Solve the System of Equations 3x+4y=6 and -3x+5y=12

$3 x + 4 y = 6$ and $- 3 x + 5 y = 12$

Solve for $x$ in $3 x + 4 y = 6$ .

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

$3 x = 6 - 4 y$

$- 3 x + 5 y = 12$

Divide each term in $3 x = 6 - 4 y$ by $3$ and simplify.

Divide each term in $3 x = 6 - 4 y$ by $3$ .

$\frac{3 x}{3} = \frac{6}{3} + \frac{- 4 y}{3}$

$- 3 x + 5 y = 12$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{6}{3} + \frac{- 4 y}{3}$

$- 3 x + 5 y = 12$

Divide $x$ by $1$ .

$x = \frac{6}{3} + \frac{- 4 y}{3}$

$- 3 x + 5 y = 12$

$x = \frac{6}{3} + \frac{- 4 y}{3}$

$- 3 x + 5 y = 12$

$x = \frac{6}{3} + \frac{- 4 y}{3}$

$- 3 x + 5 y = 12$

Simplify the right side.

Simplify each term.

Divide $6$ by $3$ .

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

$- 3 x + 5 y = 12$

Move the negative in front of the fraction.

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

$- 3 x + 5 y = 12$

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

$- 3 x + 5 y = 12$

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

$- 3 x + 5 y = 12$

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

$- 3 x + 5 y = 12$

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

$- 3 x + 5 y = 12$

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

Replace all occurrences of $x$ in $- 3 x + 5 y = 12$ with $2 - \frac{4 y}{3}$ .

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

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

Simplify the left side.

Simplify $- 3 (2 - \frac{4 y}{3}) + 5 y$ .

Simplify each term.

Apply the distributive property.

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

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

Multiply $- 3$ by $2$ .

$- 6 - 3 (- \frac{4 y}{3}) + 5 y = 12$

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

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{4 y}{3}$ into the numerator.

$- 6 - 3 \frac{- 4 y}{3} + 5 y = 12$

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

Factor $3$ out of $- 3$ .

$- 6 + 3 (- 1) (\frac{- 4 y}{3}) + 5 y = 12$

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

Cancel the common factor.

$- 6 + 3 \cdot (- 1 \frac{- 4 y}{3}) + 5 y = 12$

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

Rewrite the expression.

$- 6 - 1 (- 4 y) + 5 y = 12$

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

$- 6 - 1 (- 4 y) + 5 y = 12$

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

Multiply $- 4$ by $- 1$ .

$- 6 + 4 y + 5 y = 12$

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

$- 6 + 4 y + 5 y = 12$

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

Add $4 y$ and $5 y$ .

$- 6 + 9 y = 12$

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

$- 6 + 9 y = 12$

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

$- 6 + 9 y = 12$

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

$- 6 + 9 y = 12$

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

Solve for $y$ in $- 6 + 9 y = 12$ .

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

Add $6$ to both sides of the equation.

$9 y = 12 + 6$

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

Add $12$ and $6$ .

$9 y = 18$

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

$9 y = 18$

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

Divide each term in $9 y = 18$ by $9$ and simplify.

Divide each term in $9 y = 18$ by $9$ .

$\frac{9 y}{9} = \frac{18}{9}$

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

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 y}{9} = \frac{18}{9}$

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

Divide $y$ by $1$ .

$y = \frac{18}{9}$

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

$y = \frac{18}{9}$

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

$y = \frac{18}{9}$

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

Simplify the right side.

Divide $18$ by $9$ .

$y = 2$

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

$y = 2$

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

$y = 2$

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

$y = 2$

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

Replace all occurrences of $y$ with $2$ in each equation.

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

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

$y = 2$

Simplify the right side.

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

Multiply $4$ by $2$ .

$x = 2 - \frac{8}{3}$

$y = 2$

To write $2$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = 2 \cdot \frac{3}{3} - \frac{8}{3}$

$y = 2$

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

$x = \frac{2 \cdot 3}{3} - \frac{8}{3}$

$y = 2$

Combine the numerators over the common denominator.

$x = \frac{2 \cdot 3 - 8}{3}$

$y = 2$

Simplify the numerator.

Multiply $2$ by $3$ .

$x = \frac{6 - 8}{3}$

$y = 2$

Subtract $8$ from $6$ .

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

$y = 2$

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

$y = 2$

Move the negative in front of the fraction.

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

$y = 2$

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

$y = 2$

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

$y = 2$

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

$y = 2$

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

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

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

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

Solve the System of Equations 3x+4y=6 and -3x+5y=12

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