Solve for g (-5g+40)÷3=(2g)÷2

$(- 5 g + 40) \div 3 = 2 g \div 2$

Step 1

Multiply both sides by $3$ .

$(- 5 g + 40) \div 3 \cdot 3 = 2 g \div 2 \cdot 3$

Step 2

Simplify.

Simplify the left side.

Simplify $(- 5 g + 40) \div 3 \cdot 3$ .

Factor $5$ out of $- 5 g + 40$ .

Factor $5$ out of $- 5 g$ .

$(5 (- g) + 40) \div 3 \cdot 3 = 2 g \div 2 \cdot 3$

Factor $5$ out of $40$ .

$(5 (- g) + 5 (8)) \div 3 \cdot 3 = 2 g \div 2 \cdot 3$

Factor $5$ out of $5 (- g) + 5 (8)$ .

$5 (- g + 8) \div 3 \cdot 3 = 2 g \div 2 \cdot 3$

Rewrite the division as a fraction.

$\frac{5 (- g + 8)}{3} \cdot 3 = 2 g \div 2 \cdot 3$

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{5 (- g + 8)}{3} \cdot 3 = 2 g \div 2 \cdot 3$

Rewrite the expression.

$5 (- g + 8) = 2 g \div 2 \cdot 3$

Apply the distributive property.

$5 (- g) + 5 \cdot 8 = 2 g \div 2 \cdot 3$

Multiply.

Multiply $- 1$ by $5$ .

$- 5 g + 5 \cdot 8 = 2 g \div 2 \cdot 3$

Multiply $5$ by $8$ .

$- 5 g + 40 = 2 g \div 2 \cdot 3$

Simplify the right side.

Simplify $2 g \div 2 \cdot 3$ .

Rewrite the division as a fraction.

$- 5 g + 40 = \frac{2 g}{2} \cdot 3$

Cancel the common factor of $2$ .

Cancel the common factor.

$- 5 g + 40 = \frac{2 g}{2} \cdot 3$

Divide $g$ by $1$ .

$- 5 g + 40 = g \cdot 3$

Move $3$ to the left of $g$ .

$- 5 g + 40 = 3 g$

Step 3

Solve for $g$ .

Move all terms containing $g$ to the left side of the equation.

Subtract $3 g$ from both sides of the equation.

$- 5 g + 40 - 3 g = 0$

Subtract $3 g$ from $- 5 g$ .

$- 8 g + 40 = 0$

Subtract $40$ from both sides of the equation.

$- 8 g = - 40$

Divide each term in $- 8 g = - 40$ by $- 8$ and simplify.

Divide each term in $- 8 g = - 40$ by $- 8$ .

$\frac{- 8 g}{- 8} = \frac{- 40}{- 8}$

Simplify the left side.

Cancel the common factor of $- 8$ .

Cancel the common factor.

$\frac{- 8 g}{- 8} = \frac{- 40}{- 8}$

Divide $g$ by $1$ .

$g = \frac{- 40}{- 8}$

Simplify the right side.

Divide $- 40$ by $- 8$ .

$g = 5$

Solve for g (-5g+40)÷3=(2g)÷2

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