Higher Engineering Mathematics

INEQUALITIES 13

A

Dividing both sides of the inequality: 3x>4by3
gives:

x>

4

3

Hence all values ofxgreater than

4

3

satisfy the

inequality:

4 x+ 1 >x+ 5

Problem 3. Solve the inequality: 3− 4 t≤ 8 +t

Subtracting 3 from both sides of the inequality:
3 − 4 t≤ 8 +tgives:

− 4 t≤ 5 +t

Subtracting t from both sides of the inequality:
− 4 t≤ 5 +tgives:

− 5 t≤ 5

Dividing both sides of the inequality− 5 t≤5by− 5
gives:

t≥− 1 (remembering to reverse the

inequality)

Hence, all values oftgreater than or equal to− 1
satisfy the inequality.

Now try the following exercise.

Exercise 8 Further problems on simple

inequalities

Solve the following inequalities:

1. (a) 3t>6 (b) 2x< 10

[(a)t>2 (b)x<5]

2. (a)

x

2

>1.5 (b)x+ 2 ≥ 5

[(a)x>3 (b)x≥3]

3. (a) 4t− 1 ≤3 (b) 5−x≥− 1
   [(a)t≤1 (b)x≤6]


4. (a)

7 − 2 k

4

≤1 (b) 3z+ 2 >z+ 3

[

(a)k≥

3

2

(b)z>

1

2

]

5. (a) 5− 2 y≤ 9 +y[ (b) 1− 6 x≤ 5 + 2 x

(a)y≥−

4

3

(b)x≥−

1

2

]

2.3 Inequalities involving a modulus

Themodulusof a number is the size of the num-

ber, regardless of sign. Vertical lines enclosing the

number denote a modulus.

For example,| 4 |=4 and|− 4 |=4 (the modulus of

a number is never negative),

The inequality: |t|<1 means that all numbers

whose actual size, regardless of sign, is less than

1, i.e. any value between−1 and+1.

Thus|t|<1 means− 1 <t< 1.

Similarly,|x|>3 means all numbers whose actual

size, regardless of sign, is greater than 3, i.e. any

value greater than 3 and any value less than−3.

Thus|x|>3 meansx>3 andx<− 3.

Inequalities involving a modulus are demonstrated

in the following worked problems.

Problem 4. Solve the following inequality:

| 3 x+ 1 |< 4

Since| 3 x+ 1 |<4 then− 4 < 3 x+ 1 < 4

Now− 4 < 3 x+1 becomes− 5 < 3 x,

i.e. −

5

3

<x and 3x+ 1 <4 becomes 3x<3,

i.e. x< 1

Hence, these two results together become−

5

3

<x< 1

and mean that the inequality| 3 x+ 1 |<4 is satisfied

for any value ofxgreater than−

5

3

but less than 1.

Problem 5. Solve the inequality:| 1 + 2 t|≤ 5

Since| 1 + 2 t|≤5 then− 5 ≤ 1 + 2 t≤ 5

Now− 5 ≤ 1 + 2 tbecomes− 6 ≤ 2 t, i.e.− 3 ≤t

and 1+ 2 t≤5 becomes 2t≤4 i.e.t≤ 2

Hence, these two results together become:− 3 ≤t≤ 2

Problem 6. Solve the inequality:| 3 z− 4 |> 2