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Solve the rational equation by multiplying both sides by the LCD. Check your results for extraneous solutions.

[tex]\[
\frac{3}{x^2 + 5x + 6} + \frac{x-1}{x+2} = \frac{7}{x+3}
\][/tex]

\( x = \square \) is a solution.

[tex]\( x = \square \)[/tex] is an extraneous solution.

Sagot :

GinnyAnswer
Alright, let's solve the given rational equation step-by-step and identify the solutions and extraneous solutions.

Given equation:
[tex]\[ \frac{3}{x^2 + 5x + 6} + \frac{x-1}{x+2} = \frac{7}{x+3} \][/tex]

### Step 1: Factor the denominators

First, we factor the quadratic denominator to simplify our work with the least common denominator (LCD).

[tex]\[ x^2 + 5x + 6 = (x + 2)(x + 3) \][/tex]

### Step 2: Determine the LCD

The least common denominator (LCD) of \(x^2 + 5x + 6\), \(x + 2\), and \(x + 3\) is:

[tex]\[ \text{LCD} = (x + 2)(x + 3) \][/tex]

### Step 3: Multiply both sides by the LCD

We multiply both sides of the equation by the LCD to eliminate the denominators:

[tex]\[ (x+2)(x+3) \left(\frac{3}{(x + 2)(x + 3)} + \frac{x-1}{x + 2}\right) = (x+2)(x+3) \left(\frac{7}{x + 3}\right) \][/tex]

### Step 4: Simplify the equation

Simplify each term:

[tex]\[ 3 + (x - 1)(x + 3) = 7(x + 2) \][/tex]

Distribute and combine like terms:

[tex]\[ 3 + x^2 + 2x - 3 = 7x + 14 \][/tex]

[tex]\[ x^2 + 2x = 7x + 14 \][/tex]

### Step 5: Rearrange the equation

Move all terms to one side of the equation:

[tex]\[ x^2 + 2x - 7x - 14 = 0 \][/tex]

[tex]\[ x^2 - 5x - 14 = 0 \][/tex]

### Step 6: Solve the quadratic equation

We solve the quadratic equation by factoring, if possible, or using the quadratic formula. This factors nicely as:

[tex]\[ (x - 7)(x + 2) = 0 \][/tex]

This yields the solutions:
[tex]\[ x = 7 \quad \text{or} \quad x = -2 \][/tex]

### Step 7: Check for extraneous solutions

We substitute each solution back into the original equation to check if it is valid. Solutions that make any denominator zero are extraneous.

For \( x = 7 \):
- The denominators \( x^2 + 5x + 6 \), \( x + 2 \), and \( x + 3 \) are all non-zero.
[tex]\[ x^2 + 5x + 6 = 49 + 35 + 6 = 90 \neq 0 \][/tex]
[tex]\[ x + 2 = 7 + 2 = 9 \neq 0 \][/tex]
[tex]\[ x + 3 = 7 + 3 = 10 \neq 0 \][/tex]
So, \( x = 7 \) is a valid solution.

For \( x = -2 \):
- The denominators \( x + 2 \) is zero:
[tex]\[ x + 2 = -2 + 2 = 0 \][/tex]

This makes \( x = -2 \) an extraneous solution.

### Final Results

Thus, the solution to the equation is:
[tex]\[ x = 7 \quad \text{(valid solution)} \][/tex]
[tex]\[ x = -2 \quad \text{(extraneous solution)} \][/tex]
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