AP State Syllabus AP Board 7th Class Maths Solutions Chapter 3 Simple Equations Ex 1 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 1

Question 1.
Write L.H.S and R.H.S of the following simple equations.
(i) 2x = 10
(ii) 2x – 3 = 9
(iii) 4z + 1 = 8
(iv) 5p + 3 = 2p + 9
(v) 14 = 27 – y
(vi) 2a – 3 = 5
(vii) 7m = 14
(viii) 8 = q + 5
Solution:

 Problem / Equation L.H.S R.H.S (i) 2x= 10 2x 10 (ii) 2x – 3 = 9 2x – 3 9 (iii) 4z + 1 = 8 4z + 1 8 (iv) 5p + 3 = 2p + 9 5p + 3 2p + 9 (v) 14 = 27 – y 14 27 – y (vi) 2a – 3 = 5 2a-3 5 (vii) 7m = 14 7m 14 (viii) 8 = q + 5 8 q + 5

Question 2.
Solve the following equations by trial and error method.
(i) 2 + y = 7
(ii) a – 2 = 6
(iii) 5m = 15
(iv) 2n = 14
Solution:
(i) 2 + y = 7
if y = 1; LHS = 2 + y = 2 + 1 = 3 ≠ 7
If y = 2; LHS = 2 + 2 = 4 ≠ 7
If y = 3; LHS = 2 + 3 = 5 ≠ 7
If y = 4; LHS = 2 + 4= 6 ≠ 7
If y = 5; LHS = 2 + 5 = 7 = 7 = RHS
∴ y = 5 is the solution of 2 + y = 7

(ii) a – 2 = 6
As a – 2 = 6 ; the value of ‘a’ must be greater than 6.
If a = 7; LHS = 7 – 2 = 5 ≠ 6
If a = 8; LHS = 8 – 2 = 6 = RHS
∴ a = 8 is the solution of a – 2 = 6

(iii) 5m = 15
If m = 1 then LHS = 5 × 1 = 5 ≠ b15
m = 2 then LHS = 5 × 2 = 10 ≠ 15
m = 3 then LHS = 5 × 3 = 15 = 15
∴ m = 3 is the solution of 5m = 15

(iv) 2n = 14
If n = 1 then LHS = 2 × 1 = 2 ≠ 14
n = 2 then LHS = 2 × 2 = 4 ≠ 14
n = 3 then LHS = 2 × 3 = 6 ≠ 14
n = 4 then LHS = 2 × 4 = 8 ≠ 14
n = 5 then LHS = 2 × 5 = 10 ≠ 14
n = 6 then LHS = 2 × 6 = 12 ≠ 14
n = 7 then LHS = 2 × 7 = 14 ≠ 14
∴ n = 7 is the solution of 2n = 14