Answer :
Let's solve the inequality
[tex]\[ \frac{2}{x+1} < \frac{x}{x+3}. \][/tex]
### Step 1: Identify the domain restrictions
First, determine the values of [tex]\( x \)[/tex] that make the denominators zero, as these values are excluded from the domain of the inequality.
Denominator of the left side: [tex]\( x+1 \)[/tex]. Set it to zero:
[tex]\[ x + 1 = 0 \Rightarrow x = -1. \][/tex]
Denominator of the right side: [tex]\( x+3 \)[/tex]. Set it to zero:
[tex]\[ x + 3 = 0 \Rightarrow x = -3. \][/tex]
Thus, the values [tex]\( x = -1 \)[/tex] and [tex]\( x = -3 \)[/tex] are excluded from the domain.
### Step 2: Rewrite the inequality
Rewrite the inequality to have a common denominator:
[tex]\[ \frac{2}{x+1} - \frac{x}{x+3} < 0. \][/tex]
Combine the fractions:
[tex]\[ \frac{2(x+3) - x(x+1)}{(x+1)(x+3)} < 0. \][/tex]
Simplify the numerator:
[tex]\[ 2(x + 3) - x(x + 1) = 2x + 6 - x^2 - x = -x^2 + x + 6. \][/tex]
Thus, the inequality becomes:
[tex]\[ \frac{-x^2 + x + 6}{(x+1)(x+3)} < 0. \][/tex]
### Step 3: Solve the quadratic inequality
Factor the numerator [tex]\(-x^2 + x + 6\)[/tex]:
[tex]\[ -x^2 + x + 6 = -(x^2 - x - 6) = -(x - 3)(x + 2). \][/tex]
Now the inequality is:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} < 0. \][/tex]
### Step 4: Determine the sign changes using critical points
Identify the critical points from the factored form:
[tex]\[ -(x - 3)(x + 2) = 0 \Rightarrow x = 3 \text{ and } x = -2, \][/tex]
and from the denominators:
[tex]\[ (x + 1)(x + 3) = 0 \Rightarrow x = -1 \text{ and } x = -3. \][/tex]
The critical points divide the real number line into intervals:
[tex]\[ (-\infty, -3), (-3, -2), (-2, -1), (-1, 3), (3, \infty). \][/tex]
### Step 5: Test the signs in each interval
#### Interval [tex]\( (-\infty, -3) \)[/tex]:
Choose [tex]\( x = -4 \)[/tex]:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} = \frac{-(-4 - 3)(-4 + 2)}{(-4 + 1)(-4 + 3)} = \frac{-(-7)(-2)}{(-3)(-1)} = \frac{-14}{3} > 0. \][/tex]
The inequality is not satisfied.
#### Interval [tex]\( (-3, -2) \)[/tex]:
Choose [tex]\( x = -2.5 \)[/tex]:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} = \frac{-(-2.5 - 3)(-2.5 + 2)}{(-2.5 + 1)(-2.5 + 3)} = \frac{-(-5.5)(-0.5)}{(-1.5)(0.5)} = \frac{-2.75}{-0.75} > 0. \][/tex]
The inequality is not satisfied.
#### Interval [tex]\( (-2, -1) \)[/tex]:
Choose [tex]\( x = -1.5 \)[/tex]:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} = \frac{-(-1.5 - 3)(-1.5 + 2)}{(-1.5 + 1)(-1.5 + 3)} = \frac{-(-4.5)(0.5)}{(-0.5)(1.5)} = \frac{2.25}{-0.75} < 0. \][/tex]
The inequality is satisfied.
#### Interval [tex]\( (-1, 3) \)[/tex]:
Choose [tex]\( x = 0 \)[/tex]:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} = \frac{-(0 - 3)(0 + 2)}{(0 + 1)(0 + 3)} = \frac{-(-3)(2)}{(1)(3)} = \frac{6}{3} > 0. \][/tex]
The inequality is not satisfied.
#### Interval [tex]\( (3, \infty) \)[/tex]:
Choose [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} = \frac{-(4 - 3)(4 + 2)}{(4 + 1)(4 + 3)} = \frac{-(1)(6)}{(5)(7)} = \frac{-6}{35} < 0. \][/tex]
The inequality is satisfied.
### Step 6: Combine the solution intervals
Based on the sign test, the solutions to the inequality [tex]\(\frac{2}{x+1} < \frac{x}{x+3}\)[/tex] are:
[tex]\[ x \in (-2, -1) \cup (3, \infty), \][/tex]
along with the interval where if [tex]\((x + 3) < 0\)[/tex]:
[tex]\[ x \in (-\infty, -3). \][/tex]
Combining all intervals where the inequality is satisfied, the solution set is:
[tex]\[ x \in (-\infty, -3) \cup (-2, -1) \cup (3, \infty). \][/tex]
Hence, the solution is:
[tex]\[ (-\infty < x < -3) \cup (-2 < x < -1) \cup (3 < x < \infty). \][/tex]
[tex]\[ \frac{2}{x+1} < \frac{x}{x+3}. \][/tex]
### Step 1: Identify the domain restrictions
First, determine the values of [tex]\( x \)[/tex] that make the denominators zero, as these values are excluded from the domain of the inequality.
Denominator of the left side: [tex]\( x+1 \)[/tex]. Set it to zero:
[tex]\[ x + 1 = 0 \Rightarrow x = -1. \][/tex]
Denominator of the right side: [tex]\( x+3 \)[/tex]. Set it to zero:
[tex]\[ x + 3 = 0 \Rightarrow x = -3. \][/tex]
Thus, the values [tex]\( x = -1 \)[/tex] and [tex]\( x = -3 \)[/tex] are excluded from the domain.
### Step 2: Rewrite the inequality
Rewrite the inequality to have a common denominator:
[tex]\[ \frac{2}{x+1} - \frac{x}{x+3} < 0. \][/tex]
Combine the fractions:
[tex]\[ \frac{2(x+3) - x(x+1)}{(x+1)(x+3)} < 0. \][/tex]
Simplify the numerator:
[tex]\[ 2(x + 3) - x(x + 1) = 2x + 6 - x^2 - x = -x^2 + x + 6. \][/tex]
Thus, the inequality becomes:
[tex]\[ \frac{-x^2 + x + 6}{(x+1)(x+3)} < 0. \][/tex]
### Step 3: Solve the quadratic inequality
Factor the numerator [tex]\(-x^2 + x + 6\)[/tex]:
[tex]\[ -x^2 + x + 6 = -(x^2 - x - 6) = -(x - 3)(x + 2). \][/tex]
Now the inequality is:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} < 0. \][/tex]
### Step 4: Determine the sign changes using critical points
Identify the critical points from the factored form:
[tex]\[ -(x - 3)(x + 2) = 0 \Rightarrow x = 3 \text{ and } x = -2, \][/tex]
and from the denominators:
[tex]\[ (x + 1)(x + 3) = 0 \Rightarrow x = -1 \text{ and } x = -3. \][/tex]
The critical points divide the real number line into intervals:
[tex]\[ (-\infty, -3), (-3, -2), (-2, -1), (-1, 3), (3, \infty). \][/tex]
### Step 5: Test the signs in each interval
#### Interval [tex]\( (-\infty, -3) \)[/tex]:
Choose [tex]\( x = -4 \)[/tex]:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} = \frac{-(-4 - 3)(-4 + 2)}{(-4 + 1)(-4 + 3)} = \frac{-(-7)(-2)}{(-3)(-1)} = \frac{-14}{3} > 0. \][/tex]
The inequality is not satisfied.
#### Interval [tex]\( (-3, -2) \)[/tex]:
Choose [tex]\( x = -2.5 \)[/tex]:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} = \frac{-(-2.5 - 3)(-2.5 + 2)}{(-2.5 + 1)(-2.5 + 3)} = \frac{-(-5.5)(-0.5)}{(-1.5)(0.5)} = \frac{-2.75}{-0.75} > 0. \][/tex]
The inequality is not satisfied.
#### Interval [tex]\( (-2, -1) \)[/tex]:
Choose [tex]\( x = -1.5 \)[/tex]:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} = \frac{-(-1.5 - 3)(-1.5 + 2)}{(-1.5 + 1)(-1.5 + 3)} = \frac{-(-4.5)(0.5)}{(-0.5)(1.5)} = \frac{2.25}{-0.75} < 0. \][/tex]
The inequality is satisfied.
#### Interval [tex]\( (-1, 3) \)[/tex]:
Choose [tex]\( x = 0 \)[/tex]:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} = \frac{-(0 - 3)(0 + 2)}{(0 + 1)(0 + 3)} = \frac{-(-3)(2)}{(1)(3)} = \frac{6}{3} > 0. \][/tex]
The inequality is not satisfied.
#### Interval [tex]\( (3, \infty) \)[/tex]:
Choose [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{-(x - 3)(x + 2)}{(x + 1)(x + 3)} = \frac{-(4 - 3)(4 + 2)}{(4 + 1)(4 + 3)} = \frac{-(1)(6)}{(5)(7)} = \frac{-6}{35} < 0. \][/tex]
The inequality is satisfied.
### Step 6: Combine the solution intervals
Based on the sign test, the solutions to the inequality [tex]\(\frac{2}{x+1} < \frac{x}{x+3}\)[/tex] are:
[tex]\[ x \in (-2, -1) \cup (3, \infty), \][/tex]
along with the interval where if [tex]\((x + 3) < 0\)[/tex]:
[tex]\[ x \in (-\infty, -3). \][/tex]
Combining all intervals where the inequality is satisfied, the solution set is:
[tex]\[ x \in (-\infty, -3) \cup (-2, -1) \cup (3, \infty). \][/tex]
Hence, the solution is:
[tex]\[ (-\infty < x < -3) \cup (-2 < x < -1) \cup (3 < x < \infty). \][/tex]
