Precalculus Examples

$\frac{24}{2 x} + \frac{24 x}{5} = 4$

Step 1

Reduce the expression $\frac{24}{2 x}$ by cancelling the common factors.

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Step 1.1

Factor $2$ out of $24$ .

$\frac{2 \cdot 12}{2 x} + \frac{24 x}{5} = 4$

Step 1.2

Factor $2$ out of $2 x$ .

$\frac{2 \cdot 12}{2 (x)} + \frac{24 x}{5} = 4$

Step 1.3

Cancel the common factor.

$\frac{2 \cdot 12}{2 x} + \frac{24 x}{5} = 4$

Step 1.4

Rewrite the expression.

$\frac{12}{x} + \frac{24 x}{5} = 4$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

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

$x, 5, 1$

Step 2.2

Since $x, 5, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 5, 1$ then find LCM for the variable part $x^{1}$ .

Step 2.3

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 2.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 2.5

Since $5$ has no factors besides $1$ and $5$ .

$5$ is a prime number

Step 2.6

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 2.7

The LCM of $1, 5, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$5$

Step 2.8

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

$x^{1} = x$

$x$ occurs $1$ time.

Step 2.9

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

$x$

Step 2.10

The LCM for $x, 5, 1$ is the numeric part $5$ multiplied by the variable part.

$5 x$

Step 3

Multiply each term in $\frac{12}{x} + \frac{24 x}{5} = 4$ by $5 x$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{12}{x} + \frac{24 x}{5} = 4$ by $5 x$ .

$\frac{12}{x} (5 x) + \frac{24 x}{5} (5 x) = 4 (5 x)$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Rewrite using the commutative property of multiplication.

$5 \frac{12}{x} x + \frac{24 x}{5} (5 x) = 4 (5 x)$

Step 3.2.1.2

Multiply $5 \frac{12}{x}$ .

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Step 3.2.1.2.1

Combine $5$ and $\frac{12}{x}$ .

$\frac{5 \cdot 12}{x} x + \frac{24 x}{5} (5 x) = 4 (5 x)$

Step 3.2.1.2.2

Multiply $5$ by $12$ .

$\frac{60}{x} x + \frac{24 x}{5} (5 x) = 4 (5 x)$

Step 3.2.1.3

Cancel the common factor of $x$ .

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Step 3.2.1.3.1

Cancel the common factor.

$\frac{60}{x} x + \frac{24 x}{5} (5 x) = 4 (5 x)$

Step 3.2.1.3.2

Rewrite the expression.

$60 + \frac{24 x}{5} (5 x) = 4 (5 x)$

Step 3.2.1.4

Rewrite using the commutative property of multiplication.

$60 + 5 \frac{24 x}{5} x = 4 (5 x)$

Step 3.2.1.5

Cancel the common factor of $5$ .

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Step 3.2.1.5.1

Cancel the common factor.

$60 + 5 \frac{24 x}{5} x = 4 (5 x)$

Step 3.2.1.5.2

Rewrite the expression.

$60 + 24 x \cdot x = 4 (5 x)$

Step 3.2.1.6

Multiply $x$ by $x$ by adding the exponents.

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Step 3.2.1.6.1

Move $x$ .

$60 + 24 (x \cdot x) = 4 (5 x)$

Step 3.2.1.6.2

Multiply $x$ by $x$ .

$60 + 24 x^{2} = 4 (5 x)$

Step 3.3

Simplify the right side.

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Step 3.3.1

Multiply $5$ by $4$ .

$60 + 24 x^{2} = 20 x$

Step 4

Solve the equation.

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Step 4.1

Subtract $20 x$ from both sides of the equation.

$60 + 24 x^{2} - 20 x = 0$

Step 4.2

Factor the left side of the equation.

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Step 4.2.1

Factor $4$ out of $60 + 24 x^{2} - 20 x$ .

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Step 4.2.1.1

Factor $4$ out of $60$ .

$4 (15) + 24 x^{2} - 20 x = 0$

Step 4.2.1.2

Factor $4$ out of $24 x^{2}$ .

$4 (15) + 4 (6 x^{2}) - 20 x = 0$

Step 4.2.1.3

Factor $4$ out of $- 20 x$ .

$4 (15) + 4 (6 x^{2}) + 4 (- 5 x) = 0$

Step 4.2.1.4

Factor $4$ out of $4 (15) + 4 (6 x^{2})$ .

$4 (15 + 6 x^{2}) + 4 (- 5 x) = 0$

Step 4.2.1.5

Factor $4$ out of $4 (15 + 6 x^{2}) + 4 (- 5 x)$ .

$4 (15 + 6 x^{2} - 5 x) = 0$

Step 4.2.2

Reorder terms.

$4 (6 x^{2} - 5 x + 15) = 0$

Step 4.3

Divide each term in $4 (6 x^{2} - 5 x + 15) = 0$ by $4$ and simplify.

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Step 4.3.1

Divide each term in $4 (6 x^{2} - 5 x + 15) = 0$ by $4$ .

$\frac{4 (6 x^{2} - 5 x + 15)}{4} = \frac{0}{4}$

Step 4.3.2

Simplify the left side.

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Step 4.3.2.1

Cancel the common factor of $4$ .

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Step 4.3.2.1.1

Cancel the common factor.

$\frac{4 (6 x^{2} - 5 x + 15)}{4} = \frac{0}{4}$

Step 4.3.2.1.2

Divide $6 x^{2} - 5 x + 15$ by $1$ .

$6 x^{2} - 5 x + 15 = \frac{0}{4}$

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

Divide $0$ by $4$ .

$6 x^{2} - 5 x + 15 = 0$

Step 4.4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.5

Substitute the values $a = 6$ , $b = - 5$ , and $c = 15$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (6 \cdot 15)}}{2 \cdot 6}$

Step 4.6

Simplify.

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Step 4.6.1

Simplify the numerator.

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Step 4.6.1.1

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 6 \cdot 15}}{2 \cdot 6}$

Step 4.6.1.2

Multiply $- 4 \cdot 6 \cdot 15$ .

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Step 4.6.1.2.1

Multiply $- 4$ by $6$ .

$x = \frac{5 \pm \sqrt{25 - 24 \cdot 15}}{2 \cdot 6}$

Step 4.6.1.2.2

Multiply $- 24$ by $15$ .

$x = \frac{5 \pm \sqrt{25 - 360}}{2 \cdot 6}$

Step 4.6.1.3

Subtract $360$ from $25$ .

$x = \frac{5 \pm \sqrt{- 335}}{2 \cdot 6}$

Step 4.6.1.4

Rewrite $- 335$ as $- 1 (335)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 335}}{2 \cdot 6}$

Step 4.6.1.5

Rewrite $\sqrt{- 1 (335)}$ as $\sqrt{- 1} \cdot \sqrt{335}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{335}}{2 \cdot 6}$

Step 4.6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{5 \pm i \sqrt{335}}{2 \cdot 6}$

Step 4.6.2

Multiply $2$ by $6$ .

$x = \frac{5 \pm i \sqrt{335}}{12}$

Step 4.7

The final answer is the combination of both solutions.

$x = \frac{5 + i \sqrt{335}}{12}, \frac{5 - i \sqrt{335}}{12}$