Students can Download Maths Chapter 3 Algebra Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Question 1.

Fill in the blanks

(i) The addition of – 7b and 2b is _______

(ii) The subtraction of 5m from -3m is ______

(iii) The additive inverse of -37xyz is _____

Solution:

(i) -5b

(ii) -8m

(iii) 37xyz

Question 2.

Say True or False

(i) The expressions 8x + 3y and 7x + 2y cannot be added

(ii) If x is a natural number, then x + 1 is its predecessor.

Hint: x – 1 is its predecessor.

(iii) Sum of a – b + c and -a + b – c is zero

Solution:

(i) False

(ii) False

(iii) True

Question 3.

Add: (i) 8x, 3x

(ii) 7mn, 5mn

(iii) -9y, 11y, 2y

Solution:

(i) 8x + 3x = (8 + 3) x = 11x

(ii) 7mn + 5mn = (7 + 5)mn = 12mn

(iii) -9y + 11y + 2y =(-9 + 11 + 2 )y = (2 + 2)y = 4y

Question 4.

Subtract:

(i) 4k from 12k

(ii) 15q from 25q

(iii) 7xyz from 17xyz

Solution:

(i) 4k from 12k

12k – 4k = (12 – 4) k = 8k

(ii) 15q from 25q

25q – 15q = (25 – 15)q = 10q

(iii) 7xyz from 17xyz

17xyz – 7xyz = (17 – 7)xyz = 10xyz

Question 5.

Find the sum of the following expressions

(i) 7p + 6q, 5p – q, q + 16p

Solution:

(7p + 6q) + (5p – q) + (q + 16p) = 7p + 6q + 5p – q + q + 16p

= (7p + 5p + 16p) + (6q – q + q)

= (7 + 5 + 16) p + (6 – 1 + 1) q

= (12 + 16) p + 6q = 28p + 6q

(ii) a + 5b + 7c, 2a + 106 + 9c

Solution:

(a + 5b + 7c) + (2a + 10b + 9c) = a + 5b + 7c + 2a + 10b + 9c

= a + 2a + 5b + 10b + 7c + 9c

= (1 + 2)a + (5 + 10)b + (7 + 9)c

= 3a + 15b + 16c

(iii) mn + t, 2mn – 2t, – 3t + 3mn

Solution:

(mn + t) + (2mn – 2t) + (-3t + 3mn)

= mn + t + 2mn – 2t + (-3t) + 3mn

= (mn + 2mn + 3mn) + (t – 2t – 3t)

= (1 + 2 + 3) mn + (1 – 2 – 3) t

= 6mn + (1 – 5)t

= 6mn + (- 4) t

= 6mn – 4t

(iv) u + v, u – v, 2u + 5v, 2u – 5v

Solution:

(u + v) + (u – v) + (2u + 5v) + (2u – 5v)

= u + v + u – v + 2u + 5v + 2u – 5v

= u + u + 2u + 2u + v – v + 5v – 5v

= (1 + 1 + 2 + 2) u +(1 – 1 + 5 – 5)v = 6u + 0v

= 6u

(v) 5xyz – 3xy, 3zxy – 5yx

Solution:

5xyz – 3xy + 3zxy – 5yx = 5xyz + 3xyz – 3xy – 5xy

= (5 + 3) xyz + [(-3) + (-5)] xy = 8xyz + (-8) xy

= 8xyz – 8xy

Question 6.

Subtract

(i) 13x + 12y – 5 from 27x + 5y – 43

Solution:

27x + 5y – 43 – (13x + 12y – 5) = 27z + 5y – 43 + (-13x – 12y + 5)

= 27x + 5y – 43 – 13x – 12y + 5

= (27 – 13) x + (5 – 12)y + (- 43) + 5

= 14x + (- 7) y + (- 38) = 14x – 7y – 38

(ii) 3p + 5 from p – 2q + 7

Solution:

p – 2q + 7 – (3p + 5) = p – 2q + 7 + (- 3p – 5)

= p – 2q + 7 – 3p – 5 = p – 3p – 2q + 7 – 5

= (1 – 3)p – 2q + 2 = -2p – 2q + 2

(iii) m + n from 3m – 7n

Solution:

3m – 7n – (m + n) = 3m – 7n + (-m – n)

= 3m – 7n – m – n = (3m – m) + (-7n – n)

= (3 – 1 )m + (-7 – 1) n = 2m + (-8) n

= 2m – 8n

(iv) 2y + z from 6z – 5y

Solution:

6z – 5y – (2y + z) = 6z – 5y + (-2y – z)

= 6z – 5y – 2y – z = 6z – z – 5y – 2y

= (6 – 1) z + (-5 -2) y = 5z + (-7) y

= 5z – 7y = -7y + 5z

Question 7.

Simplify

(i) (x + y – z) + (3x – 5y + 7z) – (14x + 7y – 6z)

Solution:

(x + y – z) + (3x – 5y + 7z) – (14x – 7y – 6z)

= (x + y – z) + (3x – 5y + 7z) + (-14x – 7y + 6z)

= (x + 3x – 14x) + (y – 5y – 7y) + (-z + 7z + 6z)

= (1 + 3 – 14) x + (1 – 5 – 7)y + (-1 + 7 + 6) z

= – 10x – 11y + 12z

(ii) p + p + 2 + p + 3 + p – 4 – p – 5 + p + 10

Solution:

p + p + 2 + 3 – p – 4 – p – 5 + p + 10 = (p + p + p – p – p + p) + (2 + 3 – 4 – 5 + 10)

= (1 + 1 + 1 – 1 – 1 + 1) p + 6 = 2p + 6

(iii) n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)

Solution:

n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)

= n + m + 1 + n + 2 + m + 3 + n + 4 + m + 5

= n + n + n + m + m + m + 1 + 2 + 3 + 4 + 5

= (1 + 1 + 1)n + (1 + 1 + 1)m + 15

= 3n + 3m + 15 = 3m + 3n + 15

Objective Type Questions

Question 8.

The addition of 3mn, -5mn, 8mn and – 4mn is

(i) mn

(ii) – mn

(iii) 2mn

(iv) 3mn

Solution:

(iii) 2mn

Hint: = 3 mn + 8mn – 5 mn – 4 mn = 11 mn – 9 mn = 2 mn

Question 9.

When we subtract ‘a’ from ‘-a’, we get ______

(i) a

(ii) 2a

(iii) -2a

(iv) -a

Solution:

(iii) -2a

Hint: – a – a = – 2a

Question 10.

In an expression, we can add or subtract only _____

(i) like terms

(ii) unlike terms

(iii) all terms

(iv) None of the above

Solution:

(i) like terms