11.2.1 Linear Equations II, PT3 Focus Practice


11.2.1 Linear Equations II, PT3 Focus Practice

Question 1:
It is given that 2x = 6 and 3x + y = 10.
Calculate the value of y.

Solution:
2x = 6
= 3
Substitute = 3 into 3x + y = 10
3 (3) + y = 10
= 10 – 9
y = 1


Question 2:
It is given that = –1 and x – 3y = –10.
Calculate the value of x.

Solution:
Substitute = –1 into x – 3y = –10
– 3 (–1) = –10
+ 3 = –10
= –10 – 3
x = –13


Question 3:
It is given that 7x – 2y = 19 and x + = 13.
Calculate the value of y.

Solution:
Using Substitution method
7x – 2y = 19 ——– (1)
+ y = 13 ——- (2)
From equation (2),
x = 13 – y ——- (3)
Substitute equation (3) into equation (1),
7x – 2y = 19
7 (13 – y) – 2y = 19
91 – 7y – 2y = 19
– 9y = 19 – 91
– 9y = –72
 y = 8 


Question 4:
It is given that 2xy = 5 and 3x – 2= 8.
Calculate the value of x.

Solution:
Using Elimination method
2xy = 5 ——– (1)
3– 2y = 8 ——- (2)
(1) × 2:  4– 2y = 10 -------- (3)
   3x – 2= 8 ------- (2)
(3) – (2): x – 0 = 2
  x = 2


Question 5:
It is given that + 2y = 4 and x + 6y = –4.
Calculate the value of x.

Solution:
Using Elimination method
+ 2y = 4 -------- (1)
+ 6y = –4 ------- (2)
(1) × 3:  3x + 6y = 12 -------- (3)
   x + 6y = –4 ------- (2)
(3) – (2): 2= 12 – (–4)
2x = 16
  x = 8

Leave a Reply

Your email address will not be published. Required fields are marked *