Basic Math Examples

$\frac{1}{3} \cdot (12 f - 3) = 4 f - 1$

Step 1

Simplify $\frac{1}{3} \cdot (12 f - 3)$ .

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Step 1.1

Rewrite.

$0 + 0 + \frac{1}{3} \cdot (12 f - 3) = 4 f - 1$

Step 1.2

Simplify by adding zeros.

$\frac{1}{3} \cdot (12 f - 3) = 4 f - 1$

Step 1.3

Apply the distributive property.

$\frac{1}{3} (12 f) + \frac{1}{3} \cdot - 3 = 4 f - 1$

Step 1.4

Cancel the common factor of $3$ .

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Step 1.4.1

Factor $3$ out of $12 f$ .

$\frac{1}{3} (3 (4 f)) + \frac{1}{3} \cdot - 3 = 4 f - 1$

Step 1.4.2

Cancel the common factor.

$\frac{1}{3} (3 (4 f)) + \frac{1}{3} \cdot - 3 = 4 f - 1$

Step 1.4.3

Rewrite the expression.

$4 f + \frac{1}{3} \cdot - 3 = 4 f - 1$

Step 1.5

Cancel the common factor of $3$ .

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Step 1.5.1

Factor $3$ out of $- 3$ .

$4 f + \frac{1}{3} \cdot (3 (- 1)) = 4 f - 1$

Step 1.5.2

Cancel the common factor.

$4 f + \frac{1}{3} \cdot (3 \cdot - 1) = 4 f - 1$

Step 1.5.3

Rewrite the expression.

$4 f - 1 = 4 f - 1$

Step 2

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

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Step 2.1

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

$4 f - 1 - 4 f = - 1$

Step 2.2

Combine the opposite terms in $4 f - 1 - 4 f$ .

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Step 2.2.1

Subtract $4 f$ from $4 f$ .

$0 - 1 = - 1$

Step 2.2.2

Subtract $1$ from $0$ .

$- 1 = - 1$

Step 3

Since $- 1 = - 1$ , the equation will always be true for any value of $f$ .

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$