Solve for h 2/9h-4=-2/3h+3+5/9h

$\frac{2}{9} h - 4 = - \frac{2}{3} h + 3 + \frac{5}{9} h$

Since $h$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- \frac{2}{3} h + 3 + \frac{5}{9} h = \frac{2}{9} h - 4$

Simplify $- \frac{2}{3} h + 3 + \frac{5}{9} h$ .

Simplify each term.

Combine $h$ and $\frac{2}{3}$ .

$- \frac{h \cdot 2}{3} + 3 + \frac{5}{9} h = \frac{2}{9} h - 4$

Move $2$ to the left of $h$ .

$- \frac{2 \cdot h}{3} + 3 + \frac{5}{9} h = \frac{2}{9} h - 4$

Combine $\frac{5}{9}$ and $h$ .

$- \frac{2 h}{3} + 3 + \frac{5 h}{9} = \frac{2}{9} h - 4$

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

$- \frac{2 h}{3} \cdot \frac{3}{3} + \frac{5 h}{9} + 3 = \frac{2}{9} h - 4$

Write each expression with a common denominator of $9$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2 h}{3}$ by $\frac{3}{3}$ .

$- \frac{2 h \cdot 3}{3 \cdot 3} + \frac{5 h}{9} + 3 = \frac{2}{9} h - 4$

Multiply $3$ by $3$ .

$- \frac{2 h \cdot 3}{9} + \frac{5 h}{9} + 3 = \frac{2}{9} h - 4$

Combine the numerators over the common denominator.

$\frac{- 2 h \cdot 3 + 5 h}{9} + 3 = \frac{2}{9} h - 4$

Simplify each term.

Simplify the numerator.

Factor $h$ out of $- 2 h \cdot 3 + 5 h$ .

Factor $h$ out of $- 2 h \cdot 3$ .

$\frac{h (- 2 \cdot 3) + 5 h}{9} + 3 = \frac{2}{9} h - 4$

Factor $h$ out of $5 h$ .

$\frac{h (- 2 \cdot 3) + h \cdot 5}{9} + 3 = \frac{2}{9} h - 4$

Factor $h$ out of $h (- 2 \cdot 3) + h \cdot 5$ .

$\frac{h (- 2 \cdot 3 + 5)}{9} + 3 = \frac{2}{9} h - 4$

Multiply $- 2$ by $3$ .

$\frac{h (- 6 + 5)}{9} + 3 = \frac{2}{9} h - 4$

Add $- 6$ and $5$ .

$\frac{h \cdot - 1}{9} + 3 = \frac{2}{9} h - 4$

Move $- 1$ to the left of $h$ .

$\frac{- 1 \cdot h}{9} + 3 = \frac{2}{9} h - 4$

Move the negative in front of the fraction.

$- \frac{h}{9} + 3 = \frac{2}{9} h - 4$

Combine $\frac{2}{9}$ and $h$ .

$- \frac{h}{9} + 3 = \frac{2 h}{9} - 4$

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

Subtract $\frac{2 h}{9}$ from both sides of the equation.

$- \frac{h}{9} + 3 - \frac{2 h}{9} = - 4$

Combine the numerators over the common denominator.

$3 + \frac{- h - 2 h}{9} = - 4$

Subtract $2 h$ from $- h$ .

$3 + \frac{- 3 h}{9} = - 4$

Simplify each term.

Cancel the common factor of $- 3$ and $9$ .

Factor $3$ out of $- 3 h$ .

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

Cancel the common factors.

Factor $3$ out of $9$ .

$3 + \frac{3 (- h)}{3 (3)} = - 4$

Cancel the common factor.

$3 + \frac{3 (- h)}{3 \cdot 3} = - 4$

Rewrite the expression.

$3 + \frac{- h}{3} = - 4$

Move the negative in front of the fraction.

$3 - \frac{h}{3} = - 4$

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

Subtract $3$ from both sides of the equation.

$- \frac{h}{3} = - 4 - 3$

Subtract $3$ from $- 4$ .

$- \frac{h}{3} = - 7$

Multiply both sides of the equation by $- 3$ .

$- 3 (- \frac{h}{3}) = - 3 \cdot - 7$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- 3 (- \frac{h}{3})$ .

Cancel the common factor of $3$ .

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

$- 3 \frac{- h}{3} = - 3 \cdot - 7$

Factor $3$ out of $- 3$ .

$3 (- 1) \frac{- h}{3} = - 3 \cdot - 7$

Cancel the common factor.

$3 \cdot - 1 \frac{- h}{3} = - 3 \cdot - 7$

Rewrite the expression.

$- - h = - 3 \cdot - 7$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 h = - 3 \cdot - 7$

Multiply $h$ by $1$ .

$h = - 3 \cdot - 7$

Simplify the right side.

Multiply $- 3$ by $- 7$ .

$h = 21$

Solve for h 2/9h-4=-2/3h+3+5/9h

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