Solve by Substitution 2x+6y=32 , 5x-2y=12

$2 x + 6 y = 32$ , $5 x - 2 y = 12$

Step 1

Solve for $x$ in $2 x + 6 y = 32$ .

Subtract $6 y$ from both sides of the equation.

$2 x = 32 - 6 y$

$5 x - 2 y = 12$

Divide each term in $2 x = 32 - 6 y$ by $2$ and simplify.

Divide each term in $2 x = 32 - 6 y$ by $2$ .

$\frac{2 x}{2} = \frac{32}{2} + \frac{- 6 y}{2}$

$5 x - 2 y = 12$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{32}{2} + \frac{- 6 y}{2}$

$5 x - 2 y = 12$

Divide $x$ by $1$ .

$x = \frac{32}{2} + \frac{- 6 y}{2}$

$5 x - 2 y = 12$

$x = \frac{32}{2} + \frac{- 6 y}{2}$

$5 x - 2 y = 12$

$x = \frac{32}{2} + \frac{- 6 y}{2}$

$5 x - 2 y = 12$

Simplify the right side.

Simplify each term.

Divide $32$ by $2$ .

$x = 16 + \frac{- 6 y}{2}$

$5 x - 2 y = 12$

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6 y$ .

$x = 16 + \frac{2 (- 3 y)}{2}$

$5 x - 2 y = 12$

Cancel the common factors.

Factor $2$ out of $2$ .

$x = 16 + \frac{2 (- 3 y)}{2 (1)}$

$5 x - 2 y = 12$

Cancel the common factor.

$x = 16 + \frac{2 (- 3 y)}{2 \cdot 1}$

$5 x - 2 y = 12$

Rewrite the expression.

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

$5 x - 2 y = 12$

Divide $- 3 y$ by $1$ .

$x = 16 - 3 y$

$5 x - 2 y = 12$

$x = 16 - 3 y$

$5 x - 2 y = 12$

$x = 16 - 3 y$

$5 x - 2 y = 12$

$x = 16 - 3 y$

$5 x - 2 y = 12$

$x = 16 - 3 y$

$5 x - 2 y = 12$

$x = 16 - 3 y$

$5 x - 2 y = 12$

$x = 16 - 3 y$

$5 x - 2 y = 12$

Step 2

Replace all occurrences of $x$ with $16 - 3 y$ in each equation.

Replace all occurrences of $x$ in $5 x - 2 y = 12$ with $16 - 3 y$ .

$5 (16 - 3 y) - 2 y = 12$

$x = 16 - 3 y$

Simplify the left side.

Simplify $5 (16 - 3 y) - 2 y$ .

Simplify each term.

Apply the distributive property.

$5 \cdot 16 + 5 (- 3 y) - 2 y = 12$

$x = 16 - 3 y$

Multiply $5$ by $16$ .

$80 + 5 (- 3 y) - 2 y = 12$

$x = 16 - 3 y$

Multiply $- 3$ by $5$ .

$80 - 15 y - 2 y = 12$

$x = 16 - 3 y$

$80 - 15 y - 2 y = 12$

$x = 16 - 3 y$

Subtract $2 y$ from $- 15 y$ .

$80 - 17 y = 12$

$x = 16 - 3 y$

$80 - 17 y = 12$

$x = 16 - 3 y$

$80 - 17 y = 12$

$x = 16 - 3 y$

$80 - 17 y = 12$

$x = 16 - 3 y$

Step 3

Solve for $y$ in $80 - 17 y = 12$ .

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

Subtract $80$ from both sides of the equation.

$- 17 y = 12 - 80$

$x = 16 - 3 y$

Subtract $80$ from $12$ .

$- 17 y = - 68$

$x = 16 - 3 y$

$- 17 y = - 68$

$x = 16 - 3 y$

Divide each term in $- 17 y = - 68$ by $- 17$ and simplify.

Divide each term in $- 17 y = - 68$ by $- 17$ .

$\frac{- 17 y}{- 17} = \frac{- 68}{- 17}$

$x = 16 - 3 y$

Simplify the left side.

Cancel the common factor of $- 17$ .

Cancel the common factor.

$\frac{- 17 y}{- 17} = \frac{- 68}{- 17}$

$x = 16 - 3 y$

Divide $y$ by $1$ .

$y = \frac{- 68}{- 17}$

$x = 16 - 3 y$

$y = \frac{- 68}{- 17}$

$x = 16 - 3 y$

$y = \frac{- 68}{- 17}$

$x = 16 - 3 y$

Simplify the right side.

Divide $- 68$ by $- 17$ .

$y = 4$

$x = 16 - 3 y$

$y = 4$

$x = 16 - 3 y$

$y = 4$

$x = 16 - 3 y$

$y = 4$

$x = 16 - 3 y$

Step 4

Replace all occurrences of $y$ with $4$ in each equation.

Replace all occurrences of $y$ in $x = 16 - 3 y$ with $4$ .

$x = 16 - 3 \cdot 4$

$y = 4$

Simplify the right side.

Simplify $16 - 3 \cdot 4$ .

Multiply $- 3$ by $4$ .

$x = 16 - 12$

$y = 4$

Subtract $12$ from $16$ .

$x = 4$

$y = 4$

$x = 4$

$y = 4$

$x = 4$

$y = 4$

$x = 4$

$y = 4$

Step 5

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

$(4, 4)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 4)$

Equation Form:

$x = 4, y = 4$

Step 7

Solve by Substitution 2x+6y=32 , 5x-2y=12

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