Class 7 maths chapter 12 exercise 12.3 solutions

NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.3 Algebraic Expressions We need to identify the value of an expression in a variety of situations, such as when we want to see if a specific value of a variable satisfies a given equation or not. Thus, the NCERT solutions for Algebraic expressions help us find the value of unknown variables, and we know that this value depends on the variables that are forming that expression. The questions of ☛ Download NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.3 Exercise 12.3 Class 7 Chapter 12 Download PDF More Exercises in Class 7 Maths Chapter 12 • • • NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.3 Tips The students should learn how an expression is evaluated with the help of examples in the The substitution method is simple and would require focus from the students in order to put the correct values and further solve the calculations. Download Cuemath NCERT Solutions PDFs for free and start learning! NCERT Solutions for Class 7 Maths Video Chapter 12 Exercise 12.3 NCERT Class 7 maths Videos Chapter 12 Exercise 12.3

Class 7 Maths NCERT Solutions Chapter 12 – Algebraic Expressions comprises of the 4 Exercises • • • • This Chapter contains the Exercises relating to the following topics , which are discussed in Chapter 12 – Algebraic Expressions Class 7 NCERT book : – • 12.1 INTRODUCTION • 12.2 HOW ARE EXPRESSIONS FORMED? • 12.3 TERMS OF AN EXPRESSION • 12.4 LIKE AND UNLIKE TERMS • 12.5 MONOMIALS, BINOMIALS, TRINOMIALS AND POLYNOMIALS • 12.6 ADDITION AND SUBTRACTION OF ALGEBRAIC EXPRESSIONS • 12.7 FINDING THE VALUE OF AN EXPRESSION • 12.8 USING ALGEBRAIC EXPRESSIONS – FORMULAS AND RULES NCERT Solutions for Class 7 Maths Chapter 12 Exercise 12.1 1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations. a) Subtraction of z from y b) One-half of the sum of numbers x and y c) The number z multiplied by itself d) One-fourth of the product of numbers p and q e) Number x and y both squared and added f) Number 5 added to three times the product of numbers m and n g) Product of numbers y and z subtracted from 10 h) Sum of numbers a and b subtracted from their product Solution: a) y – z b) 1/2(x + y) c) z 2 d) (1/4)pq e) x 2 + y 2 f) 5 + 3mn g) 10 – yz h) ab – (a + b) 2. Write four more rational numbers in each of the following patterns: i. Identify the terms and their factors in the following expressions Show the terms and factors by tree diagrams. a) x – 3 b) 1 + x + x 2 c) y – y 3 d) 5xy 2 + 7x 2y e) -ab + 2b 2 – 3a 2 ii. Identify terms and fa...

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions * According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10. NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions are provided here. Students can check for the solutions whenever they are facing difficulty while solving the questions from The • How Are Expressions Formed • Terms of an Expression • Like and Unlike Terms • Monomials, Binomials, Trinomials and Polynomials • Addition and Subtraction of Algebraic Expressions • Finding the Value of An Expression • Using Algebraic Expressions – Formulas and Rules Download PDF carouselExampleControls112 Previous Next Access Exercises of NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Access answers to Maths NCERT Solutions for Class 7 Chapter 12 – Algebraic Expressions Exercise 12.1 Page: 234 1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations. (i) Subtraction of z from y. Solution:- = Y – z (ii) One-half of the sum of numbers x and y. Solution:- = ½ (x + y) = (x + y)/2 (iii) The number z multiplied by itself. Solution:- = z × z = z 2 (iv) One-fourth of the product of numbers p and q. Solution:- = ¼ (p × q) = pq/4 (v) Numbers x and y, both squared and added. Solution:- = x 2 + y 2 (vi) Number 5 added to three times the product of numbers m and n. Solution:- = 3mn + 5 (vii) Product of numbers y and z subtracted from 10. Solution:- = 10 –...

(ii) 3m – 5 = 3 (2) – 5 (m = – 2) = 6 – 5 = 1 (iii) 9 – 5m = 9 – 5(2) (m = = 2) = 9- 10 = -1 (iv) 3m 2– 2m – 7 = 3 (2) 2– 2(2) – 7 (m = 2) = 3 (4) – 4 – 7 = 12 – 4 – 7 = 12 – 11 = 1 (v) \(\frac \) – 4 (m = 2) = 5 – 4 = 1 Question 2. If p = -2, find the value of: (i) 4p + 7 (ii) – 3p 2 + 4p + 7 (iii) – 2p 2– 3p 2 + 4p + 7 Answer: (i) 4p + 7 = 4(-2) + 7 = – 8 + 7 = -1 (ii) -3p 2 + 4p + 7 = – 3 (-2) 2 + 4(-2) + 7 (when p = -2) = – 3 (4) + (- 8) + 7 = -12 – 8 + 7 = -20 +7 = -13 (iii) -2p 3– 3p 2 + 4p + 7 = -2 (-2) 3– 3 (-2) 2 + 4 (-2) + 7 (when p = -2) = – 2(- 8) — 3 (4) + (— 8) + 7 = + 16 – 12 – 8 + 7 = 23 – 20 = 3 Question 3. Find the value of the following expressions, when x = -1: (i) 2x – 7 (ii) -x + 2 (iii) x 2 + 2x + 1 (iv) 2x 2– x -2 Answer: (i) 2x – 7 = 2(-1) – 7 (whenx = -1) = -2 – 7 = -9 (ii) -x + 2 = -(-1) + 2 (When x = -1) = 1 + 2 = 3 (iii) x 2 + 2x + 1 = (-1) 2 + 2 (-1) + 1 (When x = – 1) = 1 – 2 + 1 = 2 – 2 = 0 (iv) 2x 2– x – 2 =2 (-1) 2– (-1) – 2 (When x = -1) = 2 + 1 – 2 = 3 – 2 = 1 Question 4. If a = 2, b = – 2, find the value of: (i) a 2 + b 2 (ii) a 2 + ab + b 2 (iii) a 2– b 2 Answer: (i) a 2 + b 2 = (2) 2 + (-2) 2 (When a = 2, b = -2) = 4 + 4 = 8 (ii) a 2 + ab + b 2 = (2) 2 + 2(-2) + (-2) 2 (When a = 2, b = – 2) = 4 – 4 + 4 = 8 – 4 = 4 (iii) a2 – b2 = 22 – (-2)2 (When a = 2, b = -2) = 4-(4) = 4- 4 = 0 Question 5. When a = 0, b = – 1, find the value of the given expressions: (i) 2a + 2b (ii) 2a 2 + b 2 + 1 (iii) 2a 2b + 2ab 2 + ab (iv) a 2 + ab + 2 Answer: ...

Q.1. If m = 2, find the value of: (i) m – 2 (ii) 3m – 5 (iii) 9 – 5m (iv) 3m 2 – 2m – 7 (v) $\frac-4$ ⟹ 5 – 4 =1 Q.2. If p = – 2, find the value of: (i) 4p + 7 (ii) – 3p 2 + 4p + 7 (iii) – 2p 3 – 3p 2 + 4p + 7 Ans: (i) 4p + 7 Putting p = -2 ,we get ⟹ 4p + 7 ⟹ 4(-2) + 7 = -8 + 7 = -1 (ii) – 3p 2 + 4p + 7 Putting p = -2 ,we get ⟹ – 3p 2 + 4p + 7 ⟹ – 3(-2) 2 + 4(-2) + 7 ⟹ – 3(-2 x -2) + 4(-2) + 7 ⟹ – 3(4) – 8 + 7 ⟹ -12 – 8 + 7 ⟹ -20 + 7 = – 13 (iii) – 2p 3 – 3p 2 + 4p + 7 Putting p = -2 ,we get ⟹ – 2p 3 – 3p 2 + 4p + 7 ⟹ – 2(-2) 3 – 3(-2) 2 + 4(-2) + 7 ⟹ – 2(-2 x -2 x -2) – 3(-2 x -2) + 4(-2) + 7 ⟹ – 2(-8) – 3(4) + 4(-2) + 7 ⟹ 16 – 12 – 8 + 7 ⟹ 16 + 7 – 12 – 8 ⟹ 23 – 20 = 3 Q.3. Find the value of the following expressions, when x = –1: (i) 2x – 7 (ii) – x + 2 (iii) x 2 + 2x + 1 (iv) 2x 2 – x – 2 Ans: (i) 2x – 7 Putting x = -1 , we get ⟹ 2x – 7 ⟹ 2(-1) – 7 ⟹ – 2 – 7 = – 9 (ii) – x + 2 Putting x = -1 , we get ⟹ – x + 2 ⟹ – (-1) + 2 = 1 + 2 = 3 (iii) x 2 + 2x + 1 Putting x = -1 , we get ⟹ x 2 + 2x + 1 ⟹ (-1) 2 + 2(-1) + 1 ⟹ (-1) (-1) + 2(-1) + 1 ⟹ 1 – 2 + 1 = 0 (iv) 2x 2 – x – 2 Putting x = -1 , we get ⟹ 2x 2 – x – 2 ⟹ 2(-1) 2 – (-1) – 2 ⟹ 2(-1 x -1) – (-1) – 2 ⟹ 2(1) + 1 – 2 ⟹ 2 + 1 – 2 ⟹ 3 – 2 = 1 Q.4. If a = 2, b = – 2, find the value of: (i) a 2 + b 2 (ii) a 2 + ab + b 2 (iii) a 2 – b 2 Ans: (i) a 2 + b 2 Putting a = 2 and b = -2, we get ⟹ a 2 + b 2 ⟹ (2) 2 + (-2) 2 ⟹ (2 x 2) + (-2 x -2) ⟹ 4 + 4 = 8 (ii) a 2 + ab + b 2 Putting a = 2 and b = -2, we get ⟹ a 2 + ab + b 2 ⟹ (2) ...

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ii) 3m – 5 3 × 2 – 5 = 6 – 5 = 1 iii) 9 – 5m 9 – 5 × 2 ÷ 9 – 10 = -1 iv) 3m 2– 2m – 7 = 3 × 2 2– 2 × 2 – 7 = 3 × 4 – 4 – 7 = 12 – 11 = 1 v) Solution: Question 2. If p = -2, find the value of : Solution: i) 4p + 7 = 4x – 2 + 7 = -8 + 7 = -1 ii) -3p 2 + 4p + 7 = -3 (-2) 2 + 4 (-2) + 7 = 3 × 4 – 8 + 7 = 12 – 8 + 7 = -13 iii) -2p 3– 3p 2– 4p + 7 = -2 (-2) 3– 3(2) 2 + 4 (-2) + 7 = -2(-8) – 3(4) – 8 + 7 = 16 – 12 – 8 + 7 = 16 + 7 – 12 – 8 = 23 – 20 = 3 Question 3. Find the value of the following expressions, when x = -1: i) 2x – 7 = 2(-1) – 7 = -2 – 7 = -9 ii) -x + 2 = -1 + 2 = 1 iii) x 2 + 2x + 1 = 1 – 2 + 1 = 0 iv) 2x 2– x – 2 = 2(-1) 2– (-1) – 2 = 2 × 1 + 1 – 2 = 2 + 1 – 2 = 1 Question 4. If a = 2, b = -2, find the value of: i) a 2 + b 2 = 2 2 + (-2) 2 = 4 + 4 = 8 ii) a 2 + ab + b 2 = 2 2 + 2 × -2 + (-2) = 4 – 4 + 4 = 4 iii) a 2– b 2 = (2) 2– (-2) 2 = 4 – 4 = 0 Question 5. When a = 0, b = -1, find the value of the given expressions : i) 2a + 2b = 2 × 0 + 2 × -1 = o – 2 = -2 ii) 2a 2 + 2b 2 + 1 = 2(0) 2 + (-1) 2 + 1 = 0 + 1 + 1 = 2 iii) 2a 2b + 2ab 2 + ab = 2 × 0 2 × -1 + 2 × 0 × (-1) 2 + 0 × -1 = 0 + 0 + 0 = 0 iv) a 2 + ab + 2 = (0) 2 + 0 × -1 + 2 = 0 + 0 + 2 = 2 Question 6. Simplify the expressions and find the value if x is equal to 2 Solution: i) x + 7 + 4 (x – 5) = x + 7 + 4x – 20 (re-arranging the terms) = 5x – 13 = 5(2) – 12 = 10 – 13 = -3 ii) 3(x + 2) + 5x – 7 = 3x + 6 + 5x – 1 = 8x – 1 = 8(2) – 1 = 16 – 1 = 15 iii) 6x + 5 + 5(x – 2) = 6x+ 5x – 10 = 11x – 10 = 11(2) – ...