# Lesson Notes By Weeks and Term - Senior Secondary School 2

GRAPHICAL SOLUTION OF INEQUALITY IN TWO VARIABLES

SUBJECT: MATHEMATICS

CLASS:  SS 2

DATE:

TERM: 2nd TERM

REFERENCE BOOKS

• New General Mathematics SSS2 by M.F. Macraeetal.
• Essential Mathematics SSS2 by A.J.S. Oluwasanmi.

WEEK TWO        DATE: ___________

TOPIC: GRAPHICAL SOLUTION OF INEQUALITY IN TWO VARIABLES

CONTENT

-Revision of Linear Equation in Two Variables.

-Graphical Representation of Inequalities in Two Variables.

-Graphical Solution of Simultaneous Inequality in Two Variables.

Revision of Linear Equation in Two Variables.

Examples

Solve and represent the solution on graph

1. x + y =2

Choosing values for x:let  x=0,1,2

y=2-x

 x 0 1 2 2 2 2 2 -x -0 -1 -2 y 2 1 0

1.   5x + 2y = 10             0    1    2    3        x

Using intercept method

When x = 0

5(0) + 2y = 10

2y = 10

y = 5

â¬

When y  = 0

5x + 2(0) = 10

5x     = 10

x = 2

(0,5) (2, 0)

0    1    2    3    4    x

1.   Draw the linear graph of y = 2

2

0    1    2            x

4.Draw  the  linear  graph  of  x  = 3

y

4 –

2 –

0            3

5.Draw the graph of 2x + y =3 using intercept method

When y = 0

2x=3

x = 3/2 = 1.5

When x =0

y = 3                                                                                                                                                             (0, 3) (1.5  , 0)

y

3 –     (0, 3)

(1.5, 0)

1    2            x

Evaluation

Sketch the graph  of  the functions:

1) 4x + 3y = 12

2)  y  -  x  = 5

GRAPHICAL REPRESENTATION OF INEQUALITIES IN TWO VARIABLES

Example 1:Show  on  a  graph  the  region  that  contains  the  set   of   points for  which

2x + y ≤ 3                  y

When x = 0                (0, 3)

2(0) + y = 3

y= 3

When y = 0

2x + 0 = 3

2x = 3                                (1.5. 0)

x= 3/2 = 1.5            0                            x

(0, 3) (1.5, 0)

The  unshaded  region  satisfies  the  inequalities.

Note: The continuous thick line is used in joining point when the symbols  ≥ or  is used and when < or > is used broken line or dotted line is used.

Check: When    x = 2,  y=1

2 x + y < 3

2 (2) + 1< 3

4 + 1 < 3

4 + 1 < 3

5 < 3 (No)

Therefore the other side is the region that satisfies the inequality.

Example 2                            y

2x + 3y > 6

When x = 0                        (0, 2)

3y  =  6

y=2

When y = 0

2x=6                -2    -1    0    1    2      (3, 0)        x

x = 3

( (0, 2)  (3, 0)

The  shaded  region  satisfies  the  inequality

Example 3

y< 2

y

2

0                        x

Theunshaded  region  satisfies  the  inequalities

Evaluation

Represent  the  following   functions  graphically.

1. 4x + 3y > 12
2. x +y ≥ 2

Shade the region that does not satisfy the inequality.

Graphical Solution of Simultaneous Inequality

Example I

Show on a graph the region which contains the solutions of the simultaneous inequalities

i    2x +3y < 6

ii    y – 2x ≤ 2

iii    y >  - 2                                y

Solution:                        2x + 3y < 6    6 -         y – 2x 2

2x + 3y < 6                        when  y= 0

2x + 3y < 6                          2x = 6                    4 -

When x = 0                           x=3

3y = 6                                2 – A

Y = 2

-3     -2     -1     0       1        2    3      4

Coordinates:      (0, 2) (3,0)

C        -2                    D

(ii) y – 2x ≤ 2                                            y > -2

When  x = 0            When y = 0                -4

y = 2                -2x = 2

x = -1

Coordinates;      (0, 2) (-1, 0)

(iii)    y>  - 2          (0,-2)

The unshaded region           ABC satisfies all the inequalities.

Any coordinate within the satisfied region satisfies all the inequalities e.g

(x, y) = (-1,-1) (0,-1) (1,-1) (2,-1)

(3,-1), (-1,0) (0,0) (1,0) (2,0) (0,1) (1,1)

Example 2

Solve graphically the simultaneous inequality and shade the region that does not satisfies the inequality.

-x + 5y≤ 10

3x -4y 8

and y > -1

Solution

-x + 5y10

When  x=0

5y = 10

y = 2

When y = 0

-x = 10

x =-10

x = -10

Coordinates:   (0,2) (-10, 0)

3x                                 4

Solution

-x + 5y ≤ 10                            3

5y = 10

y = 2                                2

When y = 0

-x = 10                                1

X =-10

X = -10                -10      -8     -6     -4        -2       0      2     4      6       8      10

(0,2) (-10, 0)

3x – 4y ≤ 8                        -1

When x  = 0

-4y =8                            -2

y = -2

When y = 0                                -3

3x = 8

x = 8

3

x = 2  2/3

(0, -2) (2 2/3   , 0)

(ii)     y> -1

Coodinates:   (-1,0)

Evaluation

Solve  graphically  for  integral  values  of  x  and  y

y ≥ 1 , x – y ≥ 1 and 3x + 4y ≤ 12

GENERAL EVALLUATION/REVISION QUESTIONS

Solve graphically the simultaneous inequalities

1. If  (i) x + 3y≤ 12 (ii) y ≥-1 (iii) x > -2 for integral values of x and y

2.y  is  such  that   4y – 7 ≤ 3y and 3y≤5y + 8

a)What  range  of  values  of  y  satisfies both  inequalities?

b)Hence  express  4y - 7 ≤  5y + 8 in the form a ≤ y ≤ b,where a and b are both integers

3.If 65x2+x-10=0 find the values of x

1. Solve the equations 2x+y=1 and 25x-y = 125 simultaneously

New General Mathematics SSS2, pages 98-111, exercise10e.

WEEKEND ASSIGNMENTS

Objectives

1.Which of the following number line represents the inequality 2 ≤ x < 9

(a)                  (b)                  (c)

0        9        0                      9        0        9

(d)

0        9

2.Form an inequality for a distance “d” meters which is more than 18cm but not more

than 23m.

(a) 18 ≤d ≤23     (b) 18< d ≤ 23    (c) 18 ≤ d < 23     (d) d< 18 or d > 23

1. Interprete the inequality represented on the number line

-4        0            5

(a) -4 < x d5    (b) -4 dx< 5    (c) -4 < x < 5    (d) -4 x d5

1. Solve the inequality 1 (2x-1) < 5

3

(a) x< -6     (b)  x < 7     (c) x < 8    (d) x  <  16

5.Which of the following could be the inequality illustrated on the shaded portion of  the of  the  sketched   graph  below.

y

(0, 3)

(1, 0)

x

(a)y ≤ x + 3     (b)y   3x + 2   (c) –y ≤ 3x – 3     (d) –y ≤ 3x + 3

Theory

Show  on  a  graph  the  area  which  gives  the  solution  set  of  the  inequalities  shading  the  unrequired  region.

•  y  ≤ 3, x – y     1 and    4x + 3y ≥ 12

1. y  - 2x ≤ 4, 3y + x ≥ 6  and y ≥ x-9