Trigonometry Examples

$\frac{2}{6 x} - \frac{3}{5} = \frac{2}{30 x} - \frac{3}{150}$

Step 1

Move all terms not containing $x$ to the right side of the equation.

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Step 1.1

Add $\frac{3}{5}$ to both sides of the equation.

$\frac{2}{6 x} = \frac{2}{30 x} - \frac{3}{150} + \frac{3}{5}$

Step 1.2

Simplify each term.

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Step 1.2.1

Cancel the common factor of $2$ and $30$ .

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Step 1.2.1.1

Factor $2$ out of $2$ .

$\frac{2}{6 x} = \frac{2 (1)}{30 x} - \frac{3}{150} + \frac{3}{5}$

Step 1.2.1.2

Cancel the common factors.

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Step 1.2.1.2.1

Factor $2$ out of $30 x$ .

$\frac{2}{6 x} = \frac{2 (1)}{2 (15 x)} - \frac{3}{150} + \frac{3}{5}$

Step 1.2.1.2.2

Cancel the common factor.

$\frac{2}{6 x} = \frac{2 \cdot 1}{2 (15 x)} - \frac{3}{150} + \frac{3}{5}$

Step 1.2.1.2.3

Rewrite the expression.

$\frac{2}{6 x} = \frac{1}{15 x} - \frac{3}{150} + \frac{3}{5}$

Step 1.2.2

Cancel the common factor of $3$ and $150$ .

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Step 1.2.2.1

Factor $3$ out of $3$ .

$\frac{2}{6 x} = \frac{1}{15 x} - \frac{3 (1)}{150} + \frac{3}{5}$

Step 1.2.2.2

Cancel the common factors.

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Step 1.2.2.2.1

Factor $3$ out of $150$ .

$\frac{2}{6 x} = \frac{1}{15 x} - \frac{3 \cdot 1}{3 \cdot 50} + \frac{3}{5}$

Step 1.2.2.2.2

Cancel the common factor.

$\frac{2}{6 x} = \frac{1}{15 x} - \frac{3 \cdot 1}{3 \cdot 50} + \frac{3}{5}$

Step 1.2.2.2.3

Rewrite the expression.

$\frac{2}{6 x} = \frac{1}{15 x} - \frac{1}{50} + \frac{3}{5}$

Step 1.3

To write $\frac{3}{5}$ as a fraction with a common denominator, multiply by $\frac{10}{10}$ .

$\frac{2}{6 x} = \frac{1}{15 x} - \frac{1}{50} + \frac{3}{5} \cdot \frac{10}{10}$

Step 1.4

Write each expression with a common denominator of $50$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.4.1

Multiply $\frac{3}{5}$ by $\frac{10}{10}$ .

$\frac{2}{6 x} = \frac{1}{15 x} - \frac{1}{50} + \frac{3 \cdot 10}{5 \cdot 10}$

Step 1.4.2

Multiply $5$ by $10$ .

$\frac{2}{6 x} = \frac{1}{15 x} - \frac{1}{50} + \frac{3 \cdot 10}{50}$

Step 1.5

Combine the numerators over the common denominator.

$\frac{2}{6 x} = \frac{1}{15 x} + \frac{- 1 + 3 \cdot 10}{50}$

Step 1.6

Simplify the numerator.

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Step 1.6.1

Multiply $3$ by $10$ .

$\frac{2}{6 x} = \frac{1}{15 x} + \frac{- 1 + 30}{50}$

Step 1.6.2

Add $- 1$ and $30$ .

$\frac{2}{6 x} = \frac{1}{15 x} + \frac{29}{50}$

Step 2

Reduce the expression $\frac{2}{6 x}$ by cancelling the common factors.

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Step 2.1

Factor $2$ out of $2$ .

$\frac{2 (1)}{6 x} = \frac{1}{15 x} + \frac{29}{50}$

Step 2.2

Factor $2$ out of $6 x$ .

$\frac{2 (1)}{2 (3 x)} = \frac{1}{15 x} + \frac{29}{50}$

Step 2.3

Cancel the common factor.

$\frac{2 \cdot 1}{2 (3 x)} = \frac{1}{15 x} + \frac{29}{50}$

Step 2.4

Rewrite the expression.

$\frac{1}{3 x} = \frac{1}{15 x} + \frac{29}{50}$

Step 3

Find the LCD of the terms in the equation.

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Step 3.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$3 x, 15 x, 50$

Step 3.2

Since $3 x, 15 x, 50$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $3, 15, 50$ then find LCM for the variable part $x^{1}, x^{1}$ .

Step 3.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 3.4

Since $3$ has no factors besides $1$ and $3$ .

$3$ is a prime number

Step 3.5

$15$ has factors of $3$ and $5$ .

$3 \cdot 5$

Step 3.6

The prime factors for $50$ are $2 \cdot 5 \cdot 5$ .

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Step 3.6.1

$50$ has factors of $2$ and $25$ .

$2 \cdot 25$

Step 3.6.2

$25$ has factors of $5$ and $5$ .

$2 \cdot 5 \cdot 5$

Step 3.7

The LCM of $3, 15, 50$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2 \cdot 3 \cdot 5 \cdot 5$

Step 3.8

Multiply $2 \cdot 3 \cdot 5 \cdot 5$ .

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Step 3.8.1

Multiply $2$ by $3$ .

$6 \cdot 5 \cdot 5$

Step 3.8.2

Multiply $6$ by $5$ .

$30 \cdot 5$

Step 3.8.3

Multiply $30$ by $5$ .

$150$

Step 3.9

The factor for $x^{1}$ is $x$ itself.

$x^{1} = x$

$x$ occurs $1$ time.

Step 3.10

The LCM of $x^{1}, x^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x$

Step 3.11

The LCM for $3 x, 15 x, 50$ is the numeric part $150$ multiplied by the variable part.

$150 x$

Step 4

Multiply each term in $\frac{1}{3 x} = \frac{1}{15 x} + \frac{29}{50}$ by $150 x$ to eliminate the fractions.

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Step 4.1

Multiply each term in $\frac{1}{3 x} = \frac{1}{15 x} + \frac{29}{50}$ by $150 x$ .

$\frac{1}{3 x} (150 x) = \frac{1}{15 x} (150 x) + \frac{29}{50} (150 x)$

Step 4.2

Simplify the left side.

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Step 4.2.1

Rewrite using the commutative property of multiplication.

$150 \frac{1}{3 x} x = \frac{1}{15 x} (150 x) + \frac{29}{50} (150 x)$

Step 4.2.2

Cancel the common factor of $3$ .

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Step 4.2.2.1

Factor $3$ out of $150$ .

$3 (50) \frac{1}{3 x} x = \frac{1}{15 x} (150 x) + \frac{29}{50} (150 x)$

Step 4.2.2.2

Factor $3$ out of $3 x$ .

$3 (50) \frac{1}{3 (x)} x = \frac{1}{15 x} (150 x) + \frac{29}{50} (150 x)$

Step 4.2.2.3

Cancel the common factor.

$3 \cdot 50 \frac{1}{3 x} x = \frac{1}{15 x} (150 x) + \frac{29}{50} (150 x)$

Step 4.2.2.4

Rewrite the expression.

$50 \frac{1}{x} x = \frac{1}{15 x} (150 x) + \frac{29}{50} (150 x)$

Step 4.2.3

Combine $50$ and $\frac{1}{x}$ .

$\frac{50}{x} x = \frac{1}{15 x} (150 x) + \frac{29}{50} (150 x)$

Step 4.2.4

Cancel the common factor of $x$ .

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Step 4.2.4.1

Cancel the common factor.

$\frac{50}{x} x = \frac{1}{15 x} (150 x) + \frac{29}{50} (150 x)$

Step 4.2.4.2

Rewrite the expression.

$50 = \frac{1}{15 x} (150 x) + \frac{29}{50} (150 x)$

Step 4.3

Simplify the right side.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Rewrite using the commutative property of multiplication.

$50 = 150 \frac{1}{15 x} x + \frac{29}{50} (150 x)$

Step 4.3.1.2

Cancel the common factor of $15$ .

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Step 4.3.1.2.1

Factor $15$ out of $150$ .

$50 = 15 (10) \frac{1}{15 x} x + \frac{29}{50} (150 x)$

Step 4.3.1.2.2

Factor $15$ out of $15 x$ .

$50 = 15 (10) \frac{1}{15 (x)} x + \frac{29}{50} (150 x)$

Step 4.3.1.2.3

Cancel the common factor.

$50 = 15 \cdot 10 \frac{1}{15 x} x + \frac{29}{50} (150 x)$

Step 4.3.1.2.4

Rewrite the expression.

$50 = 10 \frac{1}{x} x + \frac{29}{50} (150 x)$

Step 4.3.1.3

Combine $10$ and $\frac{1}{x}$ .

$50 = \frac{10}{x} x + \frac{29}{50} (150 x)$

Step 4.3.1.4

Cancel the common factor of $x$ .

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Step 4.3.1.4.1

Cancel the common factor.

$50 = \frac{10}{x} x + \frac{29}{50} (150 x)$

Step 4.3.1.4.2

Rewrite the expression.

$50 = 10 + \frac{29}{50} (150 x)$

Step 4.3.1.5

Cancel the common factor of $50$ .

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Step 4.3.1.5.1

Factor $50$ out of $150 x$ .

$50 = 10 + \frac{29}{50} (50 (3 x))$

Step 4.3.1.5.2

Cancel the common factor.

$50 = 10 + \frac{29}{50} (50 (3 x))$

Step 4.3.1.5.3

Rewrite the expression.

$50 = 10 + 29 (3 x)$

Step 4.3.1.6

Multiply $3$ by $29$ .

$50 = 10 + 87 x$

Step 5

Solve the equation.

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Step 5.1

Rewrite the equation as $10 + 87 x = 50$ .

$10 + 87 x = 50$

Step 5.2

Move all terms not containing $x$ to the right side of the equation.

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Step 5.2.1

Subtract $10$ from both sides of the equation.

$87 x = 50 - 10$

Step 5.2.2

Subtract $10$ from $50$ .

$87 x = 40$

Step 5.3

Divide each term in $87 x = 40$ by $87$ and simplify.

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Step 5.3.1

Divide each term in $87 x = 40$ by $87$ .

$\frac{87 x}{87} = \frac{40}{87}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $87$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{87 x}{87} = \frac{40}{87}$

Step 5.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{40}{87}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{40}{87}$

Decimal Form:

$x = 0.45977011 \dots$