The DAV Class 6 Maths Solutions and **DAV Class 6 Maths Chapter 6 Worksheet 6 **Solutions of Introduction to Algebra offer comprehensive answers to textbook questions.

## DAV Class 6 Maths Ch 6 WS 6 Solutions

Question 1.

Subtract:

(a) 3x from 5x

Solution:

3x from 5x = 5x – 3x = 2x

(b) 4a from 3a

Solution:

4a from 3a = 3a – 4a = – a

(c) 13x^{2} from x^{2}

Solution

13x^{2} from x^{2} = x^{2} – 13x^{2}

= – 12x^{2}

(d) – y from 3y

Solution:

– y from 3y = 3y – (- y)

= 3y + y

= 4y

(e) – 7a from – 3a

Solution:

– 7a from – 3a = – 3a (-7a)

= – 3a + 7a

= 4a

(f) 3abc from – 3 abc

Solution:

3abc from – 3abc = (- 3 abc) – (3abc)

= – 6abc

Question 2.

Subtract by column method:

(a) 3x + 3y from 2x + 3y

Solution:

(b) a^{2} + 2b^{2} from 3b^{2} – 4a^{2}

Solution:

(c) 5x^{2}y – 3xy^{2} from – 2x^{2}y + 4y^{2}

Solution:

(d) 2a – 3b + 5 from 5a – b – 2

Solution:

(e) p + q – r from – p – q – r

Solution:

= – 2p – 2q

(f) a^{3} + b^{3} – 3abc from – a^{3} + b^{3} + 3abc

Solution:

= – 2a^{3}

Question 3.

Subtract by horizontal method:

(a) 2x – 3y + 4z from 4x – 6y – z

Solution:

2x – 3y + 4z from 4x – 6y – z

= (4x – 6y – z) – (2x – 3y + 4z)

= 4x – 6y – z – 2x + 3y – 4z

= 2x – 3y – 5z

(b) 2x – 4y from – 4x + 3y

Solution:

2x – 4y from – 4x + 3y

= (- 4x + 3y) – (2x – 4y)

= – 4x + 3y – 2x + 4y

= – 6x + 7y

(c) 5q^{2} – p^{2} – 7r from 6p^{2} – q^{2} + 2r – 9

Solution:

5q^{2} – p^{2} – 7r from 6p^{2} – q^{2} + 2r – 9

(6p^{2} – q^{2} + 2r – 9) – (5q^{2} – p^{2} – 7r)

= 6p^{2} – q^{2} + 2r – 9 – 5q^{2} + p^{2} + 7r

= (6p^{2} + p^{2}) + (- q^{2} – 5q^{2}) + (2r + 7r) – 9

= 7p^{2} – 6q^{2} + 9r – 9

(d) – x – y from x + y

Solution:

– x – y from x + y

= (x + y) – (-x – y)

= – x + y + x + y

=> (x + x) + (y + y) = 2x + 2y

(e) 7pqr – 8p + 3 from 10 – 4pqr + 3p

Solution:

7pqr – 8p + 3 from 10 – 4pqr + 3p

= (10 – 4pqr + 3p) – (7pqr – 8p + 3)

= 10 – 4pqr + 3p – 7pqr + 8p – 3

= (10 – 3) + (- 4pqr – 7pqr) + (3p + 8p)

= 7 – 11pqr + 11p

(f) 5x^{3} + 3x^{2} – x – 3 from 5x^{3} – 3x^{2} + 2x – 3

Solution:

5x^{3} + 3x^{2} – x – 3 from 5x^{3} – 3x^{2} + 2x – 3

(5x^{2} – 3x^{2} + 2x – 3) – (5x^{3} + 3x^{2}2 – x – 3)

= 5x^{2} – 3x^{2} + 2x – 3 – 5x^{3} – 3x^{2} + x + 3

= – 5x^{3} + (5x^{2} – 3x^{2} – 3x^{2}) + (2x + x) + (- 3 + 3)

= – 5x^{3} – x^{2} + 3x + 0

= – 5x^{3} – x^{2} + 3x

Question 4.

Subtract 3x + 2y – 5xy from 0.

Solution:

0 – (3x + 2y – 5xy)

= 0 – 3x – 2y + 5x.y

= – 3x – 2y + 5 xy

Question 5.

From the sum of x^{2} + x + 1 and x^{2} – x + 1, subtract – x^{2} – x + 1.

Solution:

[(x^{2} + x + 1) + (x^{2} – x + 1)] – (- x^{2} – x + 1)

= (x^{2} + x + 1 + x^{2} – x + 1) – (- x^{2} – x + 1)

= x^{2} + x + 1 + x^{2} – x + 1 + x^{2} + x – 1

= (x^{2} + x^{2} + x^{2}) + (x – x + x) + (1 + 1 – 1)

= 3x^{2} + x + 1

Question 6.

Subtract the sum of 2a + 3b, a – 2b + c, – a + 2c and 4a + 2b – 5 from 6a – 4b + 8.

Solution:

First we find the sum of 2a + 3b, a – 2b + c, – a + 2c and 4a + 2b – 5.

= (2a + 3b) + (a – 2b + c) + (- a + 2c) + (4a + 26-5)

= 2a + 3b + a – 2b + c – a + 2c + 4a + 26-5

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

= 6a + 3b + 3c – 5

Now subtract 6a + 3b + 3c – 5 from 6a – 4b + 8

= 6a – 4b + 8

= – 7b – 3c + 13