Inequalities

and Linear Programming

some groundwork

what is the difference between : an expression an equation an inequality an identity ? [ The student should refer to the groundwork

covered upon expressions and equations]

some groundwork

An expression has no value unless arbitrary values are assigned to the variables forming the expression.

An equation has a finite ( countable ) set of values that satisfy the equation. The number of solutions is determined by the degree of the equation.

An inequality has an infinite set of values that satisfy the inequality. Even though the solutions are infinite one still cannot use “any” value for the variable to satisfy the inequality, we use the phrase “ infinite but not all”

Since the solutions are infinite it is not possible to list them and we use a Number Line or a Region to illustrate them.

An identity also has an infinite set of values that satisfy the identity. In the case of an identity one can use “any” value for the variable to satisfy the identity, we use the phrase “ infinite and all”

more groundwork

An expression 3x + 4xy – 5 has no value unless arbitrary values are assigned to the variables x and y. Choose arbitrary values to find a value to the expression.

An equation has a finite ( countable ) set of values that satisfy the equation :

Linear Quadratic Cubic

3x + 15 = 45 x2 + 5x + 6 = 0 x3 – x = 0 3x = 45 – 15 ( x + 2 )( x + 3 ) = 0 x ( x2 – 1) = 0 3x = 30 Either x + 2 = 0 x ( x + 1)(x – 1) = 0 x = 10. x = -2 Either x = 0 Or x + 3 = 0 Or x + 1 = 0 x = -3 x = -1 Or x – 1 = 0 x = 1 One Solution Two Solutions Three Solutions Degree 1 Degree 2 Degree 3

The number of solutions is determined by the degree of the equation.

more groundwork

An inequality has an infinite set of values that satisfy the inequality. Even though the solutions are infinite one still cannot use “any” value for the variable to satisfy the inequality, we use the phrase “ infinite but not all”

An inequality is defined using the inequality signs written and read as A is greater than B, A > B A is smaller than B, A < B A is smaller or equal to B, A is greater or equal to B Consider the inequality x > 4 . What values of x satisfy this inequality? Which values are greater than 4? x = 5, 6, 7, 10, 27, 9⅓, 13⅔, 11⅜, 8.5 An infinite set of values exist satisfying this inequality which is not possible to list and hence we use a

number line ! 0 4 ∞ Note that even though an infinite set

of values exist x = 2 is not a valid solution ! “ Infinite but not all ”

BA BA

0 8 -∞ ∞ -3 0

x is greater or equal to 8 x is smaller than - 3 note that the circle is “full” note that the circle is “empty” meaning that 8 is included meaning that -3 is not included i.e. x can take the any value i.e. x can take any value that is that is greater than 8 or even smaller than -3 but it cannot take equal to 8. -3.

N.B. The number line may not be to scale but numbers must be in order!

8x 3x

the number line

0 8 ∞ The number of solutions to this inequality as already stated is “infinite”, to list a few x = 8, 9, 9.1, 9.2, 9.876,…..10, 10.2…11, 11⅔ ,………….∞

What are the integer values to satisfy this Inequality ? In this manner the question restricts the range of solutions and hence all fractions and decimals must be removed, the new solution now becomes x = 8, 9, 10, 11, 12, 13, 14, 15……∞ still an infinite set ! What is the smallest integer to satisfy this inequality ? This question further restricts the range of solutions to one solution ! x = 8.

the number line

8x

-2

7

more number lines……double Inequalities

In the following examples (a) write an inequality (b) write the integer value/s that satisfy both inequalities simultaneously.

e1. e2.

and and

(a) Inequality (a) Inequality

(b) Integer Values (b) Integer values x = 6, 5, 4, 3, 2, 1, 0, -1, -2. x = 1, 0, -1, -2,…………….- ∞

Always take the common part between the number lines of the inequalities. When solving a double inequality a logical “AND” is being solved.

2x 7x

2x7

1

7

1x 7x

1x

more number lines ……..double Inequalities

In the following examples (a) write an inequality (b) write the integer value/s that satisfy both inequalities simultaneously.

e3. e4.

and and

(a) Inequality (a) Inequality No Solution for these inequalities.

(b) Integer Values (b) Integer values No Integer Values can satisfy them x = 6, 5, 4, 3, 2, 1, 0, -1, ………..-∞ both as no value can be greater than Note that “7” is not a common value for 2 and smaller than -4 at the same time both inequalities so its not taken as part There is no common part ! of the solution.

2x 7x 2-4

4x

7

7

7x

7x

Solving Inequalities……….Linear

e1. Solve the following Inequality 3x + 5 > 20 3x > 20 – 5 3x > 15 x > 5 0 5 ∞

e2. Solve the following Inequality 3(x + 4) – 2(x - 5) < 25 3x + 12 - 2x + 10 < 25 3x – 2x + 12 + 10 < 25 x < 25 – 12 – 10 x < 3 -∞ 3 0

e3. Solve the Inequality 4(2x – 3) – 6(3x -5) < -17 8x -12 – 18x + 35 < -17 8x – 18x -12 + 35 < -17 -10x + 13 < -17 -10x < -17-13 -10x < -30 x < -30 -10 x > 3

0 3 ∞

Note that the Inequality has been inverted in the last line!

Why do we invert the Inequality?

Consider the True Statement 10 > 6. The following operations are done upon

this statement. Add 5 to both sides 15 >11 TRUE Subtract 20 from both sides -10 > -14 TRUE Multiply both sides by 5 50 > 30 TRUE Divide both sides by 2 5 > 3 TRUE Multiply both sides by -3 -30 > -18 FALSE Divide both sides by -2 -5 > -3 FALSE

Note that when the inequality was multiplied or divided by a negative number the “truthfulness” of the inequality is lost. Therefore to preserve the “logic” of the inequality we reverse the inequality !

In e3 it is not the -30 that inverts the sign but the -10.

-10x < -30 x < -30 -10 x > 3.

Solving a Double Inequality……….

e4. Solve 2x – 5 15 and 3x -1 8

2x 15 + 5 3x 8 + 1 2x 20 3x 9 x 10 x 3

3

10

Ans 10 x 3

e5. Solve

4x – 2 < 4( 2x -6 ) 10x + 2

Solving Left Hand Side

Solving 4x – 12 < 4( 2x -6 )

4x - 12 < 8x – 24

4x - 8x < 24 + 12

-4x < 36

x > 36

-4

x > -9

Solving a Double Inequality……….cont

e5. Cont… Solving Right Hand Side Solving 4( 2x -6 ) 10x + 2 8x -24 10x + 2 8x – 10x 24 + 2 -2x 26 x 26 -2 x -13

When these Inequalities are drawn on the same number line we obtain the following

-13

-9

Ans x > 9

Solving Linear Inequalities …with two variables

Consider the Inequality x + y < 5 .

What values of x and y satisfy this inequality ? x = 1 ; y = 3 since 1 + 3 = 4 < 5 True. x = 2 ; y = 0 since 2 + 0 = 2 < 5 True. x = -4; y = 7 since - 4 + 7 = 3 < 5 True.

Note that x = 1 ; y = 4 is not a solution since 1 + 4 = 5 and the inequality requires values that are smaller only.

The set solutions for the “pairs” of values is infinite since one needs to consider those values that are non integers. Moreover we require two number lines, one for the variable “x” and one for “y” since we have two variables. This will result in the Cartesian Axes that we normally use to plot Graphs.

Procedure to Solve linear Inequalities with two variables

Solve x + y < 5

Replace the inequality sign by an “=“ x + y = 5.

Making ‘y’ subject y = - x + 5 This represents a straight line that has intercepts. Finding the intercepts. Y- intercept when X = 0 X – intercept when Y = 0 y = 0 + 5 0 = - x + 5 y = 5 x = 5 point ( 0, 5 ) point ( 5, 0 )

Plot the line on graph paper accurately using the above intercepts!

Procedure to Solve linear Inequalities with two variables

When joining the intercepts to draw the line, the line is to be “dashed ”------- since 5 is not included. This is the equivalent to the “empty” circle when using

the normal number line. Had the 5 been included the line would be drawn “solid “ This is the equivalent of the “full” circle on a normal number line.

Shading the unwanted region (a) Take any random point on any side of the line i.e. above it or below

it. Say (6, 7) which lies above the line. (b) Substituting these values in the original inequality results in x + y < 5 6 + 7 < 5 13 < 5 which is False. (c) Therefore the point (6, 7) lies in the False Region and we shade it. The un-shaded part of the graph is a “Region” of points all of which satisfy the

inequality. Points that lie on the line are not considered as the lines is dashed.

Shading the unwanted region

The point (6,7) defines the false region and is therefore shaded. The purple shade defines the “unwanted region”. Any point in this region makes the inequality false.

The true region is un-shaded referred to as the “wanted region”. which is in grey colour. Any point in this region makes the inequality true.

Note that the line is dashed since 5 is not included.

x + y < 5

(6,7)Defines the false Region

(0,5)

(5,0)

Shading the unwanted region

Solve y < 2x Replace the inequality sign by an

“=“ y = 2x. This represents a straight line

that passes through the origin. To plot this line we will use two

points The Origin A(0,0) and A point at random when x = 3 y = 2x y = 2(3) y = 6. B(3,6)

Using the point (4,3) which lies below the line.

On substitution in y < 2x 3 < 2(4) 3 < 8 True

Therefore (4,3) lies in the wanted region and therefore below the line is the good region. We will shade above the line the “unwanted region”

Shading the unwanted region

A(0,0)

B(3,6)

Defines the true Region

(4,3)

y < 2x The point (4,3) defines

the true region and is therefore un-shaded. The grey area defines the “wanted region”. Any point in this region makes the inequality true.

The false region is un-shaded referred to as the “unwanted region”. which is in white colour. Any point in this region makes the inequality false.

.

Shading the unwanted region

Solve x < 4 Replace the inequality sign by an “=“ x = 4. This represents a vertical line that

passes through “4” on the x-axis. The wanted region will be to the left of

this line as all the points on this side of the line have an x-value which is smaller than 4.

Solve y< 3 Replace the inequality sign by an “=“ y = 3. This represents a horizontal line that

passes through “3” on the y-axis. The wanted region will be below this

line as all the points on this side of the line have a y-value which is smaller than 3.

x < 4

y < 3

Solving a System of Inequalities

If we solve the system of Inequalities defined by

x + y < 5 y < 2x x < 4 y < 3 and y > -2 we obtain the resulting

Region “R” This Region Defines the

points which make all the inequalities true.

Use the point X(2,-1) to check.

x + y < 5y < 2x

x < 4

y < 3

y > -2

“R”

X(2,-1)

Solving a System of Inequalities

Use the point X(2,-1) to check. x + y < 5 2 + (-1) < 5 1 < 5 True. y < 2x -1< 2(2) -1< 4 True. x < 4 2 < 4 True. y < 3 -1< 3 True. y > -2 -1 > -2 True.

Note that this is not the only point that satisfies the system of Inequalities. In fact the whole Region of points R satisfy this system.

Also note that no point on the boundary lines is valid as all of them are dashed.

We will now consider a practical example of such systems of inequalities which will also include solid lines.