Solve for x 2/3(9x-15)=2(3x-5)

$\frac{2}{3} (9 x - 15) = 2 (3 x - 5)$

Simplify $\frac{2}{3} \cdot (9 x - 15)$ .

Start simplifying.

$0 + 0 + \frac{2}{3} \cdot (9 x - 15) = 2 (3 x - 5)$

Simplify by adding zeros.

$\frac{2}{3} \cdot (9 x - 15) = 2 (3 x - 5)$

Apply the distributive property.

$\frac{2}{3} (9 x) + \frac{2}{3} \cdot - 15 = 2 (3 x - 5)$

Cancel the common factor of $3$ .

Factor $3$ out of $9 x$ .

$\frac{2}{3} (3 (3 x)) + \frac{2}{3} \cdot - 15 = 2 (3 x - 5)$

Cancel the common factor.

$\frac{2}{3} (3 (3 x)) + \frac{2}{3} \cdot - 15 = 2 (3 x - 5)$

Rewrite the expression.

$2 (3 x) + \frac{2}{3} \cdot - 15 = 2 (3 x - 5)$

Multiply $3$ by $2$ .

$6 x + \frac{2}{3} \cdot - 15 = 2 (3 x - 5)$

Cancel the common factor of $3$ .

Factor $3$ out of $- 15$ .

$6 x + \frac{2}{3} \cdot (3 (- 5)) = 2 (3 x - 5)$

Cancel the common factor.

$6 x + \frac{2}{3} \cdot (3 \cdot - 5) = 2 (3 x - 5)$

Rewrite the expression.

$6 x + 2 \cdot - 5 = 2 (3 x - 5)$

Multiply $2$ by $- 5$ .

$6 x - 10 = 2 (3 x - 5)$

Simplify $2 (3 x - 5)$ .

Apply the distributive property.

$6 x - 10 = 2 (3 x) + 2 \cdot - 5$

Multiply.

Multiply $3$ by $2$ .

$6 x - 10 = 6 x + 2 \cdot - 5$

Multiply $2$ by $- 5$ .

$6 x - 10 = 6 x - 10$

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

Subtract $6 x$ from both sides of the equation.

$6 x - 10 - 6 x = - 10$

Combine the opposite terms in $6 x - 10 - 6 x$ .

Subtract $6 x$ from $6 x$ .

$0 - 10 = - 10$

Subtract $10$ from $0$ .

$- 10 = - 10$

Since $- 10 = - 10$ , the equation will always be true for any value of $x$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for x 2/3(9x-15)=2(3x-5)

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