Solve for x x+(2/3x-15)=180

$x + (\frac{2}{3} x - 15) = 180$

Step 1

Simplify $x + \frac{2}{3} x - 15$ .

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

$x + \frac{2 x}{3} - 15 = 180$

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

$x \cdot \frac{3}{3} + \frac{2 x}{3} - 15 = 180$

Combine $x$ and $\frac{3}{3}$ .

$\frac{x \cdot 3}{3} + \frac{2 x}{3} - 15 = 180$

Combine the numerators over the common denominator.

$\frac{x \cdot 3 + 2 x}{3} - 15 = 180$

Add $x \cdot 3$ and $2 x$ .

Reorder $x$ and $3$ .

$\frac{3 \cdot x + 2 x}{3} - 15 = 180$

Add $3 \cdot x$ and $2 x$ .

$\frac{5 \cdot x}{3} - 15 = 180$

$\frac{5 x}{3} - 15 = 180$

Step 2

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

Add $15$ to both sides of the equation.

$\frac{5 x}{3} = 180 + 15$

Add $180$ and $15$ .

$\frac{5 x}{3} = 195$

Step 3

Multiply both sides of the equation by $\frac{3}{5}$ .

$\frac{3}{5} \cdot \frac{5 x}{3} = \frac{3}{5} \cdot 195$

Step 4

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{3}{5} \cdot \frac{5 x}{3}$ .

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3}{5} \cdot \frac{5 x}{3} = \frac{3}{5} \cdot 195$

Rewrite the expression.

$\frac{1}{5} (5 x) = \frac{3}{5} \cdot 195$

Cancel the common factor of $5$ .

Factor $5$ out of $5 x$ .

$\frac{1}{5} (5 (x)) = \frac{3}{5} \cdot 195$

Cancel the common factor.

$\frac{1}{5} (5 x) = \frac{3}{5} \cdot 195$

Rewrite the expression.

$x = \frac{3}{5} \cdot 195$

Simplify the right side.

Simplify $\frac{3}{5} \cdot 195$ .

Cancel the common factor of $5$ .

Factor $5$ out of $195$ .

$x = \frac{3}{5} \cdot (5 (39))$

Cancel the common factor.

$x = \frac{3}{5} \cdot (5 \cdot 39)$

Rewrite the expression.

$x = 3 \cdot 39$

Multiply $3$ by $39$ .

$x = 117$

Solve for x x+(2/3x-15)=180

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