# INEQUALITIES

**WEEK 8 **

**SUBJECT: FURTHER MATHEMATICS**

**CLASS: SS 1**

**TOPIC: INEQUALITIES**

**CONTENT:**

- Notation and basic rules of inequalities
- Linear inequalities in one variable
- Absolute values
- Inequalities in two variables
- Graphs of linear inequalities in two variables.

**Notation and basic rules of inequalities**

Symbols commonly used for inequalities include;

We use the word ‘** inequal**’ or ‘

**when comparing two variables. For example 5 is not equal to 4, then we say that 5 is greater than 4. And we write with the symbol 5 > 4.**

*unequal’*3 is less than 8 and we write as 3 < 8.

In general ** a > b or b < a.** Some other times if

**is greater than**

*a***or**

*b***is equal to**

*a***then we write as**

*b***This means**

*a ≥ b or b ≤ a.***is greater or equal to**

*a***is less or equal to**

*b or b*

*a’*On the number line, 0 is called the reference point. Points to the right are positive and points to the left are negative. 5 is on the right of 3. Then 5 > 3 or 5 > 3. The information can be presented as 5 -3 > 0 or 3 -5 < 0. This can be represented on the number line.

-2

-1

0

1

2

3

4

5

The arrow represents the solution set of 5 > 3. Note that the point 3 and 5 are not included in the solution.

-2

-1

0

1

2

3

4

5

If x > 4 then this can be represented as below:

The small circles on 4 and 2 shows that both are not included.

The basic rules of inequalities are as follows:

- (a) If a > b and c > 0 then a + c > b + c

(b) If a < b and c > 0 then a + c < b + c We can add the same constant to both sides of an inequality.

- (a) If a > b and c > 0 then a – c > b – c

(b) If a < b and c > 0 then a – c < b – c We can subtract the same constant from both sides of an inequality.

- (a) If a > b and c > 0 then ac >bc and

(b) If a < b and c > 0 then ac <bc and If both sides of an inequality are multiplied or divided by the same positive number, then the sense of the inequality remains unchanged.

- (a) If a > b and c< 0 then ac <bc and

(b) If a < b and c < 0 then ac >bc and If both sides of an inequality are multiplied or divided by the same negative number, the sense of the inequality is reversed.

- (a) If a > b and c <d then a – c > b – d
- If a <b and c > d then a – c < b -d

Note that if a > b and c > d, we cannot just conclude that a – c >b – d

or that if a < b and c < d we cannot just conclude that a – c < b – d.

- (a) If a > b and n > 0 then a
^{n}>b^{n}

(b) If a > b and n < 0 then a^{n}<b^{n}

- (a) If a < b and n > 0 then a
^{n}<b^{n}

(b) If a < b and n < 0 then a^{n}>b^{n}

**Example 1** (i): 6 > 2 6 + 5 > 2 + 5

6 – 5 > 2 – 5

(ii) 2 <4 2 + 3 < 4 + 3

2 -3 < 4 -3

(iii) 6>2 6x 5 > 2 x 5

6/5 > 2/5

(iv) 3<5 3x 2 < 5 x 2

3/2 < 5/2

**Example 2:** 6 > 3, if multiplied by -2

6 x -2 < 3x -2

6/-2 < 3/-2

Find out what happens when number less than 0 is used in adding and subtracting to inequalities.

**NOTE**: Reverse the inequality sign when both sides are multiplied (or divided) by negative quantity. i.e. if 2 < 5 then -2 > -5

Reverse the inequality sign when reciprocals are taken

i.e. if

**N.B: Teachers are expected to give more example on each of the above-listed rules in order to ensure proper understanding of the concept.**

**Linear inequalities in one variable**

The number line can be used to show the graph of inequalities in one variable.

**Example 1:** Find the solution to the set of the following inequalities:

(a) x + 5 > 7 (b) 2x + 4 > x + 6 (c) 3x – 3 < 2x + 5 (d) 5x + 7 > x + 13 (e)1/2x + 3 > 1 – x

**Solution:**

- x + 5 > 7

x + 5 -5 > 7 – 5

x > 2

- 2x + 4 > x + 6

Collect like terms

2x – x > 6 – 4

X > 2

Note the rules for linear equation also holds for linear inequality

- 3x – 3 < 2x + 5

3x – 2x < 5 + 3

x < 8

-2

-1

0

1

2

3

4

5

- 5x + 7 > x + 13

5x – x > 13 – 7

4x > 6

4x/4 = 6/4

x > 3/2 0r 1^{1}/_{2}

- (½)x + 4 > 1 – x

Multiply by 2 to clear fraction

x + 8 > 2 – 2x

3x > 2 – 8 = -6

x > -2

Note that the solution set can be shown on number lines as follow:

-2

-1

0

1

2

3

4

5

-2

-1

0

1

2

3

4

5

-1

0

1

2

3

4

5

6

7

8

5

-2

-1

0

1

2

3

4

5

2

-2

-1

0

1

2

3

4

5

-3

**Example 2**

- If a and b are positive, such that a < b, show that
- Find the range of values of x for which the inequality 4x + 3 ≥ 2x – 7 is true.

**Solution:**

- If a < b show that

Let a = 5 and b = 10. Then

Show that . This is always true.

- 4x + 3 ≥ 2x – 7

4x -2x ≥ – 7 – 3

2x ≥ -10

X ≥ -5

-4

-3

-2

-1

0

1

2

-5

-6

Note that the on shows that the point is included in the solution set of the problem.

**Example 3: **Solve

**Solution**

0 1 2 3 4

Notice that the right end point x=4 is not part of the solution so the circle above is not shaded.

**Example 4: ** solve the inequality;

**Solution**

To clear the fraction, multiply through by the LCM of the denominators i.e. 12

-2 -1 0 1 2

Notice that the left end point, is part of the solution, so small circle above is shaded.

**Example 5: ** Find the range of values of x which satisfy (WAEC)

Solution;

-4 -3 -2 -1 0 1 2 3 4

**CLASS ACTIVITY**

- Choose the correct answer from the options a –d.
- 2x is greater than 10 using symbol is
- 2x ≥ 10 b. 2x < 10 c. 2x > 10 d. 2x + 10 > 0

- Find the value of x for which 3x – 1 < 0
- 1 (b) 2 (c) 3 (d) 0
- Consider the statement and say whether they are always true or not for real numbers.

If ax >bx then a > b

a/x < b/x then a < b

if 2 ≤ x ≤ 3 and 1 ≤ y ≤ 10, find the least and greatest value of x + y.

(a) 2x ≥ 10 (b) 2x < 10 (c) 2x > 10 (d) 2x + 10 > 0

- Solve the inequalities and show the interval on a number line
- Solve the inequality and represent your result on a number line;
- Find the three highest whole number that satisfy
- Solve and show on number line the values of x which satisfy

**Absolute Value function**

**If a number x is positive, the absolute value of x denoted by |x| is x itself. If the number x is negative the absolute value of x is –x. The absolute value of a number gives the magnitude of the number regardless of the sign. The absolute value function sometimes called modulus function f(x) = |x| is defined precisely as: **

*f(x) = |x| = { x if x > 0}*

*Compare the graph of y = x and that of y = |x| as shown below:*

* y y*

* y = x y = |x|*

* x x*

*Example 1:*

*Solve | x – 1 | > | 3 – x |*

*Solution*

*| x – 1 | > | 3 – x |*

* x – 1 > 3 – x or -( x – 1 ) > 3 – x *

* 2x > 4 or -x+ 1 > 3 – x*

* x > 2 or 1 > 3 (not possible)*

*Hence x >2 is the solution.*

*Example 2:*

*Solve 3| x + 2 | < | x – 2 |*

*Solution*

*3| x + 2 | < | x – 2 |*

* 3(x + 2) < x – 2 or -3( x + 2 ) < x – 2 *

* 3x + 6 < x –2 or -3x – 6 < x – 2*

*2x <-8 or -4x < 4 (not possible)*

* x < -4 or x > -1*

*Hence the solution set is*

*{x: x < -4 or x > -1}*

**CLASS ACTIVITY**

- Find the values of x satisfying:
*| x + 1 | >4**| x – 2 | < | x |*- What are the values of x satisfying:
*| 2 – x | > | x +3**| x – 3 |> 0*

**Inequalities in two variables: graphs of linear inequalities in two variables. **

For linear inequalities in two variables, first draw the corresponding straight line. Inequalities in two variables are usually plotted on plane (the Cartesian coordinate plane)

**Example 1.**

Find the region of x – y plane which satisfy the inequalities 2x – 2y > 4 + 2y – 3x

**Solution**

3x – 2y > 4 + 2y – 3x

3x + 3x – 2y – 2y > 4

6x – 4y > 4

3x – 2y > 2

Finding co-ordinate when x = 0, y = -1, y = 0 and x = 2/3

0

1

2

1

2

The region with the solution is the shaded part of the plane but not including the line 3x – 2y = 2

**Example 2:** show the region for which 2y – x > 5. Find the boundary line.

2y – x = 5

When x = 0, then 2y = 5, y = 5/2

When y = 0 then x = -5

(0, 5/2) and (-5, 0)

**Example 3:**

Show the region satisfying the inequalities x + y ≤ 4 and x + 2y 2 and x 0

for x + y = 4

The co-ordinate are (0,4) and (4,0) and x + 2y = 2. The co-ordinate are (2,0) and (0,1)

**A**

**The region labelled A satisfied the inequalities given**

**Example 4:** Show the expression on a graph.

**Solution**

First put y on one side of the inequality i.e

-3 | -1 | 0 | 3 | |

-1 | 0 | 1 | 2 |

Then draw the corresponding line

This line divides the plane into two. To find the side with the solution, we select and try out a pair of points. E.g

For

Hence, the solution set is in the region above the line

The upper part of the graph shaded satisfies the inequality

Now we shall consider range of values of combined inequalities.

**Example 5:**

If what range of x satisfies both inequalities?

Solution: solving the inequalities separately we obtain respectively.

**Example 6:** Show on a graph the region that contains the solution of the simultaneous inequalities

**Solution:** In each case put on one side of the inequality

We shall draw the lines

-2 | 0 | 2 | 3 | |

3.3 | 2 | 0.7 | 0 | |

-2 | 2 | 6 | 8 |

y-axis

8

7 y=2+2x

6

5

4

3

2

1

-3 -2 -1 0 1 2 3 4 x-axis

-1

-2 y=

-3

-4

Points are in the solution set for the three inequalities.

The shaded portion is the required region. The integral values of x & y that satisfy the inequalities simultaneously are (-1,0), (0,0), (0,1), (0,2), (1,0), (1,1), (2,0), (3,0)

**CLASS ACTIVITY**

- Show the region that satisfies the inequality

2x + 2y ≤ 4

- Show the region satisfying the following inequalities simultaneously

x 0, y 0, x + y ≤ 2 and 2x + y ≥ 2

- Show the region that satisfies the inequalities on x, y co-ordinate plane.

y ≥ 0, x + y ≥ 2 and x + 2y ≤ 6

**Simultaneous linear inequalities & Application of linear inequalities in real life.**

In solving simultaneous inequalities involving variables x & y, the expression x+y=n is called the objective function. Linear programming usually involves either maximizing or minimizing the function x+y=n. These problems are sometimes called minimax problems.

**Example 1:** A manufacturer has 120kg and 100kg of wood and plastic respectively. A product requires 2kg of wood and 3kg of plastic. Product requires 3kg of wood and 2kg of plastic. If A sells for #3500 and B for #5000. How many must be made to obtain the maximum gross income?

**Solution:**

Wood (kg/unit) | Plastic (kg/unit) | # per kg | |

Product A | 2 | 3 | 3500 |

Product B | 3 | 2 | 5000 |

Suppose there is x number of product A and suppose there is y number of product B

For the first inequality we shall draw line 2x + 3y = 120

If x=0, y=40.

**PRACTICE QUESTIONS**

Solve the following inequalities and show the solution set on number lines.

- 2x – 1 ≥ 5
- 3 – x > 7
- – ≥

**ASSIGNMENT**

- Shade the region defined by;
- Show on a graph the region which contains the solutions of the simultaneous inequalities
- Find the region common to show the region on a graph
- What range of satisfy both
- Find the range of values of such that
- Express the inequality in the form where are both integers Shade the region common to
- Show on a graph the region that contains the set of points for which
- Shade the region that satisfy the following