Page 99 - Mathematics_F1

Make the coefficients of x in (1) and 6x – 3x + y – y = 15 – 9
(2) to be the same. Multiply (2) by 3x = 6
2 and (1) by 1: 3x = 6
3 3
×
×
 1( 6x + ) y = 15 1 Hence, x = 2.
FOR ONLINE READING ONLY

 2 (3x + × y = ) 9 × 2 Therefore, x = 2 and y = 3.
y
 6x += 15 (3) Example 5.31

 6x+ 2y = 18 (4) Solve the following simultaneous
equations by elimination method:
Tanzania Institute of Education
If the signs of the variable to be
y
eliminated are the same (that is ‘+’  2x += 10

and ‘+’ or ‘–’ and ‘–’), then subtract  3x− 2y = 1
(2) from (1) or (1) from (2). But, if Solution
the signs are different (i.e. ‘–’ and Label the equations:
‘+’), then add the two equations.  (1)
2x +=
Since the signs of the variable x in  3x− y = 10
(3) and (4) are the same, subtract  2y 1 (2)

equations (4) from (3). Choose the variable to be eliminated.
y
 6x += 15 Eliminate y to get x .
− 
 6x+ 2y = 18 The coefficent of in (1) is 1y
6x – 6x + y – 2y = 15 – 18 The coefficent of in (2) is 2y
–y = –3. Make the coefficients of y in (1)

Hence, y = 3. and (2) to be the same.
Eliminate y to get x : Multiply equation (1) by 2 and
equation (2) by 1.
The coefficent of in (1) is 1y 2 (2x y+ ) = 10 2
×
y
The coefficent of in (2) is 1  1 (3x − 4x+  2y = 2 ) 1 1 × y = (3) (3)


The coefficients are the same, then
Mathematics Form One any number. 9 Add equations (3) and (4): (4) (4)
there is no need of multiplying by
20

2y =
1
3x−

Subtract equation (2) from (1):

15
6x y
+=
20
2y =
4x+ 
−

+

+=
3x y

1
3x−
2y =
92 
25/09/2025 15:01:11
Mathematics form 1.indd 92 25/09/2025 15:01:11
Mathematics form 1.indd 92