Simultaneous Linear Equation – SS1 Mathematics Lesson Note (First Term)

Methods of solving Simultaneous equation

1. Elimination method
2. Substitution method

iii. Graphical method 

ELIMINATION METHOD 

One of the unknowns with the same coefficient in the two equations is eliminated by subtracting or adding the two equations. Then the answer of the first unknown is substituted into either of the equations to get the second unknown.

Example

Solve for x and y in the equations:

2x + 5y   = 1  and 3x – 2y   =   30

Solution

To eliminate x multiply equation 1 by 3 and equation 2 by 2

2x + 5y = 1 ………. eqn 1 (x (3)

3x – 2y = 30 ………… eqn 2 (x (2)

Resulting into,

6x + 15 y = 3 ………. eqn 3

6x – 4y = 60   ……….. eqn 4

Subtract eqn 3 from eqn 4

6x – 6x + 15y – (- 4y) = 3 – 60 

19y =   –57   

 19        19

y = -3

Substitute y = – 3 into eqn 1 

2x + 5 (-3) = 1

2x = 1 + 15

2x  = 16

 2        2

x  =  8

( y =  -3  and  x = 8

Evaluation

Using the elimination method to solve the simultaneous equations.

1. 5x – 4y = 38 and x + 3y = 22
2. 2c-3d= -4 and 4c-3d= -14

SUBSTITUTION METHOD

One of the unknowns (preferably the one having 1 has its coefficient) is made the subject of the formula in one of the equations and substituted into the other equation to obtain the value of the first unknown which is then substituted into either of the equations to get the second unknown.

Example: Solve the simultaneous equation 2x + 5y = 1 and 3x – 2y = 30

Solution

2x + 5y   = 1……………. eq 1

3x – 2y   = 30 ………….. eq 2

Make x the subject in eqn 1

2x   =   1 – 5y

 2             2

x =   1 – 5y   …………    eqn 3

           2

Substitute eq3 into eqn 2

3 (1-5y)   –   2y = 30

      2

Multiply by 2 or find the LCM and cross multiply.

3 – 15y  –  4y  = 30

         2

3 – 15y – 4y = 60

3 – 19y = 60

-19y = 60 – 3

–19y  = 57   3

 -19       -19

y  =  – 3

Substitute y = -3 into eq 3

x  =1 – 5y

           2

x   =   1 – 5 (-3)=  1 + 15   =  16

    2           2             2

 x =   8

( x = 8, y = -3)

Evaluation

Solve for x and y in the equations 

1. x + 2y = 10 and 4x + 3y = 20
2. 4x-y=8 and 5x+y=19

SIMULTANEOUS EQUATIONS INVOLVING FRACTIONS

Example

1. Solve the following equations simultaneously

2 –   1 =  3   

x      y

4  +   3    =   16

x        y                             

Solution

2   –    1    =    3.     ………. i

x        y

4   + 3     =    16   ………. ii

x y

Instead of using x and y as the unknown, let the unknown be (1/x) and (1/y). 

2(1/x) –  (1/y)  =  3 ……………. 1

4 (1/x) – 3 (1/y)   =  16 …………… 2

Using the elimination method, multiply equation 1 by 2 to eliminate x.

4(1/x) – 2(1/y)  =  6 ……………..  3

4 (1/x) + 3(1/y) =  16 ……………. 4

–5 (1/y)   =   -10

  -5 -5

1   =    2

y

(   y = ½ )

Substitute (1/y) = 2   into eqn 1

2 (1/x) – (1/y) = 3

2 (1/x) – (2) = 3

2(1/x) = 3 + 2

2 (1/x) =   5     

1     =       5

x   2

( x   = 2/5

( y = ½,  x  =  2/5

WORD  PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS

Examples

1. Seven cups and eight plates cost N1750, eight cups and seven plates cost N1700. Calculate the cost of a cup and a plate.

Solution

Let   a cup be x and a plate be y

7x + 8y = 1750 ————– eq 1

8x + 7y = 1700 ————– eq 2

Multiply eq 1 by 8 and eq 2 by 7 to eliminate x (cups).

56x + 64y = 14000 ———- eq 3

56x + 49y = 11900 ———- eq 4

Subtracting eq 4 from eq 3

15y  =  2100

y =   2100

            15

Y = 140                                                                                                                                                                                                                                                                                                                                                            

Substitute y = 140 into eq 2

8x + 7y = 1700

8x + 7 (140) = 1700

8x + 980 = 1700

8x = 1700 – 980

8x = 720

x = 720

       8

x =  90

(Each cup costs N90 and each plate costs N140 

2. Find a two-digit number such that two times the tens digit is three less than thrice the unit digit and 4 times the number is 99 greater than the number obtained by reversing the digit.

Solution

Let the two-digit number be ab, where a is the tens digit and b is the unit digit.

From the first statement,

2a  + 3 = 3b

2a – 3b = -3 ………….eq1

From the second statement,

4(10a + b) – 99 = 10b + a

40a + 4b – 99 = 10b + a

40a – a + 4b – 10b = 99

39a – 6b = 99

Dividing through by 3

13a – 2b = 33 ………….eq2

S

solving both equations simultaneously,

a = 3 , b = 3

Hence, the two-digit number is 33