Pre-Algebra Examples

$\frac{1}{6} \cdot (8 t - 9) = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1

Simplify $\frac{1}{6} \cdot (8 t - 9)$ .

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Step 1.1

Apply the distributive property.

$\frac{1}{6} (8 t) + \frac{1}{6} \cdot - 9 = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.2

Cancel the common factor of $2$ .

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Step 1.2.1

Factor $2$ out of $6$ .

$\frac{1}{2 (3)} (8 t) + \frac{1}{6} \cdot - 9 = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.2.2

Factor $2$ out of $8 t$ .

$\frac{1}{2 (3)} (2 (4 t)) + \frac{1}{6} \cdot - 9 = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.2.3

Cancel the common factor.

$\frac{1}{2 \cdot 3} (2 (4 t)) + \frac{1}{6} \cdot - 9 = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.2.4

Rewrite the expression.

$\frac{1}{3} (4 t) + \frac{1}{6} \cdot - 9 = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.3

Combine $4$ and $\frac{1}{3}$ .

$\frac{4}{3} t + \frac{1}{6} \cdot - 9 = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.4

Combine $\frac{4}{3}$ and $t$ .

$\frac{4 t}{3} + \frac{1}{6} \cdot - 9 = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.5

Cancel the common factor of $3$ .

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Step 1.5.1

Factor $3$ out of $6$ .

$\frac{4 t}{3} + \frac{1}{3 (2)} \cdot - 9 = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.5.2

Factor $3$ out of $- 9$ .

$\frac{4 t}{3} + \frac{1}{3 \cdot 2} \cdot (3 \cdot - 3) = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.5.3

Cancel the common factor.

$\frac{4 t}{3} + \frac{1}{3 \cdot 2} \cdot (3 \cdot - 3) = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.5.4

Rewrite the expression.

$\frac{4 t}{3} + \frac{1}{2} \cdot - 3 = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.6

Combine $\frac{1}{2}$ and $- 3$ .

$\frac{4 t}{3} + \frac{- 3}{2} = \frac{7}{6} t - \frac{t + 9}{12}$

Step 1.7

Move the negative in front of the fraction.

$\frac{4 t}{3} - \frac{3}{2} = \frac{7}{6} t - \frac{t + 9}{12}$

Step 2

Simplify $\frac{7}{6} t - \frac{t + 9}{12}$ .

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Step 2.1

Combine $\frac{7}{6}$ and $t$ .

$\frac{4 t}{3} - \frac{3}{2} = \frac{7 t}{6} - \frac{t + 9}{12}$

Step 2.2

To write $\frac{7 t}{6}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{4 t}{3} - \frac{3}{2} = \frac{7 t}{6} \cdot \frac{2}{2} - \frac{t + 9}{12}$

Step 2.3

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.1

Multiply $\frac{7 t}{6}$ by $\frac{2}{2}$ .

$\frac{4 t}{3} - \frac{3}{2} = \frac{7 t \cdot 2}{6 \cdot 2} - \frac{t + 9}{12}$

Step 2.3.2

Multiply $6$ by $2$ .

$\frac{4 t}{3} - \frac{3}{2} = \frac{7 t \cdot 2}{12} - \frac{t + 9}{12}$

Step 2.4

Combine the numerators over the common denominator.

$\frac{4 t}{3} - \frac{3}{2} = \frac{7 t \cdot 2 - (t + 9)}{12}$

Step 2.5

Simplify the numerator.

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Step 2.5.1

Multiply $2$ by $7$ .

$\frac{4 t}{3} - \frac{3}{2} = \frac{14 t - (t + 9)}{12}$

Step 2.5.2

Apply the distributive property.

$\frac{4 t}{3} - \frac{3}{2} = \frac{14 t - t - 1 \cdot 9}{12}$

Step 2.5.3

Multiply $- 1$ by $9$ .

$\frac{4 t}{3} - \frac{3}{2} = \frac{14 t - t - 9}{12}$

Step 2.5.4

Subtract $t$ from $14 t$ .

$\frac{4 t}{3} - \frac{3}{2} = \frac{13 t - 9}{12}$

Step 3

Graph each side of the equation. The solution is the x-value of the point of intersection.

$t = 3$

Step 4