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6x – 3x + y – y = 15 – 9 4x + 3x + 2y + (–2y) = 20 + 1 3x+ 2y = 8
3x = 6 4x + 3x + 2y – 2y = 20 + 1
3x = 6 7x = 21 2x+ 3y = 12
3 3 7x 21 Solution
Hence, x = 2. 7 = 7 . Label the equations:
Therefore, x = 2 and y = 3. Hence, x = 3. Tanzania Institute of Education
Eliminate x to get y: 3x+ 2y = 8 (1)
Example 5.31 The coefficent of in (1) is 2x 2x+ 3y = 12 (2)
Solve the following simultaneous The coefficent of in (2) is 3x Choose a variable to be eliminated.
equations by elimination method: Make the coefficient of x in (1) and Eliminate x to get y:
2x += 10 (2) to be the same. The coefficent of in (1) is 3x
y
3x− = Multiply equation (1) by 3 and (2) The coefficent of in (2) is 2x
2y 1
by 2: Make the coefficients of x in (1) and
Solution 3(2x + ) y = 10 3× (2) to be the same.
Label the equations:
2 (3x − 2 ) 1 2× y = Multiply equation (1) by 2 and
2x += 10 (1) equation (2) by 3:
y
3x− = FOR ONLINE READING ONLY
30
(5)
3y =
6x +
2y 1 (2) − 6x− 4y = 2 2(3x+ 2 ) y = 8 2
×
Choose the variable to be eliminated. (6) 3(2x+ 3 ) 12 3
y =
×
Eliminate y to get x . Then, subtract (6) from (5)
6x + 3y = 30 6x + 4y = 16 (3)
The coefficent of in (1) is 1y − 6x + 9y = 36
6x−
4y =
2
The coefficent of in (2) is 2y 6x – 6x + 3y – (–4y) = 30 – 2 (4)
Make the coefficients of y in (1) Then, subtract (3) from (4).
and (2) to be the same. 6x – 6x + 3y + 4y = 30 – 2 6x + 9y = 36
Multiply equation (1) by 2 and 7y = 28 − 6x + 4y = 16
equation (2) by 1. 7y = 28 6x − 6x + 9y−+ y = −
( 4 ) 36 16
×
2 (2x y+ ) = 10 2 7 7 6x − 6x + 9y− 4y= 36 16
−
y = 4.
1 (3x − 2 ) 1 1 × y = Therefore, x = 3 and y = 4. 5y = 20 .
4x+ 2y = 20 (3) (3) 5 5
(4) Hence, y = 4.
3x− 2y = 1 (4) Example 5.32
Add equations (3) and (4): Solve the following linear Eliminate y to get x: Mathematics Form One
4x+ 2y = 20 simultaneous equations by The coefficent of in (1) is 2y
+ The coefficent of in (2) is 3y
3x− 2y = 1 elimination method:
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Mathematics form 1.indd 93
Mathematics form 1.indd 93 25/09/2025 15:01:12

