RBSE Solutions for Class 7 Maths Chapter 13 Algebraic Expression Ex 13.3 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 13 Algebraic Expression Exercise 13.3.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 7 |
| Subject | Maths |
| Chapter | Chapter 13 |
| Chapter Name | Algebraic Expression |
| Exercise | Ex 13.3 |
| Number of Questions | 4 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 7 Maths Chapter 13 Algebraic Expression Ex 13.3
Question 1
Find the value (RBSESolutions.com) of following if x = 2.
(i) – 3
(ii) 2x – 5
(iii) 9 – 6x
(iv) 3x2 – 4x – 7
(v) \(\frac { 5x }{ 2 }\) – 4
Solution:
(i) In x – 3 put x = 2
x – 3 = 2 – 3 = -1
(ii) In 2x – 5 put x = 2
2x – 5 = 2 x 2 – 5
= 4 – 5 = -1
(iii) In 9 – 6x put x = 2
9 – 6x = 9 – 6 x 2
= 9 – 12 = -3
(iv) In 3x2 – 4x – 7 put x = 2
3x2 – 4x – 7 = 3 x 2 x 2 – 4 x 2 – 7
= 12 – 8 – 7
= 12 – 15 = -3
(v) In \(\frac { 5x }{ 2 }\) – 4 put x = 2
\(\frac { 5x }{ 2 }\) – 4 = \(\frac { 5\times 2}{ 2 }\) – 4
= 5 – 4 = 1
Question 2
Find the value of (RBSESolutions.com) following, if p = – 1.
(i) 4p + 5
(ii) -3p2 + 4p + 8
(iii) 3(p – 2) + 6
Solution:
(i) In 4p + 5 put p =-1
4p + 5 = 4 x (-1) +5
= -4 + 5 = 1
(ii) In -3p2 + 4p +8 put p = -1
-3p2 + 4p + 8 = -3(-1)2+ 4(-1) + 8
= -3 x 1 – 4 + 8
=-3 – 4 + 8 = 1
(iii) In 3(p – 2) + 6 put p = -1
3(p – 2) + 6 = 3 (-1 – 2) + 6
= 3(-3) + 6 = -9 +6 = -3
Question 3
Find the value of (RBSESolutions.com) following if a = 2 and b = – 2.
(i) a2 – b2
(ii) a2 – ab + b2
(iii) a2 + b2
Solution:
(i) In (a2 – b2) put a = 2 and b = -2
a2 – b2 = (2)2 – (-2)
= 4 – 4 = 0
(ii) In (a2 – ab + b2) put a = 2 and b = -2
a2 – ab + b2 = (2)2 – 2x(-2) + (-2)2
= 4 + 4 + 4 = 12
(iii) In a2 + b2 put a = 2 and b = -2
a2 + b2 = (2)2 + (-2)2
= 4 + 4 = 8
Question 4
Find the value (RBSESolutions.com) of following if x = 1 and y = o.
(i) 2x + 2y
(ii) 2x2 + y2 +1
(iii) 2x2y + 2x2y2 +y2
(iv) x2 + xy +5
Solution:
(i) In (2x + 2) put x = 1 and y= 0
2x + 2y = 2 x 1 + 2 x 0
= 2+ 0 = 2
(ii) In (2x2 + y2 + 1) put x = 1 and y = 0
2x2 + y2 + 1 = 2(1)2 + (0)2 + 1
= 2 x 1 +0 + 1
= 2 + 1 = 3
(iii) In (2x2y+ 2x2y2 +2) put x = 1 and y = 0
2x2y + 2x2y2 + y2 = 2(1)2 x 0 + 2(1)2x (0)2+ (0)2
= 2 x 1 x 0 + 2 x 1 x 0 + 0
= 0 + 0 + 0 = 0
(iv) In (x2 + xy + 5) put x = 1 and y = 0
x2 + xy + 5 = (1)2 + 1 x 0 +5
= 1 + 0 + 5 = 6
We hope the RBSE Solutions for Class 7 Maths Chapter 13 Algebraic Expression Ex 13.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 13 Algebraic Expression Exercise 13.3, drop a comment below and we will get back to you at the earliest.