Precalculus Examples

$(a + b - \frac{2 a b}{a + b}) \div (\frac{a - b}{a + b} + \frac{b}{a})$

Step 1

Set the denominator in $\frac{2 a b}{a + b}$ equal to $0$ to find where the expression is undefined.

$a + b = 0$

Step 2

Subtract $b$ from both sides of the equation.

$a = - b$

Step 3

Set the denominator in $\frac{b}{a}$ equal to $0$ to find where the expression is undefined.

$a = 0$

Step 4

Set the denominator in $(a + b - \frac{2 a b}{a + b}) \div (\frac{a - b}{a + b} + \frac{b}{a})$ equal to $0$ to find where the expression is undefined.

$\frac{a - b}{a + b} + \frac{b}{a} = 0$

Step 5

Solve for $a$ .

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Step 5.1

Find the LCD of the terms in the equation.

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Step 5.1.1

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

$a + b, a, 1$

Step 5.1.2

Since $a + b, a, 1$ contains both numbers and variables, there are four steps to find the LCM. Find LCM for the numeric, variable, and compound variable parts. Then, multiply them all together.

Steps to find the LCM for $a + b, a, 1$ are:

1. Find the LCM for the numeric part $1, 1, 1$ .

2. Find the LCM for the variable part $a^{1}$ .

3. Find the LCM for the compound variable part $a + b$ .

4. Multiply each LCM together.

Step 5.1.3

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 5.1.4

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

Not prime

Step 5.1.5

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

$1$

Step 5.1.6

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

$a^{1} = a$

$a$ occurs $1$ time.

Step 5.1.7

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

$a$

Step 5.1.8

The factor for $a + b$ is $a + b$ itself.

$(a + b) = a + b$

$(a + b)$ occurs $1$ time.

Step 5.1.9

The LCM of $a + b$ is the result of multiplying all factors the greatest number of times they occur in either term.

$a + b$

Step 5.1.10

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$a (a + b)$

Step 5.2

Multiply each term in $\frac{a - b}{a + b} + \frac{b}{a} = 0$ by $a (a + b)$ to eliminate the fractions.

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Step 5.2.1

Multiply each term in $\frac{a - b}{a + b} + \frac{b}{a} = 0$ by $a (a + b)$ .

$\frac{a - b}{a + b} (a (a + b)) + \frac{b}{a} (a (a + b)) = 0 (a (a + b))$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Cancel the common factor of $a + b$ .

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Step 5.2.2.1.1.1

Factor $a + b$ out of $a (a + b)$ .

$\frac{a - b}{a + b} ((a + b) a) + \frac{b}{a} (a (a + b)) = 0 (a (a + b))$

Step 5.2.2.1.1.2

Cancel the common factor.

$\frac{a - b}{a + b} ((a + b) a) + \frac{b}{a} (a (a + b)) = 0 (a (a + b))$

Step 5.2.2.1.1.3

Rewrite the expression.

$(a - b) a + \frac{b}{a} (a (a + b)) = 0 (a (a + b))$

Step 5.2.2.1.2

Apply the distributive property.

$a \cdot a - b a + \frac{b}{a} (a (a + b)) = 0 (a (a + b))$

Step 5.2.2.1.3

Multiply $a$ by $a$ .

$a^{2} - b a + \frac{b}{a} (a (a + b)) = 0 (a (a + b))$

Step 5.2.2.1.4

Cancel the common factor of $a$ .

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Step 5.2.2.1.4.1

Cancel the common factor.

$a^{2} - b a + \frac{b}{a} (a (a + b)) = 0 (a (a + b))$

Step 5.2.2.1.4.2

Rewrite the expression.

$a^{2} - b a + b (a + b) = 0 (a (a + b))$

Step 5.2.2.1.5

Apply the distributive property.

$a^{2} - b a + b a + b \cdot b = 0 (a (a + b))$

Step 5.2.2.1.6

Multiply $b$ by $b$ .

$a^{2} - b a + b a + b^{2} = 0 (a (a + b))$

Step 5.2.2.2

Combine the opposite terms in $a^{2} - b a + b a + b^{2}$ .

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Step 5.2.2.2.1

Add $- b a$ and $b a$ .

$a^{2} + 0 + b^{2} = 0 (a (a + b))$

Step 5.2.2.2.2

Add $a^{2}$ and $0$ .

$a^{2} + b^{2} = 0 (a (a + b))$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Apply the distributive property.

$a^{2} + b^{2} = 0 (a \cdot a + a b)$

Step 5.2.3.2

Simplify the expression.

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Step 5.2.3.2.1

Multiply $a$ by $a$ .

$a^{2} + b^{2} = 0 (a^{2} + a b)$

Step 5.2.3.2.2

Multiply $0$ by $a^{2} + a b$ .

$a^{2} + b^{2} = 0$

Step 5.3

Solve the equation.

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Step 5.3.1

Subtract $b^{2}$ from both sides of the equation.

$a^{2} = - b^{2}$

Step 5.3.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$a = \pm \sqrt{- b^{2}}$

Step 5.3.3

Simplify $\pm \sqrt{- b^{2}}$ .

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Step 5.3.3.1

Reorder $- 1$ and $b^{2}$ .

$a = \pm \sqrt{b^{2} \cdot - 1}$

Step 5.3.3.2

Pull terms out from under the radical.

$a = \pm b \sqrt{- 1}$

Step 5.3.3.3

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

$a = \pm b i$

Step 5.3.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.3.4.1

First, use the positive value of the $\pm$ to find the first solution.

$a = b i$

Step 5.3.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$a = - b i$

Step 5.3.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$a = b i$

$a = - b i$

$a = b i$

$a = - b i$

$a = b i$

$a = - b i$

$a = b i$

$a = - b i$

Step 6

The domain is all values of $a$ that make the expression defined.

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${a | a \neq 0}$