Finite Math Examples

$\frac{\frac{1}{a} - \frac{1}{b + c}}{\frac{1}{a} + \frac{1}{b + c}} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 1

Multiply the numerator and denominator of the fraction by $a (b + c)$ .

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Step 1.1

Multiply $\frac{\frac{1}{a} - \frac{1}{b + c}}{\frac{1}{a} + \frac{1}{b + c}}$ by $\frac{a (b + c)}{a (b + c)}$ .

$\frac{a (b + c)}{a (b + c)} \cdot \frac{\frac{1}{a} - \frac{1}{b + c}}{\frac{1}{a} + \frac{1}{b + c}} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 1.2

Combine.

$\frac{a (b + c) (\frac{1}{a} - \frac{1}{b + c})}{a (b + c) (\frac{1}{a} + \frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 2

Apply the distributive property.

$\frac{a (b + c) (\frac{1}{a}) + a (b + c) (- \frac{1}{b + c})}{a (b + c) (\frac{1}{a}) + a (b + c) (\frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 3

Simplify by cancelling.

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Step 3.1

Cancel the common factor of $a$ .

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Step 3.1.1

Cancel the common factor.

$\frac{a (b + c) (\frac{1}{a}) + a (b + c) (- \frac{1}{b + c})}{a (b + c) (\frac{1}{a}) + a (b + c) (\frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 3.1.2

Rewrite the expression.

$\frac{b + c + a (b + c) (- \frac{1}{b + c})}{a (b + c) (\frac{1}{a}) + a (b + c) (\frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 3.2

Cancel the common factor of $b + c$ .

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Step 3.2.1

Move the leading negative in $- \frac{1}{b + c}$ into the numerator.

$\frac{b + c + a (b + c) (\frac{- 1}{b + c})}{a (b + c) (\frac{1}{a}) + a (b + c) (\frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 3.2.2

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

$\frac{b + c + (b + c) a (\frac{- 1}{b + c})}{a (b + c) (\frac{1}{a}) + a (b + c) (\frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 3.2.3

Cancel the common factor.

$\frac{b + c + (b + c) a (\frac{- 1}{b + c})}{a (b + c) (\frac{1}{a}) + a (b + c) (\frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 3.2.4

Rewrite the expression.

$\frac{b + c + a \cdot - 1}{a (b + c) (\frac{1}{a}) + a (b + c) (\frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 3.3

Cancel the common factor of $a$ .

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Step 3.3.1

Cancel the common factor.

$\frac{b + c + a \cdot - 1}{a (b + c) (\frac{1}{a}) + a (b + c) (\frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 3.3.2

Rewrite the expression.

$\frac{b + c + a \cdot - 1}{b + c + a (b + c) (\frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 3.4

Cancel the common factor of $b + c$ .

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Step 3.4.1

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

$\frac{b + c + a \cdot - 1}{b + c + (b + c) a (\frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 3.4.2

Cancel the common factor.

$\frac{b + c + a \cdot - 1}{b + c + (b + c) a (\frac{1}{b + c})} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 3.4.3

Rewrite the expression.

$\frac{b + c + a \cdot - 1}{b + c + a} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 4

Simplify the numerator.

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Step 4.1

Move $- 1$ to the left of $a$ .

$\frac{b + c - 1 \cdot a}{b + c + a} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 4.2

Rewrite $- 1 a$ as $- a$ .

$\frac{b + c - a}{b + c + a} \cdot (1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c}) \cdot \frac{a b c}{a - b - c}$

Step 5

Combine into one fraction.

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Step 5.1

Remove parentheses.

$\frac{b + c - a}{b + c + a} \cdot 1 + \frac{b^{2} + c^{2} - q^{2}}{2 b c} \cdot \frac{a b c}{a - b - c}$

Step 5.2

Write $1$ as a fraction with a common denominator.

$\frac{b + c - a}{b + c + a} \cdot \frac{2 b c}{2 b c} + \frac{b^{2} + c^{2} - q^{2}}{2 b c} \cdot \frac{a b c}{a - b - c}$

Step 5.3

Combine the numerators over the common denominator.

$\frac{b + c - a}{b + c + a} \cdot \frac{2 b c + b^{2} + c^{2} - q^{2}}{2 b c} \cdot \frac{a b c}{a - b - c}$

Step 6

Simplify the numerator.

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Step 6.1

Factor using the perfect square rule.

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Step 6.1.1

Rearrange terms.

$\frac{b + c - a}{b + c + a} \cdot \frac{b^{2} + 2 b c + c^{2} - q^{2}}{2 b c} \cdot \frac{a b c}{a - b - c}$

Step 6.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 b c = 2 \cdot b \cdot c$

Step 6.1.3

Rewrite the polynomial.

$\frac{b + c - a}{b + c + a} \cdot \frac{b^{2} + 2 \cdot b \cdot c + c^{2} - q^{2}}{2 b c} \cdot \frac{a b c}{a - b - c}$

Step 6.1.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = b$ and $b = c$ .

$\frac{b + c - a}{b + c + a} \cdot \frac{{(b + c)}^{2} - q^{2}}{2 b c} \cdot \frac{a b c}{a - b - c}$

Step 6.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b + c$ and $b = q$ .

$\frac{b + c - a}{b + c + a} \cdot \frac{(b + c + q) (b + c - q)}{2 b c} \cdot \frac{a b c}{a - b - c}$

Step 7

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

$b + c + a, 2 b c, a - b - c$

Step 8

Since $b + c + a, 2 b c, a - b - c$ 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 $b + c + a, 2 b c, a - b - c$ are:

1. Find the LCM for the numeric part $1, 2, 1$ .

2. Find the LCM for the variable part $b^{1}, c^{1}$ .

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

4. Multiply each LCM together.

Step 9

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 10

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

Not prime

Step 11

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

Step 12

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

Not prime

Step 13

The LCM of $1, 2, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2$

Step 14

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

$b^{1} = b$

$b$ occurs $1$ time.

Step 15

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

$c^{1} = c$

$c$ occurs $1$ time.

Step 16

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

$b \cdot c$

Step 17

Multiply $b$ by $c$ .

$b c$

Step 18

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

$(b + c + a) = b + c + a$

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

Step 19

The factor for $a - b - c$ is $a - b - c$ itself.

$(a - b - c) = a - b - c$

$(a - b - c)$ occurs $1$ time.

Step 20

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

$(b + c + a) (a - b - c)$

Step 21

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

$2 b c (b + c + a) (a - b - c)$