Higher Engineering Mathematics

14 NUMBER AND ALGEBRA

| 3 z− 4 |>2 means 3z− 4 >2 and 3z− 4 <−2,

i.e. 3z>6 and 3z<2,

i.e. the inequality:| 3 z− 4 |>2 is satisfied when

z>2 andz<

2

3

Now try the following exercise.

Exercise 9 Further problems on inequalities

involving a modulus

Solve the following inequalities:

1. |t+ 1 |<4[− 5 <t<3]


2. |y+ 3 |≤2[− 5 ≤y≤−1]


3. | 2 x− 1 |< 4

[

−

3

2

<x<

5

2

]

4. | 3 t− 5 |>4[t>3 andt<

1

3

]

5. | 1 −k|≥3[k≥4 andk≤−2]

2.4 Inequalities involving quotients

If

p

q

>0 then

p

q

must be apositivevalue.

For

p

q

to be positive,eitherpis positiveandqis

positiveorpis negativeandqis negative.

i.e.

+

+

=+and

−

−

=+

If

p

q

<0 then

p

q

must be anegativevalue.

For

p

q

to be negative,eitherpis positiveandqis

negativeorpis negativeandqis positive.

i.e.

+

−

=−and

−

+

=−

This reasoning is used when solving inequalities
involving quotients, as demonstrated in the follow-
ing worked problems.

Problem 7. Solve the inequality:

t+ 1

3 t− 6

> 0

Since

t+ 1

3 t− 6

>0 then

t+ 1

3 t− 6

must bepositive.

For

t+ 1

3 t− 6

to be positive,

either (i)t+ 1 > 0 and 3 t− 6 > 0

or (ii)t+ 1 < 0 and 3 t− 6 < 0

(i) Ift+ 1 >0 thent>−1 and if 3t− 6 >0 then

3 t>6 andt> 2

Bothof the inequalitiest>− 1 andt>2 are

only true whent>2,

i.e. the fraction

t+ 1

3 t− 6

is positive whent> 2

(ii) Ift+ 1 <0 thent<−1 and if 3t− 6 <0 then

3 t<6 andt< 2

Bothof the inequalitiest<− 1 andt<2 are

only true whent<−1,

i.e. the fraction

t+ 1

3 t− 6

is positive whent<− 1

Summarizing,

t+ 1

3 t− 6

>0 whent>2ort<− 1

Problem 8. Solve the inequality:

2 x+ 3

x+ 2

≤ 1

Since

2 x+ 3

x+ 2

≤1 then

2 x+ 3

x+ 2

− 1 ≤ 0

i.e.

2 x+ 3

x+ 2

−

x+ 2

x+ 2

≤0,

i.e.

2 x+ 3 −(x+2)

x+ 2

≤0or

x+ 1

x+ 2

≤ 0

For

x+ 1

x+ 2

to be negative or zero,

either (i)x+ 1 ≤ 0 and x+ 2 > 0

or (ii)x+ 1 ≥ 0 and x+ 2 < 0

(i) Ifx+ 1 ≤0 thenx≤−1 and ifx+ 2 >0 then

x>− 2

(Note that>is used for the denominator, not≥;

a zero denominator gives a value for the fraction

which is impossible to evaluate.)

Hence, the inequality

x+ 1

x+ 2

≤0 is true whenxis

greater than−2 and less than or equal to−1,

which may be written as− 2 <x≤− 1

(ii) Ifx+ 1 ≥0 thenx≥−1 and ifx+ 2 <0 then

x<− 2