# RD Sharma Solutions Class 7 Algebraic Expressions Exercise 7.3

## RD Sharma Solutions Class 7 Chapter 7 Exercise 7.3

#### Exercise 7.3

Q1) 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) $2a^{2}-b^{2}-3ab+6$ (vi) $a^{2}+b^{2}-c^{2}+ab-3ac$

Solution:

We have

(i) x + y – 3z + y = x + y – (3z – y)

(ii) 3x – 2y – 5z – 4 = 3x – 2y – (5z + 4)

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

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

(v) $2a^{2}-b^{2}-3ab+6$ = $2a^{2}-b^{2}-(3ab-6)$

(vi) $a^{2}+b^{2}-c^{2}+ab-3ac$ = $a^{2}+b^{2}-c^{2}-(-ab+3ac)$

Q2) 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 $x^{2}-y^{2}+4xy$ from $2x^{2}+y^{2}-3xy$ is added to $9x^{2}-3y^{2}-xy$.

Solution:

(i) 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) 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) The product of subtraction of $x^{2}-y^{2}+4xy$ from $2x^{2}+y^{2}-3xy$ is given by {($2x^{2}+y^{2}-3xy$) – ($x^{2}-y^{2}+4xy$)}

When the above equation is added to $9x^{2}-3y^{2}-xy$, we get

{($2x^{2}+y^{2}-3xy$) – ($x^{2}-y^{2}+4xy$)}+( $9x^{2}-3y^{2}-xy$))