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To solve the inequality [tex]\(\frac{4}{x-3} + 2 < x + 2\)[/tex] for [tex]\(x < 3\)[/tex], let's follow these steps:
1. Subtract 2 from both sides of the inequality:
[tex]\[ \frac{4}{x-3} + 2 - 2 < x + 2 - 2 \][/tex]
This simplifies to:
[tex]\[ \frac{4}{x-3} < x \][/tex]
2. Isolate the fraction:
We are working with the inequality [tex]\(\frac{4}{x-3} < x\)[/tex]. Note the domain here: [tex]\(x \neq 3\)[/tex] and [tex]\(x < 3\)[/tex].
3. Multiply both sides of the inequality by [tex]\(x-3\)[/tex]:
Since [tex]\(x < 3\)[/tex], [tex]\(x-3\)[/tex] is negative. When multiplying both sides of an inequality by a negative number, the inequality sign reverses.
[tex]\[ 4 < x(x-3) \][/tex]
[tex]\[ 4 < x^2 - 3x \][/tex]
4. Rewrite the inequality in standard form:
[tex]\[ x^2 - 3x - 4 > 0 \][/tex]
5. Factor the quadratic expression:
[tex]\[ x^2 - 3x - 4 = (x - 4)(x + 1) \][/tex]
So, the inequality becomes:
[tex]\[ (x - 4)(x + 1) > 0 \][/tex]
6. Determine the critical points:
The critical points for this inequality are found by setting each factor to zero:
[tex]\[ x - 4 = 0 \implies x = 4 \][/tex]
[tex]\[ x + 1 = 0 \implies x = -1 \][/tex]
7. Test the intervals determined by the critical points:
The critical points divide the number line into three intervals: [tex]\((-\infty, -1)\)[/tex], [tex]\((-1, 4)\)[/tex], and [tex]\((4, \infty)\)[/tex]. Let's test a value from each interval to determine where the inequality holds. However, we are only interested in [tex]\(x < 3\)[/tex].
- For [tex]\(x < -1\)[/tex], choose [tex]\(x = -2\)[/tex]:
[tex]\[ (-2 - 4)(-2 + 1) = (-6)(-1) = 6 \quad (\text{positive}) \][/tex]
- For [tex]\(-1 < x < 4\)[/tex], choose [tex]\(x = 0\)[/tex]:
[tex]\[ (0 - 4)(0 + 1) = (-4)(1) = -4 \quad (\text{negative}) \][/tex]
- For [tex]\(x > 4\)[/tex], choose [tex]\(x = 5\)[/tex]:
[tex]\[ (5 - 4)(5 + 1) = (1)(6) = 6 \quad (\text{positive}) \][/tex]
8. Combine results with the condition [tex]\(x < 3\)[/tex]:
The inequality [tex]\((x - 4)(x + 1) > 0\)[/tex] holds true in the intervals [tex]\((-\infty, -1)\)[/tex] and [tex]\((4, \infty)\)[/tex]. Since we are only interested in values for [tex]\(x < 3\)[/tex], we disregard [tex]\((4, \infty)\)[/tex].
So, the solution that fits both the inequality and the constraint is:
[tex]\[ \boxed{(-\infty, -1)} \][/tex]
Thus, [tex]\(x \)[/tex] must be in the interval [tex]\((-\infty, -1)\)[/tex].
1. Subtract 2 from both sides of the inequality:
[tex]\[ \frac{4}{x-3} + 2 - 2 < x + 2 - 2 \][/tex]
This simplifies to:
[tex]\[ \frac{4}{x-3} < x \][/tex]
2. Isolate the fraction:
We are working with the inequality [tex]\(\frac{4}{x-3} < x\)[/tex]. Note the domain here: [tex]\(x \neq 3\)[/tex] and [tex]\(x < 3\)[/tex].
3. Multiply both sides of the inequality by [tex]\(x-3\)[/tex]:
Since [tex]\(x < 3\)[/tex], [tex]\(x-3\)[/tex] is negative. When multiplying both sides of an inequality by a negative number, the inequality sign reverses.
[tex]\[ 4 < x(x-3) \][/tex]
[tex]\[ 4 < x^2 - 3x \][/tex]
4. Rewrite the inequality in standard form:
[tex]\[ x^2 - 3x - 4 > 0 \][/tex]
5. Factor the quadratic expression:
[tex]\[ x^2 - 3x - 4 = (x - 4)(x + 1) \][/tex]
So, the inequality becomes:
[tex]\[ (x - 4)(x + 1) > 0 \][/tex]
6. Determine the critical points:
The critical points for this inequality are found by setting each factor to zero:
[tex]\[ x - 4 = 0 \implies x = 4 \][/tex]
[tex]\[ x + 1 = 0 \implies x = -1 \][/tex]
7. Test the intervals determined by the critical points:
The critical points divide the number line into three intervals: [tex]\((-\infty, -1)\)[/tex], [tex]\((-1, 4)\)[/tex], and [tex]\((4, \infty)\)[/tex]. Let's test a value from each interval to determine where the inequality holds. However, we are only interested in [tex]\(x < 3\)[/tex].
- For [tex]\(x < -1\)[/tex], choose [tex]\(x = -2\)[/tex]:
[tex]\[ (-2 - 4)(-2 + 1) = (-6)(-1) = 6 \quad (\text{positive}) \][/tex]
- For [tex]\(-1 < x < 4\)[/tex], choose [tex]\(x = 0\)[/tex]:
[tex]\[ (0 - 4)(0 + 1) = (-4)(1) = -4 \quad (\text{negative}) \][/tex]
- For [tex]\(x > 4\)[/tex], choose [tex]\(x = 5\)[/tex]:
[tex]\[ (5 - 4)(5 + 1) = (1)(6) = 6 \quad (\text{positive}) \][/tex]
8. Combine results with the condition [tex]\(x < 3\)[/tex]:
The inequality [tex]\((x - 4)(x + 1) > 0\)[/tex] holds true in the intervals [tex]\((-\infty, -1)\)[/tex] and [tex]\((4, \infty)\)[/tex]. Since we are only interested in values for [tex]\(x < 3\)[/tex], we disregard [tex]\((4, \infty)\)[/tex].
So, the solution that fits both the inequality and the constraint is:
[tex]\[ \boxed{(-\infty, -1)} \][/tex]
Thus, [tex]\(x \)[/tex] must be in the interval [tex]\((-\infty, -1)\)[/tex].
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