Algebra Examples

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

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $4 x$ from both sides of the equation.

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

Step 1.2

Subtract $2$ from both sides of the equation.

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

Step 2

Simplify $\frac{2}{3} \cdot (6 x + 3) - 4 x - 2$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the distributive property.

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

Step 2.1.2

Cancel the common factor of $3$ .

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Step 2.1.2.1

Factor $3$ out of $6 x$ .

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

Step 2.1.2.2

Cancel the common factor.

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

Step 2.1.2.3

Rewrite the expression.

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

Step 2.1.3

Multiply $2$ by $2$ .

$4 x + \frac{2}{3} \cdot 3 - 4 x - 2 = 0$

Step 2.1.4

Cancel the common factor of $3$ .

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Step 2.1.4.1

Cancel the common factor.

$4 x + \frac{2}{3} \cdot 3 - 4 x - 2 = 0$

Step 2.1.4.2

Rewrite the expression.

$4 x + 2 - 4 x - 2 = 0$

Step 2.2

Combine the opposite terms in $4 x + 2 - 4 x - 2$ .

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Step 2.2.1

Subtract $4 x$ from $4 x$ .

$0 + 2 - 2 = 0$

Step 2.2.2

Add $0$ and $2$ .

$2 - 2 = 0$

Step 2.2.3

Subtract $2$ from $2$ .

$0 = 0$

Step 3

Since $0 = 0$ , the equation will always be true.

Always true