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Zorluk: OrtaInequalities and Absolute Value Equations

If yy is a real number satisfying the inequality ∣3y−6∣y+2<1\frac{|3y - 6|}{y + 2} < 1, which of the following represents the complete set of all possible values of yy?

  1. y<−2y < -2 or 1<y<41 < y < 4Cevap
  2. B
    1<y<41 < y < 4
  3. C
    −2<y<4-2 < y < 4
  4. D
    y<1y < 1 or y>4y > 4
  5. E
    y<−2y < -2 or y>1y > 1

Cevap

y<−2y < -2 or 1<y<41 < y < 4
To evaluate ∣3y−6∣y+2<1\frac{|3y - 6|}{y + 2} < 1, analyze the sign of the denominator y+2y + 2. If y<−2y < -2, the denominator is negative, so multiplying both sides by y+2y + 2 reverses the inequality to ∣3y−6∣>y+2|3y - 6| > y + 2. Because absolute values are always non-negative and y+2y + 2 is negative for y<−2y < -2, this inequality holds for all y<−2y < -2. If y>−2y > -2, the denominator is positive, yielding ∣3y−6∣<y+2|3y - 6| < y + 2, which expands to −y−2<3y−6<y+2-y - 2 < 3y - 6 < y + 2. Solving the left inequality gives y>1y > 1, and solving the right gives y<4y < 4, producing 1<y<41 < y < 4. Combining both valid cases gives y<−2y < -2 or 1<y<41 < y < 4.

Adım Adım Çözüm

1
Determine the domain restriction and split into cases based on the denominator's sign.
The expression is undefined when y=−2y = -2. We analyze Case 1 (y>−2y > -2) and Case 2 (y<−2y < -2).
Multiplying an inequality by an algebraic expression requires knowing its sign to preserve or reverse the inequality direction.
2
Solve Case 1 where y>−2y > -2 (positive denominator).
Multiplying by y+2y + 2 gives ∣3y−6∣<y+2|3y - 6| < y + 2, which expands to −(y+2)<3y−6<y+2-(y + 2) < 3y - 6 < y + 2. Solving −y−2<3y−6-y - 2 < 3y - 6 gives 4<4y  ⟹  y>14 < 4y \implies y > 1. Solving 3y−6<y+23y - 6 < y + 2 gives 2y<8  ⟹  y<42y < 8 \implies y < 4. Combining gives 1<y<41 < y < 4.
Since y+2>0y + 2 > 0, multiplying preserves the inequality sign.
3
Solve Case 2 where y<−2y < -2 (negative denominator).
Multiplying by y+2y + 2 flips the inequality sign to ∣3y−6∣>y+2|3y - 6| > y + 2. Since ∣3y−6∣≥0|3y - 6| \ge 0 for all real yy and y+2<0y + 2 < 0 when y<−2y < -2, a non-negative number is always strictly greater than a negative number. Thus, all y<−2y < -2 are valid solutions.
Any non-negative real value is strictly greater than any negative value.
4
Combine the valid intervals from both cases.
y<−2y < -2 or 1<y<41 < y < 4.
The complete solution set is the union of solutions from Case 1 and Case 2.

Anahtar Kavram

Solving Rational Absolute Value Inequalities via Denominator Sign Case Analysis
Tahmini Süre:2m 0s
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Konular

Inequalities and Absolute Value Equations

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