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(d) The product of 4 and y is 4y. When 6 is subtracted from 4y, the difference is 60.

Therefore, the equation is 4y – 6 = 60.
1
3
(e) The sum of 3 and one-third of a is a + .
3 Maths Talk
The sum is 30. Therefore, the equation is How can you differentiate
1 a between an expression and
3
3
a + = 30 or += 30. simple equation? Discuss.
3 3
Example 4: Convert the following equations in statement form:
2p a − 3
(a) x + 3 = 11 (b) 2y – 7 = 13 (c) = 8 (d) = 4
3 2
Solution: (a) x + 3 = 11 ⇒ 3 added to a number x gives 11.
(b) 2y – 7 = 13 ⇒ 7 subtracted from the twice of y gives 13.

2p
(c) = 8 ⇒ The quotient of double of p and 3 is 8.
3 Think and Answer

a − 3 In a Maths unit test, the highest
(d) = 4 ⇒ The difference of a and 3 divided by
2 marks obtained by a student
2 gives 4 as quotient. in the class is 7 more than the
twice of the lowest marks. The
Solution of a Simple Equation highest score is 18. Make an
equation for the given statement.

Mother to Aarav: Let us find out Pihu didi’s present age.

Let us substitute the different values of the variable x which represents the Pihu’s present age one
by one in the equation 2x – 13 = 5.

Whether the Equation is satisfied.
Variable (x) LHS = 2x – 13 RHS = 5
Yes / No. (LHS = RHS)

1 2 × 1 – 13 = –11 5 –11 ≠ 5, No
2 2 × 2 – 13 = –9 5 –9 ≠ 5, No

3 2 × 3 – 13 = –7 5 –7 ≠ 5, No
4 2 × 4 – 13 = –5 5 –5 ≠ 5, No

5 2 × 5 – 13 = –3 5 –3 ≠ 5, No
6 2 × 6 – 13 = –1 5 –1 ≠ 5, No

7 2 × 7 – 13 = 1 5 1 ≠ 5, No

8 2 × 8 – 13 = 3 5 3 ≠ 5, No
9 2 × 9 – 13 = 5 5 5 = 5, Yes

From the above table, we observe that the value of expression in LHS is equal to RHS (= 5)
only when we substitute x = 9 in the given expression of LHS, that is in 2x – 13. For all other
values LHS ≠ RHS.

131 Simple Equations