RBSE Solutions for Class 7 Maths Chapter 13 Algebraic Expression Ex 13.3

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

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