Precalculus Examples

$\frac{5 x^{2} + 3 x - 2}{2 x^{2} + 15 x - 27} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1

Factor each term.

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Step 1.1

Factor by grouping.

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Step 1.1.1

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 $a \cdot c = 5 \cdot - 2 = - 10$ and whose sum is $b = 3$ .

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Step 1.1.1.1

Factor $3$ out of $3 x$ .

$\frac{5 x^{2} + 3 (x) - 2}{2 x^{2} + 15 x - 27} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1.1.1.2

Rewrite $3$ as $- 2$ plus $5$

$\frac{5 x^{2} + (- 2 + 5) x - 2}{2 x^{2} + 15 x - 27} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1.1.1.3

Apply the distributive property.

$\frac{5 x^{2} - 2 x + 5 x - 2}{2 x^{2} + 15 x - 27} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1.1.2

Factor out the greatest common factor from each group.

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Step 1.1.2.1

Group the first two terms and the last two terms.

$\frac{(5 x^{2} - 2 x) + 5 x - 2}{2 x^{2} + 15 x - 27} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1.1.2.2

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

$\frac{x (5 x - 2) + 1 (5 x - 2)}{2 x^{2} + 15 x - 27} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1.1.3

Factor the polynomial by factoring out the greatest common factor, $5 x - 2$ .

$\frac{(5 x - 2) (x + 1)}{2 x^{2} + 15 x - 27} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1.2

Factor by grouping.

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Step 1.2.1

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 $a \cdot c = 2 \cdot - 27 = - 54$ and whose sum is $b = 15$ .

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Step 1.2.1.1

Factor $15$ out of $15 x$ .

$\frac{(5 x - 2) (x + 1)}{2 x^{2} + 15 (x) - 27} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1.2.1.2

Rewrite $15$ as $- 3$ plus $18$

$\frac{(5 x - 2) (x + 1)}{2 x^{2} + (- 3 + 18) x - 27} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1.2.1.3

Apply the distributive property.

$\frac{(5 x - 2) (x + 1)}{2 x^{2} - 3 x + 18 x - 27} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1.2.2

Factor out the greatest common factor from each group.

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Step 1.2.2.1

Group the first two terms and the last two terms.

$\frac{(5 x - 2) (x + 1)}{(2 x^{2} - 3 x) + 18 x - 27} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1.2.2.2

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

$\frac{(5 x - 2) (x + 1)}{x (2 x - 3) + 9 (2 x - 3)} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 1.2.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 3$ .

$\frac{(5 x - 2) (x + 1)}{(2 x - 3) (x + 9)} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

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

$(2 x - 3) (x + 9), 2 x - 3, x + 9$

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

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

$1$

Step 2.5

The factor for $2 x - 3$ is $2 x - 3$ itself.

$(2 x - 3) = 2 x - 3$

$(2 x - 3)$ occurs $1$ time.

Step 2.6

The factor for $x + 9$ is $x + 9$ itself.

$(x + 9) = x + 9$

$(x + 9)$ occurs $1$ time.

Step 2.7

The factor for $2 x - 3$ is $2 x - 3$ itself.

$(2 x - 3) = 2 x - 3$

$(2 x - 3)$ occurs $1$ time.

Step 2.8

The factor for $x + 9$ is $x + 9$ itself.

$(x + 9) = x + 9$

$(x + 9)$ occurs $1$ time.

Step 2.9

The LCM of $2 x - 3, x + 9, 2 x - 3, x + 9$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(2 x - 3) (x + 9)$

Step 3

Multiply each term in $\frac{(5 x - 2) (x + 1)}{(2 x - 3) (x + 9)} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$ by $(2 x - 3) (x + 9)$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{(5 x - 2) (x + 1)}{(2 x - 3) (x + 9)} - \frac{3 x + 1}{2 x - 3} = \frac{8}{x + 9}$ by $(2 x - 3) (x + 9)$ .

$\frac{(5 x - 2) (x + 1)}{(2 x - 3) (x + 9)} ((2 x - 3) (x + 9)) - \frac{3 x + 1}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Cancel the common factor of $(2 x - 3) (x + 9)$ .

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Step 3.2.1.1.1

Cancel the common factor.

$\frac{(5 x - 2) (x + 1)}{(2 x - 3) (x + 9)} ((2 x - 3) (x + 9)) - \frac{3 x + 1}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.1.2

Rewrite the expression.

$(5 x - 2) (x + 1) - \frac{3 x + 1}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.2

Expand $(5 x - 2) (x + 1)$ using the FOIL Method.

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Step 3.2.1.2.1

Apply the distributive property.

$5 x (x + 1) - 2 (x + 1) - \frac{3 x + 1}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.2.2

Apply the distributive property.

$5 x \cdot x + 5 x \cdot 1 - 2 (x + 1) - \frac{3 x + 1}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.2.3

Apply the distributive property.

$5 x \cdot x + 5 x \cdot 1 - 2 x - 2 \cdot 1 - \frac{3 x + 1}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.3

Simplify and combine like terms.

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Step 3.2.1.3.1

Simplify each term.

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Step 3.2.1.3.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 3.2.1.3.1.1.1

Move $x$ .

$5 (x \cdot x) + 5 x \cdot 1 - 2 x - 2 \cdot 1 - \frac{3 x + 1}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.3.1.1.2

Multiply $x$ by $x$ .

$5 x^{2} + 5 x \cdot 1 - 2 x - 2 \cdot 1 - \frac{3 x + 1}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.3.1.2

Multiply $5$ by $1$ .

$5 x^{2} + 5 x - 2 x - 2 \cdot 1 - \frac{3 x + 1}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.3.1.3

Multiply $- 2$ by $1$ .

$5 x^{2} + 5 x - 2 x - 2 - \frac{3 x + 1}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.3.2

Subtract $2 x$ from $5 x$ .

$5 x^{2} + 3 x - 2 - \frac{3 x + 1}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.4

Cancel the common factor of $2 x - 3$ .

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Step 3.2.1.4.1

Move the leading negative in $- \frac{3 x + 1}{2 x - 3}$ into the numerator.

$5 x^{2} + 3 x - 2 + \frac{- (3 x + 1)}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.4.2

Cancel the common factor.

$5 x^{2} + 3 x - 2 + \frac{- (3 x + 1)}{2 x - 3} ((2 x - 3) (x + 9)) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.4.3

Rewrite the expression.

$5 x^{2} + 3 x - 2 - (3 x + 1) (x + 9) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.5

Apply the distributive property.

$5 x^{2} + 3 x - 2 + (- (3 x) - 1 \cdot 1) (x + 9) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.6

Multiply $3$ by $- 1$ .

$5 x^{2} + 3 x - 2 + (- 3 x - 1 \cdot 1) (x + 9) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.7

Multiply $- 1$ by $1$ .

$5 x^{2} + 3 x - 2 + (- 3 x - 1) (x + 9) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.8

Expand $(- 3 x - 1) (x + 9)$ using the FOIL Method.

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Step 3.2.1.8.1

Apply the distributive property.

$5 x^{2} + 3 x - 2 - 3 x (x + 9) - 1 (x + 9) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.8.2

Apply the distributive property.

$5 x^{2} + 3 x - 2 - 3 x \cdot x - 3 x \cdot 9 - 1 (x + 9) = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.8.3

Apply the distributive property.

$5 x^{2} + 3 x - 2 - 3 x \cdot x - 3 x \cdot 9 - 1 x - 1 \cdot 9 = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.9

Simplify and combine like terms.

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Step 3.2.1.9.1

Simplify each term.

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Step 3.2.1.9.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 3.2.1.9.1.1.1

Move $x$ .

$5 x^{2} + 3 x - 2 - 3 (x \cdot x) - 3 x \cdot 9 - 1 x - 1 \cdot 9 = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.9.1.1.2

Multiply $x$ by $x$ .

$5 x^{2} + 3 x - 2 - 3 x^{2} - 3 x \cdot 9 - 1 x - 1 \cdot 9 = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.9.1.2

Multiply $9$ by $- 3$ .

$5 x^{2} + 3 x - 2 - 3 x^{2} - 27 x - 1 x - 1 \cdot 9 = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.9.1.3

Rewrite $- 1 x$ as $- x$ .

$5 x^{2} + 3 x - 2 - 3 x^{2} - 27 x - x - 1 \cdot 9 = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.9.1.4

Multiply $- 1$ by $9$ .

$5 x^{2} + 3 x - 2 - 3 x^{2} - 27 x - x - 9 = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.1.9.2

Subtract $x$ from $- 27 x$ .

$5 x^{2} + 3 x - 2 - 3 x^{2} - 28 x - 9 = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.2

Simplify by adding terms.

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Step 3.2.2.1

Subtract $3 x^{2}$ from $5 x^{2}$ .

$2 x^{2} + 3 x - 2 - 28 x - 9 = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.2.2

Subtract $28 x$ from $3 x$ .

$2 x^{2} - 25 x - 2 - 9 = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.2.2.3

Subtract $9$ from $- 2$ .

$2 x^{2} - 25 x - 11 = \frac{8}{x + 9} ((2 x - 3) (x + 9))$

Step 3.3

Simplify the right side.

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Step 3.3.1

Cancel the common factor of $x + 9$ .

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Step 3.3.1.1

Factor $x + 9$ out of $(2 x - 3) (x + 9)$ .

$2 x^{2} - 25 x - 11 = \frac{8}{x + 9} ((x + 9) (2 x - 3))$

Step 3.3.1.2

Cancel the common factor.

$2 x^{2} - 25 x - 11 = \frac{8}{x + 9} ((x + 9) (2 x - 3))$

Step 3.3.1.3

Rewrite the expression.

$2 x^{2} - 25 x - 11 = 8 (2 x - 3)$

Step 3.3.2

Apply the distributive property.

$2 x^{2} - 25 x - 11 = 8 (2 x) + 8 \cdot - 3$

Step 3.3.3

Multiply.

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Step 3.3.3.1

Multiply $2$ by $8$ .

$2 x^{2} - 25 x - 11 = 16 x + 8 \cdot - 3$

Step 3.3.3.2

Multiply $8$ by $- 3$ .

$2 x^{2} - 25 x - 11 = 16 x - 24$

Step 4

Solve the equation.

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Step 4.1

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

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Step 4.1.1

Subtract $16 x$ from both sides of the equation.

$2 x^{2} - 25 x - 11 - 16 x = - 24$

Step 4.1.2

Subtract $16 x$ from $- 25 x$ .

$2 x^{2} - 41 x - 11 = - 24$

Step 4.2

Move all terms to the left side of the equation and simplify.

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Step 4.2.1

Add $24$ to both sides of the equation.

$2 x^{2} - 41 x - 11 + 24 = 0$

Step 4.2.2

Add $- 11$ and $24$ .

$2 x^{2} - 41 x + 13 = 0$

Step 4.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.4

Substitute the values $a = 2$ , $b = - 41$ , and $c = 13$ into the quadratic formula and solve for $x$ .

$\frac{41 \pm \sqrt{{(- 41)}^{2} - 4 \cdot (2 \cdot 13)}}{2 \cdot 2}$

Step 4.5

Simplify.

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Step 4.5.1

Simplify the numerator.

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Step 4.5.1.1

Raise $- 41$ to the power of $2$ .

$x = \frac{41 \pm \sqrt{1681 - 4 \cdot 2 \cdot 13}}{2 \cdot 2}$

Step 4.5.1.2

Multiply $- 4 \cdot 2 \cdot 13$ .

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Step 4.5.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{41 \pm \sqrt{1681 - 8 \cdot 13}}{2 \cdot 2}$

Step 4.5.1.2.2

Multiply $- 8$ by $13$ .

$x = \frac{41 \pm \sqrt{1681 - 104}}{2 \cdot 2}$

Step 4.5.1.3

Subtract $104$ from $1681$ .

$x = \frac{41 \pm \sqrt{1577}}{2 \cdot 2}$

Step 4.5.2

Multiply $2$ by $2$ .

$x = \frac{41 \pm \sqrt{1577}}{4}$

Step 4.6

The final answer is the combination of both solutions.

$x = \frac{41 + \sqrt{1577}}{4}, \frac{41 - \sqrt{1577}}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{41 + \sqrt{1577}}{4}, \frac{41 - \sqrt{1577}}{4}$

Decimal Form:

$x = 20.17786482 \dots, 0.32213517 \dots$