Solve for x 3/(x^2+x-2)-1/(x^2-1)=7/(2x^2+6x+4)

$\frac{3}{x^{2} + x - 2} - \frac{1}{x^{2} - 1} = \frac{7}{2 x^{2} + 6 x + 4}$

Step 1

Factor each term.

Factor $x^{2} + x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

$\frac{3}{(x - 1) (x + 2)} - \frac{1}{x^{2} - 1} = \frac{7}{2 x^{2} + 6 x + 4}$

Rewrite $1$ as $1^{2}$ .

$\frac{3}{(x - 1) (x + 2)} - \frac{1}{x^{2} - 1^{2}} = \frac{7}{2 x^{2} + 6 x + 4}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\frac{3}{(x - 1) (x + 2)} - \frac{1}{(x + 1) (x - 1)} = \frac{7}{2 x^{2} + 6 x + 4}$

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

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

$\frac{3}{(x - 1) (x + 2)} - \frac{1}{(x + 1) (x - 1)} = \frac{7}{2 (x^{2}) + 6 x + 4}$

Factor $2$ out of $6 x$ .

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

Factor $2$ out of $4$ .

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

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

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

Factor $2$ out of $2 (x^{2} + 3 x) + 2 \cdot 2$ .

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

Factor.

Factor $x^{2} + 3 x + 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $2$ and whose sum is $3$ .

$1, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 2

Find the LCD of the terms in the equation.

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

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

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.

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

Not prime

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

$2$ is a prime number

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

$2$

The factor for $x - 1$ is $x - 1$ itself.

$(x - 1) = x - 1$

$(x - 1)$ occurs $1$ time.

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

$(x + 2) = x + 2$

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

The factor for $x + 1$ is $x + 1$ itself.

$(x + 1) = x + 1$

$(x + 1)$ occurs $1$ time.

The factor for $x - 1$ is $x - 1$ itself.

$(x - 1) = x - 1$

$(x - 1)$ occurs $1$ time.

The factor for $x + 1$ is $x + 1$ itself.

$(x + 1) = x + 1$

$(x + 1)$ occurs $1$ time.

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

$(x + 2) = x + 2$

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

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

$(x - 1) (x + 2) (x + 1)$

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

$2 (x - 1) (x + 2) (x + 1)$

Step 3

Multiply each term in $\frac{3}{(x - 1) (x + 2)} - \frac{1}{(x + 1) (x - 1)} = \frac{7}{2 (x + 1) (x + 2)}$ by $2 (x - 1) (x + 2) (x + 1)$ to eliminate the fractions.

Multiply each term in $\frac{3}{(x - 1) (x + 2)} - \frac{1}{(x + 1) (x - 1)} = \frac{7}{2 (x + 1) (x + 2)}$ by $2 (x - 1) (x + 2) (x + 1)$ .

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

Simplify the left side.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

Multiply $2 \frac{3}{(x - 1) (x + 2)}$ .

Combine $2$ and $\frac{3}{(x - 1) (x + 2)}$ .

$\frac{2 \cdot 3}{(x - 1) (x + 2)} ((x - 1) (x + 2) (x + 1)) - \frac{1}{(x + 1) (x - 1)} (2 (x - 1) (x + 2) (x + 1)) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

Multiply $2$ by $3$ .

$\frac{6}{(x - 1) (x + 2)} ((x - 1) (x + 2) (x + 1)) - \frac{1}{(x + 1) (x - 1)} (2 (x - 1) (x + 2) (x + 1)) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

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

Cancel the common factor.

$\frac{6}{(x - 1) (x + 2)} ((x - 1) (x + 2) (x + 1)) - \frac{1}{(x + 1) (x - 1)} (2 (x - 1) (x + 2) (x + 1)) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

Rewrite the expression.

$6 (x + 1) - \frac{1}{(x + 1) (x - 1)} (2 (x - 1) (x + 2) (x + 1)) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

Apply the distributive property.

$6 x + 6 \cdot 1 - \frac{1}{(x + 1) (x - 1)} (2 (x - 1) (x + 2) (x + 1)) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

Multiply $6$ by $1$ .

$6 x + 6 - \frac{1}{(x + 1) (x - 1)} (2 (x - 1) (x + 2) (x + 1)) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

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

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

$6 x + 6 + \frac{- 1}{(x + 1) (x - 1)} (2 (x - 1) (x + 2) (x + 1)) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

Factor $(x - 1) (x + 1)$ out of $(x + 1) (x - 1)$ .

$6 x + 6 + \frac{- 1}{(x - 1) (x + 1) (1)} (2 (x - 1) (x + 2) (x + 1)) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

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

$6 x + 6 + \frac{- 1}{(x - 1) (x + 1) (1)} ((x - 1) (x + 1) (2 (x + 2))) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

Cancel the common factor.

$6 x + 6 + \frac{- 1}{(x - 1) (x + 1) \cdot 1} ((x - 1) (x + 1) (2 (x + 2))) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

Rewrite the expression.

$6 x + 6 - (2 (x + 2)) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

Multiply $2$ by $- 1$ .

$6 x + 6 - 2 (x + 2) = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

Apply the distributive property.

$6 x + 6 - 2 x - 2 \cdot 2 = \frac{7}{2 (x + 1) (x + 2)} (2 (x - 1) (x + 2) (x + 1))$

Multiply $- 2$ by $2$ .

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

Simplify by adding terms.

Subtract $2 x$ from $6 x$ .

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

Subtract $4$ from $6$ .

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

Simplify the right side.

Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 (x + 1) (x + 2)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $(x + 2) (x + 1)$ out of $(x + 1) (x + 2)$ .

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

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

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

Cancel the common factor.

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

Rewrite the expression.

$4 x + 2 = 7 (x - 1)$

Apply the distributive property.

$4 x + 2 = 7 x + 7 \cdot - 1$

Multiply $7$ by $- 1$ .

$4 x + 2 = 7 x - 7$

Step 4

Solve the equation.

Move all terms containing $x$ to the left side of the equation.

Subtract $7 x$ from both sides of the equation.

$4 x + 2 - 7 x = - 7$

Subtract $7 x$ from $4 x$ .

$- 3 x + 2 = - 7$

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

Subtract $2$ from both sides of the equation.

$- 3 x = - 7 - 2$

Subtract $2$ from $- 7$ .

$- 3 x = - 9$

Divide each term in $- 3 x = - 9$ by $- 3$ and simplify.

Divide each term in $- 3 x = - 9$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{- 9}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{- 9}{- 3}$

Divide $x$ by $1$ .

$x = \frac{- 9}{- 3}$

Simplify the right side.

Divide $- 9$ by $- 3$ .

$x = 3$

Solve for x 3/(x^2+x-2)-1/(x^2-1)=7/(2x^2+6x+4)

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