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Problem 2.3.Determine algebraically the values of x for which 15x2

x>1.

Solution: First, if we have15x 5x

>1, then

15x2

x 1 >0

Now, if we multiply both sides by x, we need to consider two cases.

Case 1: x >0. Thus

15x2

x 1 >0 = 15x2 2 x >0 = (5x 2)(3x+ 1) >0

This means that we have to consider two sub-cases.

Case 1a: 5x 2 >0 and3x+ 1 >0. For this sub-case we have

x >2

5 and x >

1

3

Figure 8: Number line representation of x > 25

and x > 13

.

Therefore we have: x >0 and x > 25

and x > 13

. In other words x 2

5,

. See Figure

9.

Figure 9: Number line representation of x >0 and x > 25

and x > 13

.

Case 1b: 5x 2

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Figure 10: Number line representation of x < 25 and x < 13 .

However, we have that x >0. Then we cannot have

x >0 and x 0 and x < 25

and x < 13

.

In other words the intersection (0,) ,

2

5

,

1

3

is empty.

Case 2: x 0 = 15x2 2 x 25

and x < 13

.

This sub-case is rejected. There is no solution even considering that x

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Figure 13: Number line representation of x 25

and x < 13

.

This means that the intersection (, 0) 2

5,

,

1

3

is empty.

Case 2b: 5x 2 0. In this sub-case we have

x

1

3

Figure 14: Number line representation of x < 25

and x > 13

.

Thus we have that x 13

. Therefore13

< x 1provided that 13

< x 2

5. That is to say x

1

3, 0

2

5,

.

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2.1 Practice

Exercise 2.1.Determine algebraically the values of x for which 3x >5 2

x.

Exercise 2.2.Determine algebraically the values of y for which 1

y2 4

0.

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