RD Sharma Solutions for Class 7 Maths Exercise 7.3 Chapter 7 Algebraic Expressions

RD Sharma Solutions for Class 7 Maths Exercise 7.3 of Chapter 7 Algebraic Expressions are provided here. The PDF of this exercise can be easily downloaded from the below links. BYJU’S subject-matter experts have solved the exercise for students’ better understanding. By practising RD Sharma Solutions for Class 7, students can secure good marks in the annual exam. This exercise includes the use of grouping symbols or brackets in writing algebraic expressions. Students have already learned about the removal of brackets in Chapter 1; in the same way, they will perform with algebraic expressions.

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Access Answers to Maths RD Sharma Solutions for Class 7 Chapter 7 – Algebraic Expressions Exercise 7.3

1. Place the last two terms of the following expressions in parentheses preceded by a minus sign:

(i) x + y – 3z + y    

(ii) 3x – 2y – 5z – 4

(iii) 3a – 2b + 4c – 5

(iv) 7a + 3b + 2c + 4

(v) 2a² – b² – 3ab + 6

(vi) a² + b² – c² + ab – 3ac

Solution:

(i) Given x + y – 3z + y

x + y – 3z + y = x + y – (3z – y)

(ii) Given 3x – 2y – 5z – 4

3x – 2y – 5z – 4 = 3x – 2y – (5z + 4)

(iii) Given 3a – 2b + 4c – 5

3a – 2b + 4c – 5 = 3a – 2b – (–4c + 5)

(iv) Given 7a + 3b + 2c + 4

7a + 3b + 2c + 4 = 7a + 3b – (–2c – 4)

(v) Given 2a² – b² – 3ab + 6

2a² – b² – 3ab + 6 = 2a² – b² – (3ab – 6)

(vi) Given a² + b² – c² + ab – 3ac

a² + b² – c² + ab – 3ac = a² + b² – c² – (- ab + 3ac)

2. Write each of the following statements by using appropriate grouping symbols:

(i) The sum of a – b and 3a – 2b + 5 is subtracted from 4a + 2b – 7.

(ii) Three times the sum of 2x + y – [5 – (x – 3y)] and 7x – 4y + 3 is subtracted from 3x – 4y + 7

(iii) The subtraction of x² – y² + 4xy from 2x² + y² – 3xy is added to 9x² – 3y²– xy.

Solution:

(i) Given the sum of a – b and 3a – 2b + 5 = [(a – b) + (3a – 2b + 5)].

This is subtracted from 4a + 2b – 7.

Thus, the required expression is (4a + 2b – 7) – [(a – b) + (3a – 2b + 5)]

(ii) Given three times the sum of 2x + y – {5 – (x – 3y)} and 7x – 4y + 3 = 3[(2x + y – {5 – (x – 3y)}) + (7x – 4y + 3)]

This is subtracted from 3x – 4y + 7.

Thus, the required expression is (3x – 4y + 7) – 3[(2x + y – {5 – (x – 3y)}) + (7x – 4y + 3)]

(iii) Given the product of subtraction of x²– y² + 4xy from 2x² + y² – 3xy is given by {(2x² + y² – 3xy) – (x²-y² + 4xy)}

When the above equation is added to 9x² – 3y² – xy, we get

{(2x² + y² – 3xy) – (x² – y² + 4xy)} + (9x² – 3y²– xy))

This is the required expression.