Page 101 - Mathematics_F1

Make the coefficients of y in (1) Find the LCM of the denominators
and (2) to be the same. of (1) and the LCM of the
Multiply equation (1) by 3 and (2) denominators of (2). In order to
by 2: remove the denominators, multiply
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×
3(3x +  2 ) y = 8 3 equation (1) by 10 and equation
 (2) by 4:
y =
 2(2x + 3 ) 12 × 2
9x +  6y = 24 (5)   10 x + y  = 4 10×
   2 5  

 4x + 6y = 24 (6) 
 4  x + y  = 6 × 4
Then, subtract (6) from (5):    4 2  
Tanzania Institute of Education

9x +  6y = 24 5x + 2y = 40 (3)
− 
 4x + 6y = 24 x + 2y = 24 (4)
9x – 4x + 6y – 6y = 24 – 24 Eliminate y to get x:
5x = 0 . The coefficients of y in (3) and (4)

5 5 are the same.
x = 0. Subtract (4) from (3):

Therefore, x = 0 and y = 4. −  5x + 2y = 40
 x + 2y = 24
Example 5.33 5x – x + 2y – (+2y) = 40 – 24

Solve the following simultaneous 5x – x + 2y – 2y = 40 – 24
equations by elimination method: 4x =16

 x y 4x 16
 2 + 5 = 4 = .

 4 4
 x + y = 6 Hence, x = 4.
  4 2
Eliminate x to get y:
Solution The coefficient of x in (3) is 5
Mathematics Form One  2 + + 5 = = 4 (1) Multiply equation (3) by 1 and
Label the equations:
The coefficient of x in (4) is 1
y
x
equation (4) by 5:



×
40 1
2 ) y =
1(5x + 
y
x


6
2 ) y =
5 (x +
24 5
×
(2)

 
2
4
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25/09/2025 15:01:13
Mathematics form 1.indd 94 25/09/2025 15:01:13
Mathematics form 1.indd 94