Solve the System of Equations 8x+2y=6 3y=-12x+9

$8 x + 2 y = 6$ $3 y = - 12 x + 9$

Divide each term in $3 y = - 12 x + 9$ by $3$ and simplify.

Divide each term in $3 y = - 12 x + 9$ by $3$ .

$\frac{3 y}{3} = \frac{- 12 x}{3} + \frac{9}{3}$

$8 x + 2 y = 6$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{- 12 x}{3} + \frac{9}{3}$

$8 x + 2 y = 6$

Divide $y$ by $1$ .

$y = \frac{- 12 x}{3} + \frac{9}{3}$

$8 x + 2 y = 6$

$y = \frac{- 12 x}{3} + \frac{9}{3}$

$8 x + 2 y = 6$

$y = \frac{- 12 x}{3} + \frac{9}{3}$

$8 x + 2 y = 6$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 12$ and $3$ .

Factor $3$ out of $- 12 x$ .

$y = \frac{3 (- 4 x)}{3} + \frac{9}{3}$

$8 x + 2 y = 6$

Cancel the common factors.

Factor $3$ out of $3$ .

$y = \frac{3 (- 4 x)}{3 (1)} + \frac{9}{3}$

$8 x + 2 y = 6$

Cancel the common factor.

$y = \frac{3 (- 4 x)}{3 \cdot 1} + \frac{9}{3}$

$8 x + 2 y = 6$

Rewrite the expression.

$y = \frac{- 4 x}{1} + \frac{9}{3}$

$8 x + 2 y = 6$

Divide $- 4 x$ by $1$ .

$y = - 4 x + \frac{9}{3}$

$8 x + 2 y = 6$

$y = - 4 x + \frac{9}{3}$

$8 x + 2 y = 6$

$y = - 4 x + \frac{9}{3}$

$8 x + 2 y = 6$

Divide $9$ by $3$ .

$y = - 4 x + 3$

$8 x + 2 y = 6$

$y = - 4 x + 3$

$8 x + 2 y = 6$

$y = - 4 x + 3$

$8 x + 2 y = 6$

$y = - 4 x + 3$

$8 x + 2 y = 6$

Replace all occurrences of $y$ with $- 4 x + 3$ in each equation.

Replace all occurrences of $y$ in $8 x + 2 y = 6$ with $- 4 x + 3$ .

$8 x + 2 (- 4 x + 3) = 6$

$y = - 4 x + 3$

Simplify the left side.

Simplify $8 x + 2 (- 4 x + 3)$ .

Simplify each term.

Apply the distributive property.

$8 x + 2 (- 4 x) + 2 \cdot 3 = 6$

$y = - 4 x + 3$

Multiply $- 4$ by $2$ .

$8 x - 8 x + 2 \cdot 3 = 6$

$y = - 4 x + 3$

Multiply $2$ by $3$ .

$8 x - 8 x + 6 = 6$

$y = - 4 x + 3$

$8 x - 8 x + 6 = 6$

$y = - 4 x + 3$

Simplify by adding terms.

Subtract $8 x$ from $8 x$ .

$0 + 6 = 6$

$y = - 4 x + 3$

Add $0$ and $6$ .

$6 = 6$

$y = - 4 x + 3$

$6 = 6$

$y = - 4 x + 3$

$6 = 6$

$y = - 4 x + 3$

$6 = 6$

$y = - 4 x + 3$

$6 = 6$

$y = - 4 x + 3$

Remove any equations from the system that are always true.

$y = - 4 x + 3$

Solve the System of Equations 8x+2y=6 3y=-12x+9

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