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