Simultaneous Linear Equations (I) – In One Or Two Variables JSS3 Mathematics Lesson Note

A linear equation is an equation with one solution, in an equation known as linear there is only one or two unknown variables.  But in the case where we combine two equations (linear) thereby having what is known as a Simultaneous equation.  E.g 

4x + y = 8     (1)    are called simultaneous equations

3x – y = 6     (2)   

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

 ALGEBRAIC METHOD

There are two algebraic methods of solving simultaneous equations.  These are:

(a)    Substitution method

(b)    Elimination method

 Substitution Method

To use the 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 into the original equation.

WRITE ABOUT

Example 1

Solve the following simultaneous equations by substitution method.

  i.                  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 1st equation (1) and the second equation (2) for easy reference later.

Step (2)

From equation (2) make “y” the 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” the subject of the 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 the substitution method make one variable the subject and then substitute this value in the other equation.

Solve the following simultaneously using the 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