Finite Math Examples

$2 n^{- 2} + 15 n^{- 1} + 28 = 0$ , $u = n^{- 1}$

Step 1

Solve for $n$ in $2 n^{- 2} + 15 n^{- 1} + 28 = 0$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$2 (\frac{1}{n^{2}}) + 15 n^{- 1} + 28 = 0$

$u = n^{- 1}$

Step 1.1.2

Combine $2$ and $\frac{1}{n^{2}}$ .

$\frac{2}{n^{2}} + 15 n^{- 1} + 28 = 0$

$u = n^{- 1}$

Step 1.1.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{2}{n^{2}} + 15 (\frac{1}{n}) + 28 = 0$

$u = n^{- 1}$

Step 1.1.4

Combine $15$ and $\frac{1}{n}$ .

$\frac{2}{n^{2}} + \frac{15}{n} + 28 = 0$

$u = n^{- 1}$

$\frac{2}{n^{2}} + \frac{15}{n} + 28 = 0$

$u = n^{- 1}$

Step 1.2

Find the LCD of the terms in the equation.

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Step 1.2.1

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

$n^{2}, n, 1, 1$

$u = n^{- 1}$

Step 1.2.2

Since $n^{2}, n, 1, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 1, 1$ then find LCM for the variable part $n^{2}, n^{1}$ .

Since $n^{2}, n, 1, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 1, 1$ then find LCM for the variable part $n^{2}, n$ .

$u = n^{- 1}$

Step 1.2.3

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.

$u = n^{- 1}$

Step 1.2.4

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

Not prime

Step 1.2.5

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

$1$

$u = n^{- 1}$

Step 1.2.6

The factors for $n^{2}$ are $n \cdot n$ , which is $n$ multiplied by each other $2$ times.

$n^{2} = n \cdot n$

n occurs $2$ times.

$u = n^{- 1}$

Step 1.2.7

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

$n = n$

n occurs $1$ time.

$u = n^{- 1}$

Step 1.2.8

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

$n \cdot n$

$u = n^{- 1}$

Step 1.2.9

Multiply $n$ by $n$ .

$n^{2}$

$u = n^{- 1}$

$n^{2}$

$u = n^{- 1}$

Step 1.3

Multiply each term in $\frac{2}{n^{2}} + \frac{15}{n} + 28 = 0$ by $n^{2}$ to eliminate the fractions.

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Step 1.3.1

Multiply each term in $\frac{2}{n^{2}} + \frac{15}{n} + 28 = 0$ by $n^{2}$ .

$\frac{2}{n^{2}} \cdot n^{2} + \frac{15}{n} \cdot n^{2} + 28 n^{2} = 0 n^{2}$

$u = n^{- 1}$

Step 1.3.2

Simplify the left side.

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Step 1.3.2.1

Simplify each term.

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Step 1.3.2.1.1

Cancel the common factor of $n^{2}$ .

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Step 1.3.2.1.1.1

Cancel the common factor.

$\frac{2}{n^{2}} \cdot n^{2} + \frac{15}{n} \cdot n^{2} + 28 n^{2} = 0 n^{2}$

$u = n^{- 1}$

Step 1.3.2.1.1.2

Rewrite the expression.

$2 + \frac{15}{n} \cdot n^{2} + 28 n^{2} = 0 n^{2}$

$u = n^{- 1}$

$2 + \frac{15}{n} \cdot n^{2} + 28 n^{2} = 0 n^{2}$

$u = n^{- 1}$

Step 1.3.2.1.2

Cancel the common factor of $n$ .

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Step 1.3.2.1.2.1

Factor $n$ out of $n^{2}$ .

$2 + \frac{15}{n} \cdot (n \cdot n) + 28 n^{2} = 0 n^{2}$

$u = n^{- 1}$

Step 1.3.2.1.2.2

Cancel the common factor.

$2 + \frac{15}{n} \cdot (n \cdot n) + 28 n^{2} = 0 n^{2}$

$u = n^{- 1}$

Step 1.3.2.1.2.3

Rewrite the expression.

$2 + 15 n + 28 n^{2} = 0 n^{2}$

$u = n^{- 1}$

$2 + 15 n + 28 n^{2} = 0 n^{2}$

$u = n^{- 1}$

$2 + 15 n + 28 n^{2} = 0 n^{2}$

$u = n^{- 1}$

$2 + 15 n + 28 n^{2} = 0 n^{2}$

$u = n^{- 1}$

Step 1.3.3

Simplify the right side.

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Step 1.3.3.1

Multiply $0$ by $n^{2}$ .

$2 + 15 n + 28 n^{2} = 0$

$u = n^{- 1}$

$2 + 15 n + 28 n^{2} = 0$

$u = n^{- 1}$

$2 + 15 n + 28 n^{2} = 0$

$u = n^{- 1}$

Step 1.4

Solve the equation.

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Step 1.4.1

Factor by grouping.

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Step 1.4.1.1

Reorder terms.

$28 n^{2} + 15 n + 2 = 0$

$u = n^{- 1}$

Step 1.4.1.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 28 \cdot 2 = 56$ and whose sum is $b = 15$ .

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Step 1.4.1.2.1

Factor $15$ out of $15 n$ .

$28 n^{2} + 15 (n) + 2 = 0$

$u = n^{- 1}$

Step 1.4.1.2.2

Rewrite $15$ as $7$ plus $8$

$28 n^{2} + (7 + 8) n + 2 = 0$

$u = n^{- 1}$

Step 1.4.1.2.3

Apply the distributive property.

$28 n^{2} + 7 n + 8 n + 2 = 0$

$u = n^{- 1}$

$28 n^{2} + 7 n + 8 n + 2 = 0$

$u = n^{- 1}$

Step 1.4.1.3

Factor out the greatest common factor from each group.

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Step 1.4.1.3.1

Group the first two terms and the last two terms.

$(28 n^{2} + 7 n) + 8 n + 2 = 0$

$u = n^{- 1}$

Step 1.4.1.3.2

Factor out the greatest common factor (GCF) from each group.

$7 n (4 n + 1) + 2 (4 n + 1) = 0$

$u = n^{- 1}$

$7 n (4 n + 1) + 2 (4 n + 1) = 0$

$u = n^{- 1}$

Step 1.4.1.4

Factor the polynomial by factoring out the greatest common factor, $4 n + 1$ .

$(4 n + 1) (7 n + 2) = 0$

$u = n^{- 1}$

$(4 n + 1) (7 n + 2) = 0$

$u = n^{- 1}$

Step 1.4.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$4 n + 1 = 0$

$7 n + 2 = 0$

$u = n^{- 1}$

Step 1.4.3

Set $4 n + 1$ equal to $0$ and solve for $n$ .

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Step 1.4.3.1

Set $4 n + 1$ equal to $0$ .

$4 n + 1 = 0$

$u = n^{- 1}$

Step 1.4.3.2

Solve $4 n + 1 = 0$ for $n$ .

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Step 1.4.3.2.1

Subtract $1$ from both sides of the equation.

$4 n = - 1$

$u = n^{- 1}$

Step 1.4.3.2.2

Divide each term in $4 n = - 1$ by $4$ and simplify.

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Step 1.4.3.2.2.1

Divide each term in $4 n = - 1$ by $4$ .

$\frac{4 n}{4} = \frac{- 1}{4}$

$u = n^{- 1}$

Step 1.4.3.2.2.2

Simplify the left side.

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Step 1.4.3.2.2.2.1

Cancel the common factor of $4$ .

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Step 1.4.3.2.2.2.1.1

Cancel the common factor.

$\frac{4 n}{4} = \frac{- 1}{4}$

$u = n^{- 1}$

Step 1.4.3.2.2.2.1.2

Divide $n$ by $1$ .

$n = \frac{- 1}{4}$

$u = n^{- 1}$

$n = \frac{- 1}{4}$

$u = n^{- 1}$

$n = \frac{- 1}{4}$

$u = n^{- 1}$

Step 1.4.3.2.2.3

Simplify the right side.

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Step 1.4.3.2.2.3.1

Move the negative in front of the fraction.

$n = - \frac{1}{4}$

$u = n^{- 1}$

$n = - \frac{1}{4}$

$u = n^{- 1}$

$n = - \frac{1}{4}$

$u = n^{- 1}$

$n = - \frac{1}{4}$

$u = n^{- 1}$

$n = - \frac{1}{4}$

$u = n^{- 1}$

Step 1.4.4

Set $7 n + 2$ equal to $0$ and solve for $n$ .

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Step 1.4.4.1

Set $7 n + 2$ equal to $0$ .

$7 n + 2 = 0$

$u = n^{- 1}$

Step 1.4.4.2

Solve $7 n + 2 = 0$ for $n$ .

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Step 1.4.4.2.1

Subtract $2$ from both sides of the equation.

$7 n = - 2$

$u = n^{- 1}$

Step 1.4.4.2.2

Divide each term in $7 n = - 2$ by $7$ and simplify.

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Step 1.4.4.2.2.1

Divide each term in $7 n = - 2$ by $7$ .

$\frac{7 n}{7} = \frac{- 2}{7}$

$u = n^{- 1}$

Step 1.4.4.2.2.2

Simplify the left side.

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Step 1.4.4.2.2.2.1

Cancel the common factor of $7$ .

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Step 1.4.4.2.2.2.1.1

Cancel the common factor.

$\frac{7 n}{7} = \frac{- 2}{7}$

$u = n^{- 1}$

Step 1.4.4.2.2.2.1.2

Divide $n$ by $1$ .

$n = \frac{- 2}{7}$

$u = n^{- 1}$

$n = \frac{- 2}{7}$

$u = n^{- 1}$

$n = \frac{- 2}{7}$

$u = n^{- 1}$

Step 1.4.4.2.2.3

Simplify the right side.

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Step 1.4.4.2.2.3.1

Move the negative in front of the fraction.

$n = - \frac{2}{7}$

$u = n^{- 1}$

$n = - \frac{2}{7}$

$u = n^{- 1}$

$n = - \frac{2}{7}$

$u = n^{- 1}$

$n = - \frac{2}{7}$

$u = n^{- 1}$

$n = - \frac{2}{7}$

$u = n^{- 1}$

Step 1.4.5

The final solution is all the values that make $(4 n + 1) (7 n + 2) = 0$ true.

$n = - \frac{1}{4}, - \frac{2}{7}$

$u = n^{- 1}$

$n = - \frac{1}{4}, - \frac{2}{7}$

$u = n^{- 1}$

$n = - \frac{1}{4}, - \frac{2}{7}$

$u = n^{- 1}$

Step 2

Replace all occurrences of $n$ with $- \frac{1}{4}$ in each equation.

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Step 2.1

Replace all occurrences of $n$ in $u = n^{- 1}$ with $- \frac{1}{4}$ .

$u = {(- \frac{1}{4})}^{- 1}$

$n = - \frac{1}{4}$

Step 2.2

Simplify the right side.

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Step 2.2.1

Change the sign of the exponent by rewriting the base as its reciprocal.

$u = - 4$

$n = - \frac{1}{4}$

$u = - 4$

$n = - \frac{1}{4}$

$u = - 4$

$n = - \frac{1}{4}$

Step 3

Replace all occurrences of $n$ with $- \frac{2}{7}$ in each equation.

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Step 3.1

Replace all occurrences of $n$ in $u = n^{- 1}$ with $- \frac{2}{7}$ .

$u = {(- \frac{2}{7})}^{- 1}$

$n = - \frac{2}{7}$

Step 3.2

Simplify the right side.

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Step 3.2.1

Change the sign of the exponent by rewriting the base as its reciprocal.

$u = - \frac{7}{2}$

$n = - \frac{2}{7}$

$u = - \frac{7}{2}$

$n = - \frac{2}{7}$

$u = - \frac{7}{2}$

$n = - \frac{2}{7}$

Step 4

List all of the solutions.

$u = - 4, n = - \frac{1}{4}$

$u = - \frac{7}{2}, n = - \frac{2}{7}$