To solve the inequality \(\frac{x + 1}{x + 3} \leq 2\), we need to consider the following steps:

### Step 1: Move all terms to one side
First, subtract \(2\) from both sides to bring all terms to one side of the inequality:

\[
\frac{x + 1}{x + 3} - 2 \leq 0
\]

### Step 2: Combine the fractions
To combine the fractions, express \(2\) as \(\frac{2(x + 3)}{x + 3}\):

\[
\frac{x + 1 - 2(x + 3)}{x + 3} \leq 0
\]

Simplify the numerator:

\[
\frac{x + 1 - 2x - 6}{x + 3} \leq 0
\]

\[
\frac{-x - 5}{x + 3} \leq 0
\]

### Step 3: Analyze the inequality
Now, we need to determine when the fraction \(\frac{-x - 5}{x + 3}\) is less than or equal to zero.

The fraction is equal to zero when the numerator \(-x - 5 = 0\), so:

\[
-x - 5 = 0 \implies x = -5
\]

Next, we check the sign of the fraction in different intervals determined by the critical points \(x = -5\) and \(x = -3\) (where the denominator \(x + 3 = 0\)).

### Step 4: Test intervals
The intervals to test are:

1. \(x < -5\)
2. \(-5 < x < -3\)
3. \(x > -3\)

- For \(x < -5\), both \(-x - 5\) and \(x + 3\) are negative, so the fraction is positive.
- For \(-5 < x < -3\), \(-x - 5\) is negative and \(x + 3\) is positive, so the fraction is negative.
- For \(x > -3\), both \(-x - 5\) and \(x + 3\) are positive, so the fraction is negative.

### Step 5: Solution
The inequality is satisfied when the fraction is less than or equal to zero:

\[
-5 \leq x < -3
\]

Thus, the solution to the inequality is:

\[
x \in [-5, -3)
\]

### Do you have any questions, or would you like more details on any step?

Here are 5 related questions to consider:

1. How would the solution change if the inequality were strict \(\frac{x + 1}{x + 3} < 2\)?
2. How do you solve the inequality if the expression is \(\frac{x - 1}{x + 3} \leq -1\)?
3. What are the critical points for the function \(\frac{x + 1}{x + 3}\)?
4. How can the sign chart method help in solving inequalities like this?
5. What would happen if the denominator \(x + 3\) was a squared term?

**Tip:** Always check the sign of both the numerator and denominator separately when solving inequalities involving fractions. This helps determine the intervals where the inequality holds.

Explore the step-by-step solution to the rational inequality (x + 1)/(x + 3) <= 2, using critical point analysis and sign chart methods. This problem is ideal for students in Grades 10-12, focusing on key algebraic concepts and rational function inequalities.

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