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We have learned about the simple linear inequality like an x + b > c but sometimes the inequality may be like d < an x + b < d.

**Examples of compound inequalities: - **

**·** 1 < 2 x + 3 < 3

**·** -2 ≥ 4 x + 5 ≥ 5

**·** x + 3 > 4 x > 5 x – 5

**·** -2 ≤ 5 x + 9 ≤ 9

**How to solve compound inequalities: -**

**Question 1**: - If 1 < 2 x + 3 < 3, then find x.

**Solution**: -

i) Separate the inequality like

1 < 2 x + 3 and 2 x + 3 < 3

ii) Solve each of these inequalities separately like a simple linear inequality.

1 < 2 x + 3

1 – 3 < 2 x + 3 – 3

– 2 < 2 xi) Separate the inequality like

1 < 2 x + 3 and 2 x + 3 < 3

ii) Solve each of these inequalities separately like a simple linear inequality.

1 < 2 x + 3

1 – 3 < 2 x + 3 – 3

– 2 / 2 < 2 x / 2

– 1 < x

And 2 x + 3 < 3

2 x + 3 – 3 < 3 – 3

2 x < 0

2 x / 2 < 0 / 2

x < 0

Therefore, –1< x<0.

i) x + 3 > 4 x

x + 3 – x > 4 x – x

3 > 3 x

3 / 3 > 3 x / 3

1 > x

ii) 4 x > 5 x - 5

4 x – 5 x > 5 x - 5 – 5x

x < 5

Therefore, 1

Therefore, 1