Precalculus Examples

$\frac{2 b}{2 b + 3} + \frac{5}{3 - 2 b} - \frac{4 b^{2} + 9}{4 b^{2} - 9}$

Step 1

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

$2 b + 3 = 0$

Step 2

Solve for $b$ .

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Step 2.1

Subtract $3$ from both sides of the equation.

$2 b = - 3$

Step 2.2

Divide each term in $2 b = - 3$ by $2$ and simplify.

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Step 2.2.1

Divide each term in $2 b = - 3$ by $2$ .

$\frac{2 b}{2} = \frac{- 3}{2}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $2$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{2 b}{2} = \frac{- 3}{2}$

Step 2.2.2.1.2

Divide $b$ by $1$ .

$b = \frac{- 3}{2}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Move the negative in front of the fraction.

$b = - \frac{3}{2}$

Step 3

Set the denominator in $\frac{5}{3 - 2 b}$ equal to $0$ to find where the expression is undefined.

$3 - 2 b = 0$

Step 4

Solve for $b$ .

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Step 4.1

Subtract $3$ from both sides of the equation.

$- 2 b = - 3$

Step 4.2

Divide each term in $- 2 b = - 3$ by $- 2$ and simplify.

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Step 4.2.1

Divide each term in $- 2 b = - 3$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{- 3}{- 2}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $- 2$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{- 3}{- 2}$

Step 4.2.2.1.2

Divide $b$ by $1$ .

$b = \frac{- 3}{- 2}$

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.

$b = \frac{3}{2}$

Step 5

Set the denominator in $\frac{4 b^{2} + 9}{4 b^{2} - 9}$ equal to $0$ to find where the expression is undefined.

$4 b^{2} - 9 = 0$

Step 6

Solve for $b$ .

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Step 6.1

Add $9$ to both sides of the equation.

$4 b^{2} = 9$

Step 6.2

Divide each term in $4 b^{2} = 9$ by $4$ and simplify.

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Step 6.2.1

Divide each term in $4 b^{2} = 9$ by $4$ .

$\frac{4 b^{2}}{4} = \frac{9}{4}$

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Cancel the common factor of $4$ .

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Step 6.2.2.1.1

Cancel the common factor.

$\frac{4 b^{2}}{4} = \frac{9}{4}$

Step 6.2.2.1.2

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

$b^{2} = \frac{9}{4}$

Step 6.3

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

$b = \pm \sqrt{\frac{9}{4}}$

Step 6.4

Simplify $\pm \sqrt{\frac{9}{4}}$ .

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Step 6.4.1

Rewrite $\sqrt{\frac{9}{4}}$ as $\frac{\sqrt{9}}{\sqrt{4}}$ .

$b = \pm \frac{\sqrt{9}}{\sqrt{4}}$

Step 6.4.2

Simplify the numerator.

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Step 6.4.2.1

Rewrite $9$ as $3^{2}$ .

$b = \pm \frac{\sqrt{3^{2}}}{\sqrt{4}}$

Step 6.4.2.2

Pull terms out from under the radical, assuming positive real numbers.

$b = \pm \frac{3}{\sqrt{4}}$

Step 6.4.3

Simplify the denominator.

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Step 6.4.3.1

Rewrite $4$ as $2^{2}$ .

$b = \pm \frac{3}{\sqrt{2^{2}}}$

Step 6.4.3.2

Pull terms out from under the radical, assuming positive real numbers.

$b = \pm \frac{3}{2}$

Step 6.5

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

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Step 6.5.1

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

$b = \frac{3}{2}$

Step 6.5.2

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

$b = - \frac{3}{2}$

Step 6.5.3

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

$b = \frac{3}{2}, - \frac{3}{2}$

Step 7

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

Interval Notation:

$(- \infty, - \frac{3}{2}) \cup (- \frac{3}{2}, \frac{3}{2}) \cup (\frac{3}{2}, \infty)$

Set-Builder Notation:

${b | b \neq \frac{3}{2}, - \frac{3}{2}}$

Step 8