All the values of x satisfying the inequality 1/(x + 4) ≤ 1/(3x − 2) are:
Answer and solution
Correct answer: A
Solution
Answer: A — (−∞, −4) ∪ (2/3, 3]
1. First, note the values that are not allowed:
x ≠ −4 and x ≠ 2/3, because they make a denominator zero.
2. Bring both fractions to the left:
1/(x + 4) − 1/(3x − 2) ≤ 0
3. Take the common denominator:
[(3x − 2) − (x + 4)] / [(x + 4)(3x − 2)] ≤ 0
2(x − 3) / [(x + 4)(3x − 2)] ≤ 0
4. The critical values are −4, 2/3 and 3. They divide the number line into four intervals.
5. Now check the sign in each interval:
1. For x < −4, take x = −5. The expression is negative, so this interval is included.
2. For −4 < x < 2/3, take x = 0. The expression is positive, so this interval is excluded.
3. For 2/3 < x < 3, take x = 1. The expression is negative, so this interval is included.
4. For x > 3, take x = 4. The expression is positive, so this interval is excluded.
6. At x = 3, the numerator becomes zero while the denominators remain non-zero. Since the inequality includes equality (≤), x = 3 is included.
x = −4 and x = 2/3 are excluded because the original expression is undefined there.
7. Therefore, the solution is:
(−∞, −4) ∪ (2/3, 3]
8. Answer: Option A.
Why this method works:
Do not cross-multiply an inequality by (x+4)(3x−2) before checking its sign: the inequality reverses where that product is negative. The sign chart handles all intervals and denominator exclusions at once.
Independent check and common trap:
The expression is zero only at 3, which is allowed because the inequality is non-strict. The points −4 and 2/3 are never included. These endpoint checks distinguish option A from the distractor that closes an undefined endpoint.
Apply: Combine to 2(x−3)/[(x+4)(3x−2)]≤0. Critical points −4,2/3,3; test−5,0,1,4 gives−,+,−,+. Include 3, exclude poles.
Check: The expression is zero only at 3, which is allowed because the inequality is non-strict. The points −4 and 2/3 are never included. These endpoint checks distinguish option A from the distractor that closes an undefined endpoint.
1. First, note the values that are not allowed:
x ≠ −4 and x ≠ 2/3, because they make a denominator zero.
2. Bring both fractions to the left:
1/(x + 4) − 1/(3x − 2) ≤ 0
3. Take the common denominator:
[(3x − 2) − (x + 4)] / [(x + 4)(3x − 2)] ≤ 0
2(x − 3) / [(x + 4)(3x − 2)] ≤ 0
4. The critical values are −4, 2/3 and 3. They divide the number line into four intervals.
5. Now check the sign in each interval:
1. For x < −4, take x = −5. The expression is negative, so this interval is included.
2. For −4 < x < 2/3, take x = 0. The expression is positive, so this interval is excluded.
3. For 2/3 < x < 3, take x = 1. The expression is negative, so this interval is included.
4. For x > 3, take x = 4. The expression is positive, so this interval is excluded.
6. At x = 3, the numerator becomes zero while the denominators remain non-zero. Since the inequality includes equality (≤), x = 3 is included.
x = −4 and x = 2/3 are excluded because the original expression is undefined there.
7. Therefore, the solution is:
(−∞, −4) ∪ (2/3, 3]
8. Answer: Option A.
Why this method works:
Do not cross-multiply an inequality by (x+4)(3x−2) before checking its sign: the inequality reverses where that product is negative. The sign chart handles all intervals and denominator exclusions at once.
Independent check and common trap:
The expression is zero only at 3, which is allowed because the inequality is non-strict. The points −4 and 2/3 are never included. These endpoint checks distinguish option A from the distractor that closes an undefined endpoint.
G Strategy shortcut
Spot: Algebra – Inequalities.Apply: Combine to 2(x−3)/[(x+4)(3x−2)]≤0. Critical points −4,2/3,3; test−5,0,1,4 gives−,+,−,+. Include 3, exclude poles.
Check: The expression is zero only at 3, which is allowed because the inequality is non-strict. The points −4 and 2/3 are never included. These endpoint checks distinguish option A from the distractor that closes an undefined endpoint.
G Strategy: Spot: Algebra – Inequalities. Apply: Combine to 2(x−3)/[(x+4)(3x−2)]≤0. Critical points −4,2/3,3; test−5,0,1,4 gives−,+,−,+. Include 3, exclude poles. Check: The expression is zero only at 3, which is allowed because the inequality is non-strict. The points −4 and 2/3 are never included. These endpoint checks distinguish option A from the distractor that closes an undefined endpoint.