For the rational equation, [tex]\frac{x^2+5x+6}{x+3}=1[/tex], what is a valid value of [tex]x[/tex]?

A. -2
B. -1
C. 2
D. 3



Answer :

To solve the rational equation [tex]\(\frac{x^2 + 5x + 6}{x + 3} = 1\)[/tex], we need to eliminate the fraction by multiplying both sides of the equation by [tex]\(x + 3\)[/tex]. Here are the steps in detail:

1. Start with the given equation:
[tex]\[ \frac{x^2 + 5x + 6}{x + 3} = 1 \][/tex]

2. Multiply both sides by [tex]\(x + 3\)[/tex] to eliminate the denominator:
[tex]\[ x^2 + 5x + 6 = 1 \cdot (x + 3) \][/tex]
Simplifying the right side:
[tex]\[ x^2 + 5x + 6 = x + 3 \][/tex]

3. Bring all terms to one side to set the equation to zero:
[tex]\[ x^2 + 5x + 6 - x - 3 = 0 \][/tex]
Combine like terms:
[tex]\[ x^2 + 4x + 3 = 0 \][/tex]

4. Solve the quadratic equation [tex]\(x^2 + 4x + 3 = 0\)[/tex]:
To solve the quadratic equation, we can factor it.
The quadratic [tex]\(x^2 + 4x + 3\)[/tex] factors into:
[tex]\[ (x + 1)(x + 3) = 0 \][/tex]

5. Set each factor equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ x + 1 = 0 \quad \text{or} \quad x + 3 = 0 \][/tex]
Solving these equations, we get:
[tex]\[ x = -1 \quad \text{or} \quad x = -3 \][/tex]

6. Check for any restrictions or invalid values:
Since the original equation has a denominator [tex]\(x + 3\)[/tex], we must ensure we don't choose a value that makes the denominator zero.
If [tex]\(x = -3\)[/tex], the denominator becomes zero, which is undefined. Therefore, [tex]\(x = -3\)[/tex] is not a valid solution.

Thus, the only valid solution is:
[tex]\[ x = -1 \][/tex]

So, the correct answer is:
[tex]\[ \boxed{-1} \][/tex]

Among the choices, this corresponds to:
[tex]\[ \text{B. -1} \][/tex]