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Question Solve the equation 3x2−2x−15−1x2−25=3x2+8x+15\frac{3}{x^{2}-2x-15}-\frac{1}{x^{2}-25}=\frac{3}{x^{2}+8x+15} and determine the solution set.

Studdy Solution

STEP 1

Assumptions
1. We are given the equation: 3x2−2x−15−1x2−25=3x2+8x+15\frac{3}{x^{2}-2x-15}-\frac{1}{x^{2}-25}=\frac{3}{x^{2}+8x+15}
2. We need to solve for the variable xx.
3. The solution set can be all real numbers, the empty set, or a finite set.

STEP 2

First, we need to factor the denominators in the equation to simplify the expression.
x2−2x−15=(x−5)(x+3)x^{2}-2x-15 = (x-5)(x+3) x2−25=(x−5)(x+5)x^{2}-25 = (x-5)(x+5) x2+8x+15=(x+3)(x+5)x^{2}+8x+15 = (x+3)(x+5)

STEP 3

Rewrite the equation with the factored denominators.
3(x−5)(x+3)−1(x−5)(x+5)=3(x+3)(x+5)\frac{3}{(x-5)(x+3)}-\frac{1}{(x-5)(x+5)}=\frac{3}{(x+3)(x+5)}

STEP 4

Identify the common denominator, which is the product of all distinct linear factors from the denominators.
Common denominator=(x−5)(x+3)(x+5)Common\, denominator = (x-5)(x+3)(x+5)

STEP 5

Rewrite each fraction with the common denominator.
3(x+5)(x−5)(x+3)(x+5)−(x+3)(x−5)(x+3)(x+5)=3(x−5)(x+3)(x+5)(x−5)\frac{3(x+5)}{(x-5)(x+3)(x+5)}-\frac{(x+3)}{(x-5)(x+3)(x+5)}=\frac{3(x-5)}{(x+3)(x+5)(x-5)}

STEP 6

Since the denominators are now the same, we can combine the numerators and set them equal to each other.
3(x+5)−(x+3)=3(x−5)3(x+5)-(x+3)=3(x-5)

STEP 7

Expand the terms in the numerators.
3x+15−x−3=3x−153x+15-x-3=3x-15

STEP 8

Combine like terms on the left side of the equation.
(3x−x)+(15−3)=3x−15(3x-x)+(15-3)=3x-15

STEP 9

Simplify the left side of the equation.
2x+12=3x−152x+12=3x-15

STEP 10

Subtract 2x2x from both sides of the equation to get all the xx terms on one side.
2x+12−2x=3x−15−2x2x+12-2x=3x-15-2x

STEP 11

Simplify both sides of the equation.
12=x−1512=x-15

STEP 12

Add 1515 to both sides of the equation to solve for xx.
12+15=x−15+1512+15=x-15+15

STEP 13

Simplify both sides of the equation.
27=x27=x

STEP 14

We have found a single solution for xx, which is x=27x=27.

STEP 15

Check the solution by substituting x=27x=27 back into the original equation to ensure it does not make any denominator zero.
3272−2(27)−15−1272−25=3272+8(27)+15\frac{3}{27^{2}-2(27)-15}-\frac{1}{27^{2}-25}=\frac{3}{27^{2}+8(27)+15}

STEP 16

Calculate the values of the denominators to ensure they are not zero.
(272−2(27)−15)=(729−54−15)=660≠0(27^{2}-2(27)-15) = (729-54-15) = 660 \neq 0 (272−25)=(729−25)=704≠0(27^{2}-25) = (729-25) = 704 \neq 0 (272+8(27)+15)=(729+216+15)=960≠0(27^{2}+8(27)+15) = (729+216+15) = 960 \neq 0

STEP 17

Since none of the denominators are zero, the solution x=27x=27 is valid.

STEP 18

Since we have found a single valid solution, the solution set is a finite set.

STEP 19

Write the solution set.
The solution set is {27}\{27\}.

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