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

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## 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.
(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

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