Precalculus Examples

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

Step 1

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

$b = 0$

Step 2

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

$a + b = 0$

Step 3

Subtract $b$ from both sides of the equation.

$a = - b$

Step 4

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

$a = 0$

Step 5

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

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

Step 6

Solve for $a$ .

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Step 6.1

Find the LCD of the terms in the equation.

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Step 6.1.1

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

$a, a + b, 1$

Step 6.1.2

Since $a, a + b, 1$ 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 $a, a + b, 1$ 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 6.1.3

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 6.1.4

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

Not prime

Step 6.1.5

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 6.1.6

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

$a^{1} = a$

$a$ occurs $1$ time.

Step 6.1.7

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 6.1.8

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

$(a + b) = a + b$

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

Step 6.1.9

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 6.1.10

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

$a (a + b)$

Step 6.2

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

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Step 6.2.1

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

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

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Simplify each term.

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Step 6.2.2.1.1

Cancel the common factor of $a$ .

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Step 6.2.2.1.1.1

Cancel the common factor.

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

Step 6.2.2.1.1.2

Rewrite the expression.

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

Step 6.2.2.1.2

Raise $a + b$ to the power of $1$ .

${(a + b)}^{1} (a + b) - \frac{b}{a + b} (a (a + b)) = 0 (a (a + b))$

Step 6.2.2.1.3

Raise $a + b$ to the power of $1$ .

${(a + b)}^{1} {(a + b)}^{1} - \frac{b}{a + b} (a (a + b)) = 0 (a (a + b))$

Step 6.2.2.1.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${(a + b)}^{1 + 1} - \frac{b}{a + b} (a (a + b)) = 0 (a (a + b))$

Step 6.2.2.1.5

Add $1$ and $1$ .

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

Step 6.2.2.1.6

Cancel the common factor of $a + b$ .

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Step 6.2.2.1.6.1

Move the leading negative in $- \frac{b}{a + b}$ into the numerator.

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

Step 6.2.2.1.6.2

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

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

Step 6.2.2.1.6.3

Cancel the common factor.

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

Step 6.2.2.1.6.4

Rewrite the expression.

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

Step 6.2.3

Simplify the right side.

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Step 6.2.3.1

Apply the distributive property.

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

Step 6.2.3.2

Simplify the expression.

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Step 6.2.3.2.1

Multiply $a$ by $a$ .

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

Step 6.2.3.2.2

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

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

Step 6.3

Solve the equation.

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Step 6.3.1

Rewrite ${(a + b)}^{2}$ as $(a + b) (a + b)$ .

$(a + b) (a + b) - b a = 0$

Step 6.3.2

Expand $(a + b) (a + b)$ using the FOIL Method.

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Step 6.3.2.1

Apply the distributive property.

$a (a + b) + b (a + b) - b a = 0$

Step 6.3.2.2

Apply the distributive property.

$a \cdot a + a b + b (a + b) - b a = 0$

Step 6.3.2.3

Apply the distributive property.

$a \cdot a + a b + b a + b \cdot b - b a = 0$

Step 6.3.3

Simplify and combine like terms.

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Step 6.3.3.1

Simplify each term.

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Step 6.3.3.1.1

Multiply $a$ by $a$ .

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

Step 6.3.3.1.2

Multiply $b$ by $b$ .

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

Step 6.3.3.2

Add $a b$ and $b a$ .

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Step 6.3.3.2.1

Reorder $b$ and $a$ .

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

Step 6.3.3.2.2

Add $a b$ and $a b$ .

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

Step 6.3.4

Subtract $b a$ from $2 a b$ .

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Step 6.3.4.1

Move $b$ .

$a^{2} + b^{2} + 2 a b - 1 a b = 0$

Step 6.3.4.2

Subtract $a b$ from $2 a b$ .

$a^{2} + b^{2} + 1 a b = 0$

Step 6.3.5

Multiply $a$ by $1$ .

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

Step 6.3.6

Use the quadratic formula to find the solutions.

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

Step 6.3.7

Substitute the values $a = 1$ , $b = b$ , and $c = b^{2}$ into the quadratic formula and solve for $a$ .

$\frac{- b \pm \sqrt{b^{2} - 4 \cdot (1 \cdot b^{2})}}{2 \cdot 1}$

Step 6.3.8

Simplify.

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Step 6.3.8.1

Simplify the numerator.

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Step 6.3.8.1.1

Factor $b^{2}$ out of $b^{2} - 4 \cdot 1 \cdot b^{2}$ .

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Step 6.3.8.1.1.1

Multiply by $1$ .

$a = \frac{- b \pm \sqrt{b^{2} \cdot 1 - 4 \cdot 1 \cdot b^{2}}}{2 \cdot 1}$

Step 6.3.8.1.1.2

Factor $b^{2}$ out of $- 4 \cdot 1 \cdot b^{2}$ .

$a = \frac{- b \pm \sqrt{b^{2} \cdot 1 + b^{2} \cdot (- 4 \cdot 1)}}{2 \cdot 1}$

Step 6.3.8.1.1.3

Factor $b^{2}$ out of $b^{2} \cdot 1 + b^{2} \cdot (- 4 \cdot 1)$ .

$a = \frac{- b \pm \sqrt{b^{2} (1 - 4 \cdot 1)}}{2 \cdot 1}$

Step 6.3.8.1.2

Multiply $- 4$ by $1$ .

$a = \frac{- b \pm \sqrt{b^{2} (1 - 4)}}{2 \cdot 1}$

Step 6.3.8.1.3

Subtract $4$ from $1$ .

$a = \frac{- b \pm \sqrt{b^{2} \cdot - 3}}{2 \cdot 1}$

Step 6.3.8.1.4

Pull terms out from under the radical.

$a = \frac{- b \pm b \sqrt{- 3}}{2 \cdot 1}$

Step 6.3.8.1.5

Rewrite $- 3$ as $- 1 (3)$ .

$a = \frac{- b \pm b \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 6.3.8.1.6

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$a = \frac{- b \pm b (\sqrt{- 1} \cdot \sqrt{3})}{2 \cdot 1}$

Step 6.3.8.1.7

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

$a = \frac{- b \pm b i \sqrt{3}}{2 \cdot 1}$

Step 6.3.8.2

Multiply $2$ by $1$ .

$a = \frac{- b \pm b i \sqrt{3}}{2}$

Step 6.3.9

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 6.3.9.1

Simplify the numerator.

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Step 6.3.9.1.1

Factor $b^{2}$ out of $b^{2} - 4 \cdot 1 \cdot b^{2}$ .

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Step 6.3.9.1.1.1

Multiply by $1$ .

$a = \frac{- b \pm \sqrt{b^{2} \cdot 1 - 4 \cdot 1 \cdot b^{2}}}{2 \cdot 1}$

Step 6.3.9.1.1.2

Factor $b^{2}$ out of $- 4 \cdot 1 \cdot b^{2}$ .

$a = \frac{- b \pm \sqrt{b^{2} \cdot 1 + b^{2} \cdot (- 4 \cdot 1)}}{2 \cdot 1}$

Step 6.3.9.1.1.3

Factor $b^{2}$ out of $b^{2} \cdot 1 + b^{2} \cdot (- 4 \cdot 1)$ .

$a = \frac{- b \pm \sqrt{b^{2} (1 - 4 \cdot 1)}}{2 \cdot 1}$

Step 6.3.9.1.2

Multiply $- 4$ by $1$ .

$a = \frac{- b \pm \sqrt{b^{2} (1 - 4)}}{2 \cdot 1}$

Step 6.3.9.1.3

Subtract $4$ from $1$ .

$a = \frac{- b \pm \sqrt{b^{2} \cdot - 3}}{2 \cdot 1}$

Step 6.3.9.1.4

Pull terms out from under the radical.

$a = \frac{- b \pm b \sqrt{- 3}}{2 \cdot 1}$

Step 6.3.9.1.5

Rewrite $- 3$ as $- 1 (3)$ .

$a = \frac{- b \pm b \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 6.3.9.1.6

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$a = \frac{- b \pm b (\sqrt{- 1} \cdot \sqrt{3})}{2 \cdot 1}$

Step 6.3.9.1.7

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

$a = \frac{- b \pm b i \sqrt{3}}{2 \cdot 1}$

Step 6.3.9.2

Multiply $2$ by $1$ .

$a = \frac{- b \pm b i \sqrt{3}}{2}$

Step 6.3.9.3

Change the $\pm$ to $+$ .

$a = \frac{- b + b i \sqrt{3}}{2}$

Step 6.3.9.4

Factor $b$ out of $- b + b i \sqrt{3}$ .

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Step 6.3.9.4.1

Factor $b$ out of $- b$ .

$a = \frac{b \cdot - 1 + b i \sqrt{3}}{2}$

Step 6.3.9.4.2

Factor $b$ out of $b i \sqrt{3}$ .

$a = \frac{b \cdot - 1 + b (i \sqrt{3})}{2}$

Step 6.3.9.4.3

Factor $b$ out of $b \cdot - 1 + b (i \sqrt{3})$ .

$a = \frac{b (- 1 + i \sqrt{3})}{2}$

Step 6.3.9.5

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

$a = \frac{b (- 1 \cdot 1 + i \sqrt{3})}{2}$

Step 6.3.9.6

Factor $- 1$ out of $i \sqrt{3}$ .

$a = \frac{b (- 1 \cdot 1 - (- i \sqrt{3}))}{2}$

Step 6.3.9.7

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{3})$ .

$a = \frac{b (- 1 (1 - i \sqrt{3}))}{2}$

Step 6.3.9.8

Move the negative in front of the fraction.

$a = - \frac{b (1 - i \sqrt{3})}{2}$

Step 6.3.10

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 6.3.10.1

Simplify the numerator.

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Step 6.3.10.1.1

Factor $b^{2}$ out of $b^{2} - 4 \cdot 1 \cdot b^{2}$ .

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Step 6.3.10.1.1.1

Multiply by $1$ .

$a = \frac{- b \pm \sqrt{b^{2} \cdot 1 - 4 \cdot 1 \cdot b^{2}}}{2 \cdot 1}$

Step 6.3.10.1.1.2

Factor $b^{2}$ out of $- 4 \cdot 1 \cdot b^{2}$ .

$a = \frac{- b \pm \sqrt{b^{2} \cdot 1 + b^{2} \cdot (- 4 \cdot 1)}}{2 \cdot 1}$

Step 6.3.10.1.1.3

Factor $b^{2}$ out of $b^{2} \cdot 1 + b^{2} \cdot (- 4 \cdot 1)$ .

$a = \frac{- b \pm \sqrt{b^{2} (1 - 4 \cdot 1)}}{2 \cdot 1}$

Step 6.3.10.1.2

Multiply $- 4$ by $1$ .

$a = \frac{- b \pm \sqrt{b^{2} (1 - 4)}}{2 \cdot 1}$

Step 6.3.10.1.3

Subtract $4$ from $1$ .

$a = \frac{- b \pm \sqrt{b^{2} \cdot - 3}}{2 \cdot 1}$

Step 6.3.10.1.4

Pull terms out from under the radical.

$a = \frac{- b \pm b \sqrt{- 3}}{2 \cdot 1}$

Step 6.3.10.1.5

Rewrite $- 3$ as $- 1 (3)$ .

$a = \frac{- b \pm b \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 6.3.10.1.6

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$a = \frac{- b \pm b (\sqrt{- 1} \cdot \sqrt{3})}{2 \cdot 1}$

Step 6.3.10.1.7

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

$a = \frac{- b \pm b i \sqrt{3}}{2 \cdot 1}$

Step 6.3.10.2

Multiply $2$ by $1$ .

$a = \frac{- b \pm b i \sqrt{3}}{2}$

Step 6.3.10.3

Change the $\pm$ to $-$ .

$a = \frac{- b - b i \sqrt{3}}{2}$

Step 6.3.10.4

Simplify the numerator.

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Step 6.3.10.4.1

Factor $b$ out of $- b - b i \sqrt{3}$ .

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Step 6.3.10.4.1.1

Factor $b$ out of $- b$ .

$a = \frac{b \cdot - 1 - b i \sqrt{3}}{2}$

Step 6.3.10.4.1.2

Factor $b$ out of $- b i \sqrt{3}$ .

$a = \frac{b \cdot - 1 + b (- 1 i \sqrt{3})}{2}$

Step 6.3.10.4.1.3

Factor $b$ out of $b \cdot - 1 + b (- 1 i \sqrt{3})$ .

$a = \frac{b (- 1 - 1 i \sqrt{3})}{2}$

Step 6.3.10.4.2

Rewrite $- 1 i$ as $- i$ .

$a = \frac{b (- 1 - i \sqrt{3})}{2}$

Step 6.3.10.5

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

$a = \frac{b (- 1 \cdot 1 - i \sqrt{3})}{2}$

Step 6.3.10.6

Factor $- 1$ out of $- i \sqrt{3}$ .

$a = \frac{b (- 1 \cdot 1 - (i \sqrt{3}))}{2}$

Step 6.3.10.7

Factor $- 1$ out of $- 1 (1) - (i \sqrt{3})$ .

$a = \frac{b (- 1 (1 + i \sqrt{3}))}{2}$

Step 6.3.10.8

Move the negative in front of the fraction.

$a = - \frac{b (1 + i \sqrt{3})}{2}$

Step 6.3.11

The final answer is the combination of both solutions.

$a = - \frac{b (1 - i \sqrt{3})}{2}$

$a = - \frac{b (1 + i \sqrt{3})}{2}$

$a = - \frac{b (1 - i \sqrt{3})}{2}$

$a = - \frac{b (1 + i \sqrt{3})}{2}$

$a = - \frac{b (1 - i \sqrt{3})}{2}$

$a = - \frac{b (1 + i \sqrt{3})}{2}$

Step 7

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

Interval Notation:

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

Set-Builder Notation:

${a | a \neq 0}$