x = 1 - \frac{- 4 (\frac{3}{2})}{3}$

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

Combine $- 4$ and $\frac{3}{2}$ .

$x = 1 - \frac{\frac{- 4 \cdot 3}{2}}{3}$

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

$x = 1 - \frac{\frac{- 4 \cdot 3}{2}}{3}$

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

Multiply $- 4$ by $3$ .

$x = 1 - \frac{\frac{- 12}{2}}{3}$

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

Divide $- 12$ by $2$ .

$x = 1 - \frac{- 6}{3}$

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

Divide $- 6$ by $3$ .

$x = 1 + 2$

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

Multiply $- 1$ by $- 2$ .

$x = 1 + 2$

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

$x = 1 + 2$

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

Add $1$ and $2$ .

$x = 3$

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

$x = 3$

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

$x = 3$

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

$x = 3$

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

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

$(3, - \frac{3}{2})$

The result can be shown in multiple forms.

Point Form:

$(3, - \frac{3}{2})$

Equation Form:

$x = 3, y = - \frac{3}{2}$

Solve the System of Equations 3x+4y=3 and 2x-4y=12

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