Page 149 - ICSE Math 6

(b) y + 3 = 1 – y
⇒ y + y = 1 – 3 (Collecting variables on one side and
⇒ 2y = –2 constant terms on the other side)
2y –2
⇒ = (Dividing both the sides by 2)
2 2
⇒ y = –1

10x
(c) + 3 = 13
3
10x
⇒ = 13 – 3 (Transposing 3 to the other side)
3
10x
⇒ = 10
3
 10  x  10   10  10
⇒   ÷   = (10) ÷   (Dividing both the sides by )
 3   3   3  3
 10 3  3
⇒  ×  x = 10 ×
 3 10  10
⇒ x = 3

Verification of the solution

The solution of an equation can be verified by finding the values of both the sides of the equation
after substituting the value of the variable. If both the values are equal, then the solution is correct.

Example 4: Solve the following and verify the solution.
y 1
(a) 3(x – 1) = 6 (b) 4(x + 2) = 2x + 5 (c) +1 2=
7 7
Solution: (a) 3(x – 1) = 6
⇒ 3x – 3 = 6

⇒ 3x = 6 + 3 (Transposing 3 to the other side)
⇒ 3x = 9
3x 9
⇒ = (Dividing both the sides by 3)
3 3
⇒ x = 3
Verification: Substitute x = 3 in LHS
3(x – 1) = 3(3 – 1) = 3(2) = 6 = RHS
As LHS = RHS, therefore the solution x = 3 is correct.

(b) 4(x + 2) = 2x + 5
⇒ 4x + 8 = 2x + 5
⇒ 4x – 2x = 5 – 8 (Collecting variables on one side and
⇒ 2x = –3 constant terms on the other side)
2x –3
⇒ = (Dividing both the sides by 2)
2 2
–3
⇒ x =
2

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