Precalculus Examples

$\frac{c}{b - c} + \frac{b^{2} - 3 b c}{b^{2} - c^{2}}$

Step 1

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

$b - c = 0$

Step 2

Solve for $c$ .

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Step 2.1

Subtract $b$ from both sides of the equation.

$- c = - b$

Step 2.2

Divide each term in $- c = - b$ by $- 1$ and simplify.

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Step 2.2.1

Divide each term in $- c = - b$ by $- 1$ .

$\frac{- c}{- 1} = \frac{- b}{- 1}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

$\frac{c}{1} = \frac{- b}{- 1}$

Step 2.2.2.2

Divide $c$ by $1$ .

$c = \frac{- b}{- 1}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Dividing two negative values results in a positive value.

$c = \frac{b}{1}$

Step 2.2.3.2

Divide $b$ by $1$ .

$c = b$

Step 3

Set the denominator in $\frac{b^{2} - 3 b c}{b^{2} - c^{2}}$ equal to $0$ to find where the expression is undefined.

$b^{2} - c^{2} = 0$

Step 4

Solve for $c$ .

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Step 4.1

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

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

Step 4.2

Divide each term in $- c^{2} = - b^{2}$ by $- 1$ and simplify.

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Step 4.2.1

Divide each term in $- c^{2} = - b^{2}$ by $- 1$ .

$\frac{- c^{2}}{- 1} = \frac{- b^{2}}{- 1}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Dividing two negative values results in a positive value.

$\frac{c^{2}}{1} = \frac{- b^{2}}{- 1}$

Step 4.2.2.2

Divide $c^{2}$ by $1$ .

$c^{2} = \frac{- b^{2}}{- 1}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Dividing two negative values results in a positive value.

$c^{2} = \frac{b^{2}}{1}$

Step 4.2.3.2

Divide $b^{2}$ by $1$ .

$c^{2} = b^{2}$

Step 4.3

Since the exponents are equal, the bases of the exponents on both sides of the equation must be equal.

$| c | = | b |$

Step 4.4

Solve for $c$ .

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Step 4.4.1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$c = \pm | b |$

Step 4.4.2

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

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Step 4.4.2.1

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

$c = | b |$

Step 4.4.2.2

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

$c = - | b |$

Step 4.4.2.3

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

$c = | b |$

$c = - | b |$

$c = | b |$

$c = - | b |$

$c = | b |$

$c = - | b |$

$c = | b |$

$c = - | b |$

Step 5

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${c | c \in ℝ}$