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To solve the rational inequality [tex]\[ \frac{x-6}{x+4} > 0 \][/tex], we need to determine when the fraction [tex]\(\frac{x-6}{x+4}\)[/tex] is greater than zero. Here's a detailed step-by-step solution:
### Step 1: Determine the critical points
The critical points occur where the numerator [tex]\(x-6\)[/tex] and the denominator [tex]\(x+4\)[/tex] are zero.
1. Numerator Zero:
[tex]\[ x - 6 = 0 \implies x = 6 \][/tex]
2. Denominator Zero:
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
These points divide the real number line into three intervals: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, 6) \)[/tex], and [tex]\( (6, \infty) \)[/tex].
### Step 2: Test intervals
We need to test a point from each interval to determine where [tex]\(\frac{x-6}{x+4}\)[/tex] is positive.
1. Interval [tex]\( (-\infty, -4) \)[/tex]:
Choose [tex]\( x = -5 \)[/tex]
[tex]\[ \frac{-5 - 6}{-5 + 4} = \frac{-11}{-1} = 11 \quad (\text{positive}) \][/tex]
2. Interval [tex]\( (-4, 6) \)[/tex]:
Choose [tex]\( x = 0 \)[/tex]
[tex]\[ \frac{0 - 6}{0 + 4} = \frac{-6}{4} = -1.5 \quad (\text{negative}) \][/tex]
3. Interval [tex]\( (6, \infty) \)[/tex]:
Choose [tex]\( x = 7 \)[/tex]
[tex]\[ \frac{7 - 6}{7 + 4} = \frac{1}{11} \quad (\text{positive}) \][/tex]
### Step 3: Determine the solution intervals
From the interval testing, we conclude:
- The function is positive in [tex]\( (-\infty, -4) \)[/tex] and [tex]\( (6, \infty) \)[/tex].
- The function changes signs at [tex]\( x = -4 \)[/tex] and [tex]\( x = 6 \)[/tex].
### Step 4: Consider the boundary points
- At [tex]\( x = -4 \)[/tex], the denominator is zero, so the fraction is undefined. Thus, [tex]\( x = -4 \)[/tex] is not included in the solution set.
- At [tex]\( x = 6 \)[/tex], the numerator is zero. Since the inequality is strict ([tex]\(> 0\)[/tex]), [tex]\( x = 6 \)[/tex] is not included in the solution set.
### Step 5: Express the solution set in interval notation
Combining the results from the above steps:
- The solution set is [tex]\( (-\infty, -4) \cup (6, \infty) \)[/tex].
### Step 6: Graph the solution set on a real number line
1. Draw a number line.
2. Mark the critical points [tex]\( x = -4 \)[/tex] and [tex]\( x = 6 \)[/tex] with open circles (since they are not included in the solution set).
3. Shade the intervals [tex]\( (-\infty, -4) \)[/tex] and [tex]\( (6, \infty) \)[/tex].
The final graph should show open circles at [tex]\( -4 \)[/tex] and [tex]\( 6 \)[/tex] with shading to the left of [tex]\( -4 \)[/tex] and to the right of [tex]\( 6 \)[/tex].
Solution in interval notation:
[tex]\[ (-\infty, -4) \cup (6, \infty) \][/tex]
This solution set shows where the inequality [tex]\(\frac{x-6}{x+4} > 0\)[/tex] holds true.
### Step 1: Determine the critical points
The critical points occur where the numerator [tex]\(x-6\)[/tex] and the denominator [tex]\(x+4\)[/tex] are zero.
1. Numerator Zero:
[tex]\[ x - 6 = 0 \implies x = 6 \][/tex]
2. Denominator Zero:
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
These points divide the real number line into three intervals: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, 6) \)[/tex], and [tex]\( (6, \infty) \)[/tex].
### Step 2: Test intervals
We need to test a point from each interval to determine where [tex]\(\frac{x-6}{x+4}\)[/tex] is positive.
1. Interval [tex]\( (-\infty, -4) \)[/tex]:
Choose [tex]\( x = -5 \)[/tex]
[tex]\[ \frac{-5 - 6}{-5 + 4} = \frac{-11}{-1} = 11 \quad (\text{positive}) \][/tex]
2. Interval [tex]\( (-4, 6) \)[/tex]:
Choose [tex]\( x = 0 \)[/tex]
[tex]\[ \frac{0 - 6}{0 + 4} = \frac{-6}{4} = -1.5 \quad (\text{negative}) \][/tex]
3. Interval [tex]\( (6, \infty) \)[/tex]:
Choose [tex]\( x = 7 \)[/tex]
[tex]\[ \frac{7 - 6}{7 + 4} = \frac{1}{11} \quad (\text{positive}) \][/tex]
### Step 3: Determine the solution intervals
From the interval testing, we conclude:
- The function is positive in [tex]\( (-\infty, -4) \)[/tex] and [tex]\( (6, \infty) \)[/tex].
- The function changes signs at [tex]\( x = -4 \)[/tex] and [tex]\( x = 6 \)[/tex].
### Step 4: Consider the boundary points
- At [tex]\( x = -4 \)[/tex], the denominator is zero, so the fraction is undefined. Thus, [tex]\( x = -4 \)[/tex] is not included in the solution set.
- At [tex]\( x = 6 \)[/tex], the numerator is zero. Since the inequality is strict ([tex]\(> 0\)[/tex]), [tex]\( x = 6 \)[/tex] is not included in the solution set.
### Step 5: Express the solution set in interval notation
Combining the results from the above steps:
- The solution set is [tex]\( (-\infty, -4) \cup (6, \infty) \)[/tex].
### Step 6: Graph the solution set on a real number line
1. Draw a number line.
2. Mark the critical points [tex]\( x = -4 \)[/tex] and [tex]\( x = 6 \)[/tex] with open circles (since they are not included in the solution set).
3. Shade the intervals [tex]\( (-\infty, -4) \)[/tex] and [tex]\( (6, \infty) \)[/tex].
The final graph should show open circles at [tex]\( -4 \)[/tex] and [tex]\( 6 \)[/tex] with shading to the left of [tex]\( -4 \)[/tex] and to the right of [tex]\( 6 \)[/tex].
Solution in interval notation:
[tex]\[ (-\infty, -4) \cup (6, \infty) \][/tex]
This solution set shows where the inequality [tex]\(\frac{x-6}{x+4} > 0\)[/tex] holds true.
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