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RBSE Solutions for Class 7 Maths Chapter 14 सरल समीकरण Ex 14.1

March 16, 2019 by Fazal Leave a Comment

RBSE Solutions for Class 7 Maths Chapter 14 सरल समीकरण Ex 14.1 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 14 सरल समीकरण Exercise 14.1.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 7
Subject Maths
Chapter Chapter 14
Chapter Name सरल समीकरण
Exercise Ex 14.1
Number of Questions 1
Category RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 14 सरल समीकरण Ex 14.1

प्रश्न
नीचे दिए गए समीकरण हल कीजिए एवं उत्तर की जाँच कीजिए।
1. 2x + 1 = 9
2. 5x – 4 = 26
3. 5x – 2x + 7 = 31
4. 5x + 8 = 12 + 6
5. 12x + 3x = 60
6. \(\frac { 7x }{ 9 }\) = 21
7. \(\frac { 2x }{ 3 }\) – \(\frac { x }{ 2 }\) = 3
8. \(\frac { 3x }{ 4 }\) – \(\frac { 2x }{ 5 }\) = 7
9. \(\frac { 7x+1 }{ 2 }\) +1=11
10. \(\frac { 3l }{ 2 }\) = \(\frac { 2 }{ 3 }\)
11. 7m + \(\frac { 19 }{ 2 }\) = 13
12. 6z + 10 = -2
13. \(\frac { q }{ 4 }\) + 7 = 5
14. 4(2 – x) = 8
15. 3(n – 5) = 21
16. 4 = 5(t – 2)
17. 0 = 16+4(m – 6)
हल:
1. 2x + 1 = 9
⇒ 2x = 9 – 1 = 8 (1 का RHS में पक्षान्तर करनेपर)
⇒ \(\frac { 2x }{ 2 }\) = \(\frac { 8 }{ 2 }\) ⇒ x = 4
उत्तर की जाँच LHS = 2x +1
= 2 x 4 + 1
= 8 + 1
= 9 = RHS
अतः x = 4 सही है।

RBSE Solutions

2. 5x – 4 = 26
⇒ 5x = 26 + 4 (पक्षान्तर करने पर)
⇒ 5x = 30
⇒ \(\frac { 5x }{ 5 }\) = \(\frac { 30 }{ 5 }\) (दोनों पक्षों में 5 का भाग दने पर)
⇒ x = 6
उत्तर की जाँच
LHS = 5x – 4
= 5 x 6 – 4 = 30 – 4
= 26 = RHS
अतः = 6 सही है।

3. 5x – 2x + 7 = 31
3x + 7 = 31
⇒ 3x = 31 – 7 (पक्षान्तर करने पर)
3x = 24
⇒ \(\frac { 3x }{ 3 }\) = \(\frac { 24 }{ 3 }\) (दोनों पक्षों में 3 का भाग दने पर)
x = 8
उत्तर की जाँच
LHS = 5x – 2x + 7
= 5 x 8 – 2 x 8 + 7
= 40 – 16 +7 = 31 = RHS
अतः = 8 सही है।

RBSE Solutions

4. 5x + 8 = 12 + 6
⇒ 5x = 18 – 8 = 10 (8 का पक्षान्तर करने पर)
⇒ \(\frac { 5x }{ 5 }\) = \(\frac { 10 }{ 5 }\)
(दोनों पक्षों में 5 का भाग देने पर)
⇒ x = 2
उत्तर की जाँच
LHS = 5x + 8
= 5 x 2 + 8 = 18
RHS = 12 + 6 = 18
अतः x = 2 सही है।

5. 12x + 3x = 60
⇒ 15 = 60
⇒ \(\frac { 15x }{ 15 }\) = \(\frac { 60 }{ 15 }\) = 4 (15 का भाग दोनों पक्षों में भाग देने पर)
⇒ x = 4
उत्तर की जाँच
LHS = 12x + 3x = 15x
= 15 x 4 = 60 = RHS
अतः x = 4 सही है।

RBSE Solutions for Class 7 Maths Chapter 14 सरल समीकरण Ex 14.
RBSE Solutions for Class 7 Maths Chapter 14 सरल समीकरण Ex 14.
RBSE Solutions for Class 7 Maths Chapter 14 सरल समीकरण Ex 14.
RBSE Solutions for Class 7 Maths Chapter 14 सरल समीकरण Ex 14.

RBSE Solutions

12. 6z + 10 = -2
⇒ 6z = -2 – 10
⇒ 6z = -12
z = – \(\frac { 12 }{ 6 }\) = -2
उत्तर की जाँच
LHS = 6z + 10
= 6 x (-2) + 10
= -12 + 10
= -2 = RHS
अत: z = -2 सही है।

13. \(\frac { q }{ 4 }\) + 7 = 5
⇒ \(\frac { q }{ 4 }\) = 5 – 7 = -2
q = -2 x 4 = -8
उत्तर की जाँच
LHS = \(\frac { q }{ 4 }\) + 7 = \(\frac { -8 }{ 4 }\) + 7
= -2 +7 = 5 = RHS
अतः q = -8 सही है।

14. 4(2 – x) = 8
⇒ 2 – x = \(\frac { 8 }{ 4 }\)
⇒ 2 – x = 2
⇒ x = 2 – 2 = 0
उत्तर की जाँच
LHS = 4(2 – x)
= 4(2 – 0) = 8 = RHS
अतः x = 0 सही है।

RBSE Solutions

15. 3(n -5) = 21
⇒ n – 5 = \(\frac { 21 }{ 3 }\) = 7
⇒ n = 7 + 5 = 12
उत्तर की जाँच
LHS = 3(n – 5)
= 3(12 – 5) = 3(7)
= 21 = RHS
अतः n = 12 सही है।

16. 4 = 5(t – 2)
RBSE Solutions for Class 7 Maths Chapter 14 सरल समीकरण Ex 14.1 Q5

17. 0 = 16 + 4(m – 6)
⇒ 16 + 4(m – 6) = 0
⇒ 4(m – 6) = -16
⇒ m – 6 = \(\frac { -16 }{ 4 }\) = -4
⇒ m = -4 + 6 = 2
उत्तर की जाँच
RHS = 16 + 4(m – 6)
= 16 + 4(2 – 6)
= 16 + 4(-4)
= 16 – 16 = 0 = LHS
अतः m = 2 सही है।

RBSE Solutions

We hope the RBSE Solutions for Class 7 Maths Chapter 14 सरल समीकरण Ex 14.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 14 सरल समीकरण Exercise 14.1, drop a comment below and we will get back to you at the earliest.

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