SOLVING COMPOUND INEQUALITITES

There are two operators in inequality. Those are 

i)  And

ii)  or

How we use the operator and ?

The region that satisfies both conditions can be fixed as solution. n - and

How we use the operator Or ?

The region that satisfies any one of the conditions can be fixed as solution. u - or

Problem 1 :

(-∞, 1/2) ∩ [-3, 4)

To find the intersection, graph each interval separately. Then find the real numbers common to both intervals.

Problem 2 :

(-∞, -2) ∪ [-4, 3)

Solve the compound inequalities . Write the solutions in interval notation.

Problem 1 :

4m > -11 and 4m – 3 ≤ 13

Solution : 

4m > -11

Divide by 4 on both sides.

4/4m > -11/4

m > -11/4

By graphing the inequality m > -11/4, we get the graph given below.  

4m – 3 ≤ 13

Add 3 on both sides.

4m – 3 + 3 ≤ 13 + 3

4m ≤ 16

Divide by 4 on both sides.

4m/4 ≤ 16/4

m ≤ 4

By combining the above two graphs, we get the common region between -11/4 and 4.

Problem 2 :

4n – 7 < 1 and 7 + 3n ≥ -8

Solution : 

4n – 7 < 1

Add 7 on both sides.

4n – 7 + 7 < 1 + 7

4n < 8

Divide by 4 on both sides.

4n/4 < 8/4

n < 2

7 + 3n ≥ -8

Subtract by 7 on both sides.

7 + 3n - 7 ≥ -8 – 7

3n ≥ -15

Divide by 3 on both sides.

3/3n ≥ -15/3

n ≥ -5  

Problem 3 :

-3y + 1 ≥ 10 and -2y – 5 ≤ -15

Solution : 

Solving -3y + 1 ≥ 10 :

-3y + 1 ≥ 10

Subtract by 1 on both sides.

-3y + 1 - 1 ≥ 10 – 1

-3y ≥ 9

Divide by -3 on both sides.

y ≤ -3

Solving -2y – 5 ≤ -15 :

-2y – 5 ≤ -15

Add 5 on both sides.

-2y – 5 + 5 ≤ -15 + 5

-2y ≤ -10

Divide by -2 on both sides.

y ≥ 5

It should satisfy the conditions y ≤ -3 and y ≥ 5. No such values in the number line. So, there is no solution.

Problem 4 :

(1/2) – (h/12) ≤ -7/12 and (1/2) - (h/10) > -1/5

Solution : 

Solving 1/2 - h/10 > -1/5 :

1/2 - h/10 > -1/5

Subtract by 1/2 on both sides.

-h/10 > -1/5 – 1/2

-h/10 > -1/5 × (2/2) – 1/2 × (5/5)

-h/10 > -2/10 – 5/10

-h/10 > -7/10

Multiply 10 on each sides.

(-h/10) × 10 > (-7/10) × 10

-h > -7

h < 7

So, no solution.

Problem 5 :

(2/3)t – 3 ≤ 1 or (3/4)t – 2 > 7

Solution : 

Solving (2/3)t – 3 ≤ 1 :

2t – 9 ≤ 3

Add 9 on both sides.

2t – 9 + 9 ≤ 3 + 9

2t ≤ 12

Divide by 2 on both sides.

2/2t ≤ 12/2

t ≤  6

Solving for (3/4)t – 2 > 7 :

(3/4)t – 2 > 7

3t – 8 > 28

Add 8 on both sides.

3t – 8 + 8 > 28 + 8

3t < 36

Divide by 3 on both sides.

3/3t < 36/3

t < 12

Problem 6 :

2(3x + 1) < -10 or 3(2x – 4) ≥ 0

Solution : 

Solving 2(3x + 1) < -10 :

2(3x + 1) < -10

6x + 2 < -10

Subtract by 2 on both sides.

6x + 2 - 2 < -10 – 2

6x < -12

Divide by 6 on both sides.

6x/6 < -12/6

x < -2

Solving 3(2x – 4) ≥ 0 :

3(2x – 4) ≥ 0

6x – 12 ≥ 0

Add 12 on both sides.

6x ≥ 12

Divide by 6 on both sides.

6x/6 ≥ 12/6

x ≥ 2

Problem 7 :

-7 < -7(2w + 3) or -2 < -4(3w – 1)

Solution : 

Solving -7 < -7(2w + 3) :

-7 < -7(2w + 3)

Multiplying by negative -1, we get

1 > (2w + 3)

Subtracting 3

1 - 3 > 2w

-2 > 2w

-1 > w

Solving -2 < -4(3w – 1) :

-2 < -4(3w – 1)

Dividing by -2, we get

1 > 2(3w - 1)

1/2 > 3w - 1

Add 1, we get

1/2 + 1 > 3w

3/2 > 3w

1/2 > w

Problem 8 :

5(p + 3) + 4 > p -1 or 4(p – 1) + 2 > p + 8

Solution :

Solving 5(p + 3) + 4 > p -1 :

5(p + 3) + 4 > p -1

5p + 15 + 4 > p -1

5p + 19 > p – 1

Subtract by 19 on both sides.

5p > p – 1 – 19

5p > p – 20

Subtract by p on both sides.

4p > -20

Divide by 4 on both sides.

4p/4 > -20/4

p > -5

Solving 4(p – 1) + 2 > p + 8 :

4(p – 1) + 2 > p + 8

4p – 4 + 2 > p + 8

4p - 2 > p + 8

Subtract by p on both sides.

4p – p – 2 > 8

3p – 2 > 8

Add 2 on both sides.

3p > 8 + 2

3p > 10

Divide by 3 on both sides.

3p/p > 10/3

p > 10/3