Solve the System of Equations 3x+2y=4 and 5x-2y=12

$3 x + 2 y = 4$ and $5 x - 2 y = 12$

Step 1

Solve for $x$ in $3 x + 2 y = 4$ .

Subtract $2 y$ from both sides of the equation.

$3 x = 4 - 2 y$

$5 x - 2 y = 12$

Divide each term in $3 x = 4 - 2 y$ by $3$ and simplify.

Divide each term in $3 x = 4 - 2 y$ by $3$ .

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

$5 x - 2 y = 12$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

$5 x - 2 y = 12$

Divide $x$ by $1$ .

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

$5 x - 2 y = 12$

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

$5 x - 2 y = 12$

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

$5 x - 2 y = 12$

Simplify the right side.

Move the negative in front of the fraction.

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

$5 x - 2 y = 12$

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

$5 x - 2 y = 12$

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

$5 x - 2 y = 12$

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

$5 x - 2 y = 12$

Step 2

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

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

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

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

Simplify the left side.

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

Simplify each term.

Apply the distributive property.

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

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

Multiply $5 (\frac{4}{3})$ .

Combine $5$ and $\frac{4}{3}$ .

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

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

Multiply $5$ by $4$ .

$\frac{20}{3} + 5 (- \frac{2 y}{3}) - 2 y = 12$

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

$\frac{20}{3} + 5 (- \frac{2 y}{3}) - 2 y = 12$

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

Multiply $5 (- \frac{2 y}{3})$ .

Multiply $- 1$ by $5$ .

$\frac{20}{3} - 5 \frac{2 y}{3} - 2 y = 12$

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

Combine $- 5$ and $\frac{2 y}{3}$ .

$\frac{20}{3} + \frac{- 5 (2 y)}{3} - 2 y = 12$

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

Multiply $2$ by $- 5$ .

$\frac{20}{3} + \frac{- 10 y}{3} - 2 y = 12$

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

$\frac{20}{3} + \frac{- 10 y}{3} - 2 y = 12$

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

Move the negative in front of the fraction.

$\frac{20}{3} - \frac{10 y}{3} - 2 y = 12$

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

$\frac{20}{3} - \frac{10 y}{3} - 2 y = 12$

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

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

$\frac{20}{3} - \frac{10 y}{3} - 2 y \cdot \frac{3}{3} = 12$

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

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

$\frac{20}{3} - \frac{10 y}{3} + \frac{- 2 y \cdot 3}{3} = 12$

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

Combine the numerators over the common denominator.

$\frac{20}{3} + \frac{- 10 y - 2 y \cdot 3}{3} = 12$

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

Combine the numerators over the common denominator.

$\frac{20 - 10 y - 2 y \cdot 3}{3} = 12$

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

Multiply $3$ by $- 2$ .

$\frac{20 - 10 y - 6 y}{3} = 12$

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

Subtract $6 y$ from $- 10 y$ .

$\frac{20 - 16 y}{3} = 12$

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

Factor $4$ out of $20 - 16 y$ .

Factor $4$ out of $20$ .

$\frac{4 (5) - 16 y}{3} = 12$

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

Factor $4$ out of $- 16 y$ .

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

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

Factor $4$ out of $4 (5) + 4 (- 4 y)$ .

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

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

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

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

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

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

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

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

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

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

Step 3

Solve for $y$ in $\frac{4 (5 - 4 y)}{3} = 12$ .

Multiply both sides by $3$ .

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

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

Simplify.

Simplify the left side.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

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

Rewrite the expression.

$4 (5 - 4 y) = 12 \cdot 3$

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

$4 (5 - 4 y) = 12 \cdot 3$

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

Apply the distributive property.

$4 \cdot 5 + 4 (- 4 y) = 12 \cdot 3$

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

Simplify the expression.

Multiply $4$ by $5$ .

$20 + 4 (- 4 y) = 12 \cdot 3$

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

Multiply $- 4$ by $4$ .

$20 - 16 y = 12 \cdot 3$

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

Reorder $20$ and $- 16 y$ .

$- 16 y + 20 = 12 \cdot 3$

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

$- 16 y + 20 = 12 \cdot 3$

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

$- 16 y + 20 = 12 \cdot 3$

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

$- 16 y + 20 = 12 \cdot 3$

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

Simplify the right side.

Multiply $12$ by $3$ .

$- 16 y + 20 = 36$

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

$- 16 y + 20 = 36$

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

$- 16 y + 20 = 36$

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

Solve for $y$ .

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

Subtract $20$ from both sides of the equation.

$- 16 y = 36 - 20$

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

Subtract $20$ from $36$ .

$- 16 y = 16$

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

$- 16 y = 16$

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

Divide each term in $- 16 y = 16$ by $- 16$ and simplify.

Divide each term in $- 16 y = 16$ by $- 16$ .

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

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

Simplify the left side.

Cancel the common factor of $- 16$ .

Cancel the common factor.

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

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

Divide $y$ by $1$ .

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

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

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

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

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

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

Simplify the right side.

Divide $16$ by $- 16$ .

$y = - 1$

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

$y = - 1$

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

$y = - 1$

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

$y = - 1$

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

$y = - 1$

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

Step 4

Replace all occurrences of $y$ with $- 1$ in each equation.

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

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

$y = - 1$

Simplify the right side.

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

Combine the numerators over the common denominator.

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

$y = - 1$

Simplify the expression.

Multiply $- 2$ by $- 1$ .

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

$y = - 1$

Add $4$ and $2$ .

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

$y = - 1$

Divide $6$ by $3$ .

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

Step 5

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

$(2, - 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(2, - 1)$

Equation Form:

$x = 2, y = - 1$

Step 7

Solve the System of Equations 3x+2y=4 and 5x-2y=12

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