Solve the System of Equations 3x=12-2y 12x+8y-48=0

$3 x = 12 - 2 y$ $12 x + 8 y - 48 = 0$

Step 1

Divide each term in $3 x = 12 - 2 y$ by $3$ and simplify.

Divide each term in $3 x = 12 - 2 y$ by $3$ .

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

$12 x + 8 y - 48 = 0$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

$12 x + 8 y - 48 = 0$

Divide $x$ by $1$ .

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

$12 x + 8 y - 48 = 0$

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

$12 x + 8 y - 48 = 0$

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

$12 x + 8 y - 48 = 0$

Simplify the right side.

Simplify each term.

Divide $12$ by $3$ .

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

$12 x + 8 y - 48 = 0$

Move the negative in front of the fraction.

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

$12 x + 8 y - 48 = 0$

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

$12 x + 8 y - 48 = 0$

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

$12 x + 8 y - 48 = 0$

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

$12 x + 8 y - 48 = 0$

Step 2

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

Replace all occurrences of $x$ in $12 x + 8 y - 48 = 0$ with $4 - \frac{2 y}{3}$ .

$12 (4 - \frac{2 y}{3}) + 8 y - 48 = 0$

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

Simplify the left side.

Simplify $12 (4 - \frac{2 y}{3}) + 8 y - 48$ .

Simplify each term.

Apply the distributive property.

$12 \cdot 4 + 12 (- \frac{2 y}{3}) + 8 y - 48 = 0$

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

Multiply $12$ by $4$ .

$48 + 12 (- \frac{2 y}{3}) + 8 y - 48 = 0$

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

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{2 y}{3}$ into the numerator.

$48 + 12 (\frac{- 2 y}{3}) + 8 y - 48 = 0$

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

Factor $3$ out of $12$ .

$48 + 3 (4) (\frac{- 2 y}{3}) + 8 y - 48 = 0$

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

Cancel the common factor.

$48 + 3 \cdot (4 (\frac{- 2 y}{3})) + 8 y - 48 = 0$

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

Rewrite the expression.

$48 + 4 (- 2 y) + 8 y - 48 = 0$

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

$48 + 4 (- 2 y) + 8 y - 48 = 0$

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

Multiply $- 2$ by $4$ .

$48 - 8 y + 8 y - 48 = 0$

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

$48 - 8 y + 8 y - 48 = 0$

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

Combine the opposite terms in $48 - 8 y + 8 y - 48$ .

Add $- 8 y$ and $8 y$ .

$48 + 0 - 48 = 0$

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

Add $48$ and $0$ .

$48 - 48 = 0$

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

Subtract $48$ from $48$ .

$0 = 0$

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

$0 = 0$

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

$0 = 0$

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

$0 = 0$

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

$0 = 0$

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

Step 3

Remove any equations from the system that are always true.

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

Step 4

Solve the System of Equations 3x=12-2y 12x+8y-48=0

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