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Zorluk: OrtaInequalities and Absolute Value Equations

If xx is a real number satisfying the inequality ∣2x−5∣x+1≤1\frac{|2x - 5|}{x + 1} \le 1, which of the following represents the complete set of all possible values of xx?

  1. x<−1x < -1 or 43≤x≤6\frac{4}{3} \le x \le 6Cevap
  2. B
    43≤x≤6\frac{4}{3} \le x \le 6
  3. C
    −1<x≤6-1 < x \le 6
  4. D
    x<−1x < -1 or x≥43x \ge \frac{4}{3}
  5. E
    x≤−1x \le -1 or 43≤x≤6\frac{4}{3} \le x \le 6

Cevap

x<−1x < -1 or 43≤x≤6\frac{4}{3} \le x \le 6
The correct answer accounts for both possible sign states of the denominator x+1x + 1. When x>−1x > -1, multiplying gives ∣2x−5∣≤x+1|2x - 5| \le x + 1, yielding 43≤x≤6\frac{4}{3} \le x \le 6. When x<−1x < -1, multiplying flips the inequality to ∣2x−5∣≥x+1|2x - 5| \ge x + 1, which is trivially satisfied by all x<−1x < -1 because an absolute value is non-negative and x+1x + 1 is negative. Combining both cases yields x<−1x < -1 or 43≤x≤6\frac{4}{3} \le x \le 6.

Adım Adım Çözüm

1
Determine domain restrictions and break the inequality into cases based on the denominator's sign.
The expression is undefined at x=−1x = -1, so x≠−1x \neq -1. We evaluate Case 1 (x>−1x > -1) and Case 2 (x<−1x < -1).
Multiplying an inequality by an algebraic expression requires knowing whether that expression is positive or negative to maintain or flip the inequality sign.
2
Solve Case 1 where x+1>0x + 1 > 0 (x>−1x > -1).
∣2x−5∣≤x+1  ⟹  −(x+1)≤2x−5≤x+1|2x - 5| \le x + 1 \implies -(x + 1) \le 2x - 5 \le x + 1. Solving −x−1≤2x−5-x - 1 \le 2x - 5 yields x≥43x \ge \frac{4}{3}. Solving 2x−5≤x+12x - 5 \le x + 1 yields x≤6x \le 6. Combining gives 43≤x≤6\frac{4}{3} \le x \le 6.
For positive denominators, multiplying both sides by x+1x + 1 preserves the inequality direction.
3
Solve Case 2 where x+1<0x + 1 < 0 (x<−1x < -1).
∣2x−5∣≥x+1|2x - 5| \ge x + 1. Since ∣2x−5∣≥0|2x - 5| \ge 0 for all real xx and x+1<0x + 1 < 0, the left-hand side is non-negative while the right-hand side is strictly negative, which is always true for all x<−1x < -1.
Multiplying by a negative expression flips the inequality sign, and any non-negative quantity is strictly greater than any negative quantity.
4
Combine solutions from both cases.
The final solution set is x<−1x < -1 or 43≤x≤6\frac{4}{3} \le x \le 6.
Taking the union of valid solution regions from both disjoint cases gives the complete solution.

Anahtar Kavram

Solving Rational Inequalities with Absolute Values by Case Analysis
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Konular

Inequalities and Absolute Value Equations

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