Week 2 – Jss 3 Second Term Mathematics Notes

WEEK 2

SOLVING OF SIMULTANEOUS LINEAR EQUATIONS

A linear equation is an equation with one solution, in equation known as linear there is only one or two variable unknown variables. But in the case where we combine two equations (linear) thereby having what is known as Simultaneous equation. E.g
4x + y = 8    (1)    are called simultaneous equations
3x – y = 6    (2)    

 Simultaneous Linear equations can be solved, graphically, algebraically. But in today’s class we shall be considering the algebraically method of solution.

ALGEBRAIC METHOD

There are two algebraic methods of solving simultaneous equations. These are:
(a)    Substitution method
(b)    Elimination method

Substitution Method

To use substitution method

1. Re-arrange one of the equations so that one variable is made the subject of the formula of the equation.
2. Substitute this into the other equations.
3. Solve the resulting equation to obtain one variable.
4. The other variable is found by substituting your answer into the original equation.
5. Check the solutions by substituting the two answers back into the original equation.

 WRITE ABOUT
Example 1
Solve the following simultaneous equations by substitution method.

1. y = 5x + 2            ii.    2x + 3y = 5

   x + 2y = 15                3x + y = 4

iii.    4m – 3n = 0            iv.    x + 6y = -2
    m + 2n = 3                3x + 2y = 10

 Solution
II.    2x + 3y = 5    ………….    (1)
    3x + y = 4    ………….    (2)

 Step (1)
Label the 1^st equation (1) and the second equation (2) for easy reference later on.
Step (2)
From equation (2) make “y” subject of formulae     3x + y = 4
                            Y = 4 – 3x …………… (3)
Step (3)
Substitute y = 4 – 3x into equation (1)
2x + 3y = 5
2x + 3 (4 – 3x) = 5
Step (4)
Open the brackets and solve for x.
2x + 12 – 9x = 5
12 – 7x = 5
12 – 5 – 7x = 10
=
x = 1

 Step 5
Substitute for x = 1 into equation…………. (3)
y = 4 – 3x; y = 4 – 3(1), y = 4 – 3
y = 4 – 3                        check
y = 1                            
Hence: x = 1, y = 1                    2(1) + 3(1)
is the solution to the equation            2 + 3 = 5
                            In Equation (2)
                            3(1) + (1)
                            3 + 1 = 4
Example II
4m – 3n = 0
m + 2n = 3
Step (1)
Label the equations
4m – 3n = 0    ……….    (1)
m + 2n = 3    ……….    (2)
Step 2
Make “m” subject of formula in equations (2)
m + 2n = 3
m = 3 – 2n     ……….    (3)
Step 3
Substitute m = 3 – 2n into equation ………. (1)
4m – 3n = 0
4(3 – 2n) – 3n = 0
Step 4
Open the bracket and solve for “n”
12 – 8n – 3n = 0
12 – 11n = 0
=
n = 1

 Step 5
Substitute the value on n = into equation………….. (3)
m = 3 – 2
m =
=
Hence m =, n =

 WRAP UP AND ASSESSMENT
Two equations are called simultaneous equations if they are to be solved at the same time. In substitution method make one variable the subject and then substitute this value in the other equation.
Solve the following simultaneously using substitution method.
(1)    x + 6y = -2                (2)    -2 = 5x – y
    3x + 2y = 10                    15 = x + 2y

 (3)    4x + 7y = 20
    3x + y = -2

 TICKET OUT
Solve the following simultaneous Equation by substitution method.
Exercise 16.3 pg 149 No 11 – 1