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Exercise 9 · Q5

Q.Solve the following inequality:

(i) (−2z−6)<10(-2z - 6) < 10
(ii) 2a<a−4≤3a+82a < a - 4 \le 3a + 8
(iii) (y−1)3+4<(y−5)5−2\dfrac{(y-1)}{3} + 4 < \dfrac{(y-5)}{5} - 2
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Solving the three linear inequalities gives z>−8z>-8; −6≤a<−4-6\le a<-4; and y<−50y<-50.

Isolate the variable. Dividing/multiplying by a negative number reverses the inequality sign. For a double inequality, solve both halves and take the common region.

(i) −2z−6<10-2z - 6 < 10

  1. Add 66: −2z<16-2z < 16.
  2. Divide by −2-2 (negative → flip sign): z>−8z > -8.

(ii) 2a<a−4≤3a+82a < a - 4 \le 3a + 8 — split into two parts.

  1. Left: 2a<a−4⇒2a−a<−4⇒a<−42a < a - 4 \Rightarrow 2a - a < -4 \Rightarrow a < -4.
  2. Right: a−4≤3a+8⇒−4−8≤3a−a⇒−12≤2a⇒a≥−6a - 4 \le 3a + 8 \Rightarrow -4 - 8 \le 3a - a \Rightarrow -12 \le 2a \Rightarrow a \ge -6.
  3. Common region: −6≤a<−4-6 \le a < -4.

(iii) y−13+4<y−55−2\dfrac{y-1}{3} + 4 < \dfrac{y-5}{5} - 2 …

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