Evaluate 2/3(9x+6)=6x+4

$\frac{2}{3} (9 x + 6) = 6 x + 4$

Step 1

Simplify $\frac{2}{3} \cdot (9 x + 6)$ .

Rewrite.

$0 + 0 + \frac{2}{3} \cdot (9 x + 6) = 6 x + 4$

Simplify by adding zeros.

$\frac{2}{3} \cdot (9 x + 6) = 6 x + 4$

Apply the distributive property.

$\frac{2}{3} (9 x) + \frac{2}{3} \cdot 6 = 6 x + 4$

Cancel the common factor of $3$ .

Factor $3$ out of $9 x$ .

$\frac{2}{3} (3 (3 x)) + \frac{2}{3} \cdot 6 = 6 x + 4$

Cancel the common factor.

$\frac{2}{3} (3 (3 x)) + \frac{2}{3} \cdot 6 = 6 x + 4$

Rewrite the expression.

$2 (3 x) + \frac{2}{3} \cdot 6 = 6 x + 4$

Multiply $3$ by $2$ .

$6 x + \frac{2}{3} \cdot 6 = 6 x + 4$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

$6 x + \frac{2}{3} \cdot (3 (2)) = 6 x + 4$

Cancel the common factor.

$6 x + \frac{2}{3} \cdot (3 \cdot 2) = 6 x + 4$

Rewrite the expression.

$6 x + 2 \cdot 2 = 6 x + 4$

Multiply $2$ by $2$ .

$6 x + 4 = 6 x + 4$

Step 2

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

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

$6 x + 4 - 6 x = 4$

Combine the opposite terms in $6 x + 4 - 6 x$ .

Subtract $6 x$ from $6 x$ .

$0 + 4 = 4$

Add $0$ and $4$ .

$4 = 4$

Step 3

Since $4 = 4$ , the equation will always be true.

Always true

Step 4

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$