Finite Math Examples

$\frac{2}{n + 2} - \frac{4}{4 n - 2} = - \frac{7}{4 n^{2} + 6 n - 4}$

Step 1

Factor each term.

Tap for more steps...

Step 1.1

Factor $2$ out of $4 n - 2$ .

Tap for more steps...

Step 1.1.1

Factor $2$ out of $4 n$ .

$\frac{2}{n + 2} - \frac{4}{2 (2 n) - 2} = - \frac{7}{4 n^{2} + 6 n - 4}$

Step 1.1.2

Factor $2$ out of $- 2$ .

$\frac{2}{n + 2} - \frac{4}{2 (2 n) + 2 (- 1)} = - \frac{7}{4 n^{2} + 6 n - 4}$

Step 1.1.3

Factor $2$ out of $2 (2 n) + 2 (- 1)$ .

$\frac{2}{n + 2} - \frac{4}{2 (2 n - 1)} = - \frac{7}{4 n^{2} + 6 n - 4}$

Step 1.2

Reduce the expression $\frac{4}{2 (2 n - 1)}$ by cancelling the common factors.

Tap for more steps...

Step 1.2.1

Factor $2$ out of $4$ .

$\frac{2}{n + 2} - \frac{2 \cdot 2}{2 (2 n - 1)} = - \frac{7}{4 n^{2} + 6 n - 4}$

Step 1.2.2

Cancel the common factor.

$\frac{2}{n + 2} - \frac{2 \cdot 2}{2 (2 n - 1)} = - \frac{7}{4 n^{2} + 6 n - 4}$

Step 1.2.3

Rewrite the expression.

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{4 n^{2} + 6 n - 4}$

Step 1.3

Factor $2$ out of $4 n^{2} + 6 n - 4$ .

Tap for more steps...

Step 1.3.1

Factor $2$ out of $4 n^{2}$ .

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (2 n^{2}) + 6 n - 4}$

Step 1.3.2

Factor $2$ out of $6 n$ .

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (2 n^{2}) + 2 (3 n) - 4}$

Step 1.3.3

Factor $2$ out of $- 4$ .

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (2 n^{2}) + 2 (3 n) + 2 (- 2)}$

Step 1.3.4

Factor $2$ out of $2 (2 n^{2}) + 2 (3 n)$ .

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (2 n^{2} + 3 n) + 2 (- 2)}$

Step 1.3.5

Factor $2$ out of $2 (2 n^{2} + 3 n) + 2 (- 2)$ .

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (2 n^{2} + 3 n - 2)}$

Step 1.4

Factor.

Tap for more steps...

Step 1.4.1

Factor by grouping.

Tap for more steps...

Step 1.4.1.1

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 $a \cdot c = 2 \cdot - 2 = - 4$ and whose sum is $b = 3$ .

Tap for more steps...

Step 1.4.1.1.1

Factor $3$ out of $3 n$ .

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (2 n^{2} + 3 (n) - 2)}$

Step 1.4.1.1.2

Rewrite $3$ as $- 1$ plus $4$

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (2 n^{2} + (- 1 + 4) n - 2)}$

Step 1.4.1.1.3

Apply the distributive property.

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (2 n^{2} - 1 n + 4 n - 2)}$

Step 1.4.1.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 1.4.1.2.1

Group the first two terms and the last two terms.

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 ((2 n^{2} - 1 n) + 4 n - 2)}$

Step 1.4.1.2.2

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

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (n (2 n - 1) + 2 (2 n - 1))}$

Step 1.4.1.3

Factor the polynomial by factoring out the greatest common factor, $2 n - 1$ .

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 ((2 n - 1) (n + 2))}$

Step 1.4.2

Remove unnecessary parentheses.

$\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (2 n - 1) (n + 2)}$

Step 2

Find the LCD of the terms in the equation.

Tap for more steps...

Step 2.1

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

$n + 2, 2 n - 1, 2 (2 n - 1) (n + 2)$

Step 2.2

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 2.3

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

Not prime

Step 2.4

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

$2$ is a prime number

Step 2.5

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

$2$

Step 2.6

The factor for $n + 2$ is $n + 2$ itself.

$(n + 2) = n + 2$

$(n + 2)$ occurs $1$ time.

Step 2.7

The factor for $2 n - 1$ is $2 n - 1$ itself.

$(2 n - 1) = 2 n - 1$

$(2 n - 1)$ occurs $1$ time.

Step 2.8

The factor for $n + 2$ is $n + 2$ itself.

$(n + 2) = n + 2$

$(n + 2)$ occurs $1$ time.

Step 2.9

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

$(n + 2) (2 n - 1)$

Step 2.10

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

$2 (n + 2) (2 n - 1)$

Step 3

Multiply each term in $\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (2 n - 1) (n + 2)}$ by $2 (n + 2) (2 n - 1)$ to eliminate the fractions.

Tap for more steps...

Step 3.1

Multiply each term in $\frac{2}{n + 2} - \frac{2}{2 n - 1} = - \frac{7}{2 (2 n - 1) (n + 2)}$ by $2 (n + 2) (2 n - 1)$ .

$\frac{2}{n + 2} (2 (n + 2) (2 n - 1)) - \frac{2}{2 n - 1} (2 (n + 2) (2 n - 1)) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2

Simplify the left side.

Tap for more steps...

Step 3.2.1

Simplify each term.

Tap for more steps...

Step 3.2.1.1

Rewrite using the commutative property of multiplication.

$2 \frac{2}{n + 2} ((n + 2) (2 n - 1)) - \frac{2}{2 n - 1} (2 (n + 2) (2 n - 1)) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.2

Multiply $2 \frac{2}{n + 2}$ .

Tap for more steps...

Step 3.2.1.2.1

Combine $2$ and $\frac{2}{n + 2}$ .

$\frac{2 \cdot 2}{n + 2} ((n + 2) (2 n - 1)) - \frac{2}{2 n - 1} (2 (n + 2) (2 n - 1)) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.2.2

Multiply $2$ by $2$ .

$\frac{4}{n + 2} ((n + 2) (2 n - 1)) - \frac{2}{2 n - 1} (2 (n + 2) (2 n - 1)) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.3

Cancel the common factor of $n + 2$ .

Tap for more steps...

Step 3.2.1.3.1

Cancel the common factor.

$\frac{4}{n + 2} ((n + 2) (2 n - 1)) - \frac{2}{2 n - 1} (2 (n + 2) (2 n - 1)) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.3.2

Rewrite the expression.

$4 (2 n - 1) - \frac{2}{2 n - 1} (2 (n + 2) (2 n - 1)) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.4

Apply the distributive property.

$4 (2 n) + 4 \cdot - 1 - \frac{2}{2 n - 1} (2 (n + 2) (2 n - 1)) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.5

Multiply $2$ by $4$ .

$8 n + 4 \cdot - 1 - \frac{2}{2 n - 1} (2 (n + 2) (2 n - 1)) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.6

Multiply $4$ by $- 1$ .

$8 n - 4 - \frac{2}{2 n - 1} (2 (n + 2) (2 n - 1)) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.7

Cancel the common factor of $2 n - 1$ .

Tap for more steps...

Step 3.2.1.7.1

Move the leading negative in $- \frac{2}{2 n - 1}$ into the numerator.

$8 n - 4 + \frac{- 2}{2 n - 1} (2 (n + 2) (2 n - 1)) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.7.2

Factor $2 n - 1$ out of $2 (n + 2) (2 n - 1)$ .

$8 n - 4 + \frac{- 2}{2 n - 1} ((2 n - 1) (2 (n + 2))) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.7.3

Cancel the common factor.

$8 n - 4 + \frac{- 2}{2 n - 1} ((2 n - 1) (2 (n + 2))) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.7.4

Rewrite the expression.

$8 n - 4 - 2 (2 (n + 2)) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.8

Multiply $2$ by $- 2$ .

$8 n - 4 - 4 (n + 2) = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.9

Apply the distributive property.

$8 n - 4 - 4 n - 4 \cdot 2 = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.1.10

Multiply $- 4$ by $2$ .

$8 n - 4 - 4 n - 8 = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.2

Simplify by adding terms.

Tap for more steps...

Step 3.2.2.1

Subtract $4 n$ from $8 n$ .

$4 n - 4 - 8 = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.2.2.2

Subtract $8$ from $- 4$ .

$4 n - 12 = - \frac{7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.3

Simplify the right side.

Tap for more steps...

Step 3.3.1

Cancel the common factor of $2 (n + 2) (2 n - 1)$ .

Tap for more steps...

Step 3.3.1.1

Move the leading negative in $- \frac{7}{2 (2 n - 1) (n + 2)}$ into the numerator.

$4 n - 12 = \frac{- 7}{2 (2 n - 1) (n + 2)} (2 (n + 2) (2 n - 1))$

Step 3.3.1.2

Factor $2 (n + 2) (2 n - 1)$ out of $2 (2 n - 1) (n + 2)$ .

$4 n - 12 = \frac{- 7}{2 (n + 2) (2 n - 1) (1)} (2 (n + 2) (2 n - 1))$

Step 3.3.1.3

Cancel the common factor.

$4 n - 12 = \frac{- 7}{2 (n + 2) (2 n - 1) \cdot 1} (2 (n + 2) (2 n - 1))$

Step 3.3.1.4

Rewrite the expression.

$4 n - 12 = - 7$

Step 4

Solve the equation.

Tap for more steps...

Step 4.1

Move all terms not containing $n$ to the right side of the equation.

Tap for more steps...

Step 4.1.1

Add $12$ to both sides of the equation.

$4 n = - 7 + 12$

Step 4.1.2

Add $- 7$ and $12$ .

$4 n = 5$

Step 4.2

Divide each term in $4 n = 5$ by $4$ and simplify.

Tap for more steps...

Step 4.2.1

Divide each term in $4 n = 5$ by $4$ .

$\frac{4 n}{4} = \frac{5}{4}$

Step 4.2.2

Simplify the left side.

Tap for more steps...

Step 4.2.2.1

Cancel the common factor of $4$ .

Tap for more steps...

Step 4.2.2.1.1

Cancel the common factor.

$\frac{4 n}{4} = \frac{5}{4}$

Step 4.2.2.1.2

Divide $n$ by $1$ .

$n = \frac{5}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$n = \frac{5}{4}$

Decimal Form:

$n = 1.25$

Mixed Number Form:

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