Second Term Lesson Note for Week Three.
Class : JSS 3 / Basic 9
Subject : Mathematics
Topic : Elimination Method of Simultaneous Equation.
Duration : 80 Minutes
Period : Double Period
Reference Book :
Essential Mathematics for Junior Secondary School, JSS 3.
Lagos State Unified schemes of work for Junior Secondary School, JSS 1 – 3.
Online Resources
Instructional Material : Chart showing the steps applied in solving problems of elimination method of Simultaneous Equation.
Learning Objectives : By the end of the lesson learners will be able to :
i. Explain a Simultaneous Equation
ii. Identify the methods of solving Simultaneous Equation.
iii. Solve problems on simultaneous Equation using elimination method.
Content :
This method is very useful to solve simultaneous equations especially when none of the coefficients of the unknown is 1.
Example I :
Solve the following simultaneous equations by elimination method.
(a) 6x + 5y = 15… (1) (b) 4c – 4d = 9… (1)
3x + 5y = 12 … (2) 5c + 4d = 18… (2)
One of the unknown “Y” has equal coefficient and with the same signs so we subtract the two equations to eliminate y terms.
Eqn. (1) – Eqn. (2)
6x + 5y – (3x + 5y) = 15 – 12
6x + 5y – 3x – 5y = 3
6x – 3x + 5y – 5y = 3
3x + 0 = 3 Divide both sides by coefficient of x, we have
x = 1
To find y, substitute x = 1 in either Eqn. (1) or Eqn. (2)
Using equation (1) : 6x + 5y = 15, then substitute x = 1
6(1) + 5y = 15
6 + 5y = 15
5y = 15 – 6
5y = 9
y = 9/5 or 1.8
(b) 4c – 4d = 9 ……. (1)
5c + 4d = 18 …… (2)
One of the unknown “d” has equal coefficient but with different sign so we add the two equations to eliminate “d”
4c – 4d + 5c + 4d = 9 + 18
4c + 5c – 4d + 4d = 27
9c = 27
c = 3
To find ‘d’ substitute c = 3 into Eqn. (1)
4c – 4d = 9
4(3) – 4d = 9
4d = 12 – 9
4d = 4
d = 1
WRAP UP AND ASSESSMENT
In elimination method you may need to multiply one or both of the equations by a number in order to obtain a variable with e same coefficient in both equations. Then add both equations when the signs of the variables you want to eliminate are opposite but subtract them when the signs are the same.
Exercise: 16 4 No 2, 3, 7 – 11.
Use elimination method to solve the following simultaneous equations.
2. 6x + 7y = 15 3) 4x + 3y = 10
6x – 9y = 31 4x + 5y = 8
7) 2x + 3y = 8 8) 3x + 4y = 10
3x + 2y = 7 2x + 5y = 9
9) 4x + 3y = 11 10) 4a + 3b = 3
3x – 4y = 2 3a + 2b = 1
TICKET OUT
Solve the following Simultaneous equation by Elimination method. Exercise 16.4 No 12 -1