# Solved Examples

**Illustration 1 :**

If 2 â‰¤ 8 x - 1 â‰¤ 6, then x lies between:

1. 2 â‰¤ x â‰¤ 15 2. - 2 â‰¤ x â‰¤ - 15 3. 3/8 â‰¤ x â‰¤ 7/8 4. 2/8 â‰¤ x â‰¤ 15/8

Transpose the function to get:

3 â‰¤ 8x â‰¤ 7, divide by 8, hence 3/8 â‰¤ x â‰¤ 7/8 (dividing by positive quantities)

**Illustration 2:**

If a > 0, b > 0, c < 0, then which of the following is necessarily true?

1. a^{-3} . b^{- 3} . c^{-3} > 0 2. c^{2}/a^{b} < 0 3. a ^{c} b > 0 4. c^{c} . a^{3} < 0

Best solved by eliminating choices.

If c < 0, then the first choice is clearly wrong, the second choice c^{2} /a^{b} must be positive as both quantities are positive; the fourth choice may be positive even if c is negative. Hence (3) is the correct answer: a raised to the power of a negative quantity is positive.

**Illustration 3 :**

x^{2} + 4x > 2x + 8 is true when:

1. x > 4 or x < - 2 2. x > 4 > or x < 2 3. x > 2 or x < - 4 4. x > - 2 or x < 4

x^{2} + 4x > 2x + 8, so x^{2} + 2x - 8 > 0

Factorise to get: (x + 4) (x - 2) > 0.

Either both quantities must be positive or both quantities must be negative, which is true if x < - 4 or x > 2

**Illustration 4**:

Solve this inequality |y + 9| < 15 for y.

Take both the sides y + 9 < 15 and y + 9 > - 15

y < 15 â€“ 9 and y > â€“ 15 â€“ 9 â‡’ y < 6 and y > - 24. - 24 < y < 6.

**Illustration 5**:

Solve this inequalities x^{2} â€“ 7x + 16 < 4 for x.

Take all the terms given on one side.

x^{2} â€“ 7x + 16 â€“ 4 < 0 â‡’ x^{2} â€“ 7x + 12 < 0.

x^{2} â€“ 4x â€“ 3x + 12 < 0 â‡’ (x â€“ 4) (x â€“ 3) < 0.

Now because their product is less than zero, implies one of these is negative and the other is positive. Now out of the factors, it can be seen that x â€“ 3 is greater than x â€“ 4. So it can be concluded that x â€“ 3 is positive and x â€“ 4 is negative.

x â€“ 4 < 0 and x â€“ 3 > 0 â‡’ x < 4 and x > 3.

The solution set of the inequality is 3 < x < 4.

**Illustration 6**:

x^{2} â€“ 11x + 28 > 0

x^{2} â€“ 11x + 28 > 0. â‡’ (x â€“ 7)(x â€“ 4) > 0.

Since their product is positive means, either both are positive or both are negative.

Taking both to be positive x â€“ 7 > 0 and x â€“ 4 > 0 â‡’ x > 7 and x > 4.

Now choose that one out of the two, which includes the other also. Anything that is greater than 7, will definitely be greater than 4 also. So x > 7.

Now taking both to be negative x â€“ 7 < 0 and x â€“ 4 < 0 â‡’ x < 7 and x < 4.

Now choose that one out of these two, which includes the other also. Anything that is lesser than 4, will definitely be smaller than 7 also. So x < 4.

So the solution set is x < 4 or x > 7.