Samacheer Class 7 Maths Algebra Intext Questions Book Back Answers & Solutions

Q.1Identify the variable and constants among the following terms. a, 11 – 3x, xy, -89, -m, -n, 5, 5ab, -5 3y, 8pqr, 18, -9t, -1, -8v

Solution

Variable : a, -3x, xy, -m, -n, 5ab, 3y, -9t, 8pqr
Constants : 11, -89, 5, -5, 18, -1, -8

Answer:

Variable : a, -3x, xy, -m, -n, 5ab, 3y, -9t, 8pqr
Constants : 11, -89, 5, -5, 18, -1, -8

Q.1Fill in the blanks (i) The variable in the expression 16x – 7 is _____ (ii) The constant term of the expression 2y – 6 is _____ (iii) In the expression 25m + 14M, the type of the terms are ______ terms (iv) The number of terms in the expression 3ab + 4c – 9 is _____ Hint: Terms are 3ab, 4c – 9. (v) The numerical co-efficient of the term -xy is ______ Hint: -x,y = (- 1 )xy.v

Solution

(i) x
(ii) -6
(iii) unlike
(iv) three
(v) -1

Answer:

(i) x
(ii) -6
(iii) unlike
(iv) three
(v) -1

Q.2Say true or False (i) x + (-x) = 0. (ii) The co-efficient of ab in the term 15 abc is 15. Hint: Coefficient of ab is 15c (iii) 2pq and – 7qp are like terms. (iv) When y = -1, the value of the expression 2y – 1 is 3. Hint: 2(-1) – 1 = -2 – 1 = – 3v

Solution

(i) True
(ii) False
(iii) True
(iv) False

Answer:

(i) True
(ii) False
(iii) True
(iv) False

Q.3Fing the numerical co-efficient of each of the following terms: -3yx, 12k, y, 121bc, -x, 9pq, 2ab.v

Solution

(i) Numerical co-efficient of-3yx is – 3
(ii) Numerical co-efficient of 12k is 12
(iii) Numerical coefficient of y is 1
(iv) Numerical co-efficient of 1216c is 121
(v) Numerical co-efficient of – x is – 1
(vi) Numerical co-efficient of 9pq is 9
(vii) Numerical co-efficient of 2ab is 2
SamacheerKalvi.Guru

Answer:

(i) Numerical co-efficient of-3yx is – 3
(ii) Numerical co-efficient of 12k is 12
(iii) Numerical coefficient of y is 1
(iv) Numerical co-efficient of 1216c is 121
(v) Numerical co-efficient of – x is – 1
(vi) Numerical co-efficient of 9pq is 9
(vii) Numerical co-efficient of 2ab is 2
SamacheerKalvi.Guru

Q.4Write the variables, constants and terms of the following expressions, (i) 18 + x – y (ii) 7p – 4q + 5 (iii) 29x + 13y (iv) b + 2v

Solution

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1 1

Answer:

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1 1

Q.5Identify the like terms among the following 7x, 5y, -8x, 12y, 6z, z, -12x, -9y, 11 zv

Solution

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1 2

Answer:

Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1 2

Q.6If x = 2 andy = 3, then find the value of the following expressions, (i) 2x – 3y (ii) x + y (iii) 4y – x (iv) x + 1 – yv

Solution

Given x = 2; y = 3.
(i) 2x – 3y = 2 (2) – 3 (3) = 4 – 9
= 4 + (Additive inverse of 9)
= 4 +(-9) = -5
(ii) x + y = 2 + 3 = 5
(iii) 4y – x = 4 (3) – 2 = 12 – 2 = 10
(iv) x + 1 – y = 2 + 1 – 3 = 3 – 3 = 0
Objective Type Questions

Answer:

Given x = 2; y = 3.
(i) 2x – 3y = 2 (2) – 3 (3) = 4 – 9
= 4 + (Additive inverse of 9)
= 4 +(-9) = -5
(ii) x + y = 2 + 3 = 5
(iii) 4y – x = 4 (3) – 2 = 12 – 2 = 10
(iv) x + 1 – y = 2 + 1 – 3 = 3 – 3 = 0
Objective Type Questions

Q.1An algebraic statement which is equivalent to the verbal statement “Three times the sum of ‘x’ and ‘y’ is (i) 3 (x + y) (ii) 3 + x + y (iii) 3x + y (iv) 3 + xyv

Solution

(i) 3 [(x + y)]

Answer:

(i) 3 [(x + y)]

Q.2The numerical co-efficient of -7mn is (i) 7 (ii) -7 (iii) p (iv) -pv

Solution

(ii) -7

Answer:

(ii) -7

Q.3Choose the pair of like terms (i) 7p, 7x (ii) 7r, 7x (iii) – 4x, 4 (iv) – 4x, 7xv

Solution

(iv) -4x, 7x
SamacheerKalvi.Guru

Answer:

(iv) -4x, 7x
SamacheerKalvi.Guru

Q.4The value of 7a – 4b when a = 3, b = 2 is (i) 21 (ii) 13 (iii) 8 (iv) 32v

Solution

(ii) 13
Hint: 7(3) – 4(2) = 21 – 8 = 13
About Us
Privacy Policy
Disclaimer
Contact Us
Facebook
Twitter
Pinterest
LinkedIn
Copyright © 2026 Samacheer Kalvi Guru

Answer:

(ii) 13
Hint: 7(3) – 4(2) = 21 – 8 = 13
About Us
Privacy Policy
Disclaimer
Contact Us
Facebook
Twitter
Pinterest
LinkedIn
Copyright © 2026 Samacheer Kalvi Guru

Q.1Fill 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 _____v

Solution

(i) -5b
(ii) -8m
(iii) 37xyz

Answer:

(i) -5b
(ii) -8m
(iii) 37xyz

Q.2Say 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 zerov

Solution

(i) False
(ii) False
(iii) True

Answer:

(i) False
(ii) False
(iii) True

Q.3Add: (i) 8x, 3x (ii) 7mn, 5mn (iii) -9y, 11y, 2yv

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

Answer:

(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

Q.4Subtract: (i) 4k from 12k (ii) 15q from 25q (iii) 7xyz from 17xyzv

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
SamacheerKalvi.Guru

Answer:

(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
SamacheerKalvi.Guru

Q.5Find the sum of the following expressions (i) 7p + 6q, 5p – q, q + 16pv

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

Answer:

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

Q.6Subtract (i) 13x + 12y – 5 from 27x + 5y – 43v

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
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
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
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

Answer:

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
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
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
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

Q.7Simplify (i) (x + y – z) + (3x – 5y + 7z) – (14x + 7y – 6z)v

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
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)
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

Answer:

(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
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)
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

Q.8The addition of 3mn, -5mn, 8mn and – 4mn is (i) mn (ii) – mn (iii) 2mn (iv) 3mnv

Solution

(iii) 2mn
Hint: = 3 mn + 8mn – 5 mn – 4 mn = 11 mn – 9 mn = 2 mn
SamacheerKalvi.Guru

Answer:

(iii) 2mn
Hint: = 3 mn + 8mn – 5 mn – 4 mn = 11 mn – 9 mn = 2 mn
SamacheerKalvi.Guru

Q.9When we subtract ‘a’ from ‘-a’, we get ______ (i) a (ii) 2a (iii) -2a (iv) -av

Solution

(iii) -2a
Hint: – a – a = – 2a

Answer:

(iii) -2a
Hint: – a – a = – 2a

Q.10In an expression, we can add or subtract only _____ (i) like terms (ii) unlike terms (iii) all terms (iv) None of the abovev

Solution

(i) like terms
About Us
Privacy Policy
Disclaimer
Contact Us
Facebook
Twitter
Pinterest
LinkedIn
Copyright © 2026 Samacheer Kalvi Guru

Answer:

(i) like terms
About Us
Privacy Policy
Disclaimer
Contact Us
Facebook
Twitter
Pinterest
LinkedIn
Copyright © 2026 Samacheer Kalvi Guru

Q.1Fill in the blanks. (i) An expressions equated to another expression is called _______. (ii) If a = 5, the value of 2a + 5 is _______. (iii) The sum of twice and four times of the variable x is ______.v

Solution

(i) an equation
(ii) 15
(iii) 6x
Question 2:
Say True or False
(i) Every algebraic expression is an equation.
(ii) The expression 7x + 1 cannot be reduced without knowing the value of x.
(iii) To add two like terms, its coefficients can be added.
(i) False
(ii) True
(iii) True

Answer:

(i) an equation
(ii) 15
(iii) 6x
Question 2:
Say True or False
(i) Every algebraic expression is an equation.
(ii) The expression 7x + 1 cannot be reduced without knowing the value of x.
(iii) To add two like terms, its coefficients can be added.
(i) False
(ii) True
(iii) True

Q.3Solve (i) x + 5 = 8 (ii) p – 3 = 1 (iii) 2x = 30 (iv) \(\frac{m}{6}\) = 5 (v) 7x + 10 = 80v

Solution

(i) Given x + 5 = 8 ; Subtracting 5 on both the sides
x + 5 – 5 = 8 – 5
x = 3
(ii) Given p – 3 = 7 ; Adding 3 on both the sides,
p – 3 + 3 = 7 + 3
p = 10
(iii) Given 2x = 30 ; Dividing both the sides by 2,
\(\frac{2 x}{2}=\frac{30}{2}\)
x = 15
(iv) Given \(\frac{m}{6}\) = 5 ; Multiplying both the sides by 6,
\(\frac{m}{6}\) × 6 = 5 × 6
m = 30
(v) Given 7x + 10 = 80 ; Subtracting 10 from both the sides,
7x + 10 – 10 = 80 – 10
7x = 70
Dividing both sides by 7,
\(\frac{7 x}{7}=\frac{70}{7}\)
x = 10

Answer:

(i) Given x + 5 = 8 ; Subtracting 5 on both the sides
x + 5 – 5 = 8 – 5
x = 3
(ii) Given p – 3 = 7 ; Adding 3 on both the sides,
p – 3 + 3 = 7 + 3
p = 10
(iii) Given 2x = 30 ; Dividing both the sides by 2,
\(\frac{2 x}{2}=\frac{30}{2}\)
x = 15
(iv) Given \(\frac{m}{6}\) = 5 ; Multiplying both the sides by 6,
\(\frac{m}{6}\) × 6 = 5 × 6
m = 30
(v) Given 7x + 10 = 80 ; Subtracting 10 from both the sides,
7x + 10 – 10 = 80 – 10
7x = 70
Dividing both sides by 7,
\(\frac{7 x}{7}=\frac{70}{7}\)
x = 10

Q.4What should be added to 3x + 6y to get 5x + 8y?v

Solution

To get the expression we should subtract 3x + 6y from 5x + 8y
5x + 8y – (3x + 6y) = 5x + 8y + (-3x – 6y)
= 5x + 8y – 3x – 6y = (5 – 3) x + (8 – 6) y
= 2x + 2y
So 2x + 2y should be added.
SamacheerKalvi.Guru

Answer:

To get the expression we should subtract 3x + 6y from 5x + 8y
5x + 8y – (3x + 6y) = 5x + 8y + (-3x – 6y)
= 5x + 8y – 3x – 6y = (5 – 3) x + (8 – 6) y
= 2x + 2y
So 2x + 2y should be added.
SamacheerKalvi.Guru

Q.5Nine added to thrice a whole number gives 45. Find the numberv

Solution

Let the whole number required be x.
Thrice the whole number = 3x
Nine added to it = 3x + 9
Given 3x + 9 = 45
3x + 9 – 9 = 45 – 9 [Subtracting 9 on both sides]
3x = 36
\(\frac{3 x}{3}=\frac{36}{3}\)
x = 12
∴ The required whole number is 12

Answer:

Let the whole number required be x.
Thrice the whole number = 3x
Nine added to it = 3x + 9
Given 3x + 9 = 45
3x + 9 – 9 = 45 – 9 [Subtracting 9 on both sides]
3x = 36
\(\frac{3 x}{3}=\frac{36}{3}\)
x = 12
∴ The required whole number is 12

Q.6Find the two consecutive odd numbers whose sum is 200v

Solution

Let the two consecutive odd numbers be x and x + 2
∴ Their sum = 200
x + (x + 2) = 200
x + x + 2 = 200
2x + 2 = 200
2x + 2 – 2 = 200 – 2 [∵ Subtracting 2 from both sides]
2x = 198
\(\frac{2 x}{2}=\frac{198}{2}\) [Dividing both sides by 2]
x = 99
The numbers will be 99 and 99 + 2.
∴ The numbers will be 99 and 101.

Answer:

Let the two consecutive odd numbers be x and x + 2
∴ Their sum = 200
x + (x + 2) = 200
x + x + 2 = 200
2x + 2 = 200
2x + 2 – 2 = 200 – 2 [∵ Subtracting 2 from both sides]
2x = 198
\(\frac{2 x}{2}=\frac{198}{2}\) [Dividing both sides by 2]
x = 99
The numbers will be 99 and 99 + 2.
∴ The numbers will be 99 and 101.

Q.7The taxi charges in a city comprise of a fixed charge of ₹ 100 for 5 kms and ₹ 16 per km for ever additional km. If the amount paid at the end of the trip was ₹ 740, find the distance traveled.v

Solution

Let the distance travelled by taxi be ‘x’ km
For the first 5 km the charge = ₹ 100
For additional kms the charge = ₹ 16(x – 5)
∴ For x kms the charge = 100 + 16(x – 5)
Amount paid = ₹ 740
∴ 100 + 16 (x – 5) = 740
100 + 16 (x – 5) – 100 = 740- 100
16 (x – 5) = 640
\(\frac{16(x-5)}{16}=\frac{640}{16}\)
x – 5 = 40
x – 5 + 5 = 45 + 5
x = 45
x = 45 km
∴ Total distance travelled = 45 km
Objective Type Questions

Answer:

Let the distance travelled by taxi be ‘x’ km
For the first 5 km the charge = ₹ 100
For additional kms the charge = ₹ 16(x – 5)
∴ For x kms the charge = 100 + 16(x – 5)
Amount paid = ₹ 740
∴ 100 + 16 (x – 5) = 740
100 + 16 (x – 5) – 100 = 740- 100
16 (x – 5) = 640
\(\frac{16(x-5)}{16}=\frac{640}{16}\)
x – 5 = 40
x – 5 + 5 = 45 + 5
x = 45
x = 45 km
∴ Total distance travelled = 45 km
Objective Type Questions

Q.8The generalization of the number pattern 3, 6, 9, 12, …………. is (i) n (ii) 2n (iii) 3n (iv) 4nv

Solution

(iii) 3n

Answer:

(iii) 3n

Q.9The solution of 3x + 5 = x + 9 is t (i) 2 (ii) 3 (iii) 5 (iv)4v

Solution

(i) 2
Hint: 3x + 5 = x + 9 ⇒ 3x – x = 9 – 5 ⇒ 2x = 4 ⇒ x = 2
SamacheerKalvi.Guru

Answer:

(i) 2
Hint: 3x + 5 = x + 9 ⇒ 3x – x = 9 – 5 ⇒ 2x = 4 ⇒ x = 2
SamacheerKalvi.Guru

Q.10The equation y + 1 = 0 is true only when y is (i) 0 (ii) -1 (iii) 1 (iv) – 2v

Solution

Answer:

Q.1Subtract – 3ab – 8 from 3ab – 8. Also subtract 3ab + 8 from -3ab – 8.v

Solution

Subtracting -3ab – 8 from 3ab + 8
= 3ab + 8 – (-3ab – 8) = 3ab + 8 + (3ab + 8)
= 3ab + 8 + 3ab + 8 = (3 + 3) ab + (8 + 8)
= 6ab + 16
Also subtracting 3 ab + 8 from – 3ab – 8
= – 3ab – 8 – (3ab + 8) = – 3ab – 8 + (-3ab – 8) = – 3ab – 8 – 3 ab – 8
= [(-3) + (- 3)] ab + [(-8) + (-8)] = – 6ab + (- 16)
= -6ab – 16

Answer:

Subtracting -3ab – 8 from 3ab + 8
= 3ab + 8 – (-3ab – 8) = 3ab + 8 + (3ab + 8)
= 3ab + 8 + 3ab + 8 = (3 + 3) ab + (8 + 8)
= 6ab + 16
Also subtracting 3 ab + 8 from – 3ab – 8
= – 3ab – 8 – (3ab + 8) = – 3ab – 8 + (-3ab – 8) = – 3ab – 8 – 3 ab – 8
= [(-3) + (- 3)] ab + [(-8) + (-8)] = – 6ab + (- 16)
= -6ab – 16

Q.2Find the perimeter of a triangle whose sides are x + 3y, 2x + y, x – y.v

Solution

Perimeter of a triangle = Sum of three sides
= (x + 3y) + (2x + y) + (x – y)
= x + 3y + 2x + y + x – y
= (1 + 2 + 1)x + (3 + 1 + (-1))y = 4x + 3y
∴ Perimeter of the triangle = 4x + 3y

Answer:

Perimeter of a triangle = Sum of three sides
= (x + 3y) + (2x + y) + (x – y)
= x + 3y + 2x + y + x – y
= (1 + 2 + 1)x + (3 + 1 + (-1))y = 4x + 3y
∴ Perimeter of the triangle = 4x + 3y

Q.3Thrice a number when increased by 5 gives 44. Find the number.v

Solution

Let the required number be x.
Thrice the number = 3x.
Thrice the number increased by 4 = 3x + 5
Given 3x + 5 = 44
3x + 5 – 5 = 44 – 5
3x = 39
\(\frac{3 x}{3}=\frac{39}{3}\)
x = 13
∴ The required number = 13

Answer:

Let the required number be x.
Thrice the number = 3x.
Thrice the number increased by 4 = 3x + 5
Given 3x + 5 = 44
3x + 5 – 5 = 44 – 5
3x = 39
\(\frac{3 x}{3}=\frac{39}{3}\)
x = 13
∴ The required number = 13

Q.4How much smaller is 2ab + 4b – c than 5ab – 3b + 2c.v

Solution

To find the answer we have to find the difference.
Here greater number 5ab – 3ab + 2c.
∴ Difference = 5ab – 3b + 2c – (2ab + 4b – c) = 5ab – 3b + 2c + (- 2ab -4b + c)
= 5ab – 3b + 2c – 2ab – 4b + c
= (5 – 2) ab + (-3 – 4) b + (2 + 1) c = 3ab + (-7)b + 3c
= 3ab – 7b + 3c
It is 3ab – 7b + 3c smaller.
SamacheerKalvi.Guru

Answer:

To find the answer we have to find the difference.
Here greater number 5ab – 3ab + 2c.
∴ Difference = 5ab – 3b + 2c – (2ab + 4b – c) = 5ab – 3b + 2c + (- 2ab -4b + c)
= 5ab – 3b + 2c – 2ab – 4b + c
= (5 – 2) ab + (-3 – 4) b + (2 + 1) c = 3ab + (-7)b + 3c
= 3ab – 7b + 3c
It is 3ab – 7b + 3c smaller.
SamacheerKalvi.Guru

Q.5Six times a number subtracted from 40 gives – 8. Find the number.v

Solution

Let the required number be x. Six times the number = 6x.
Given 40 – 6x = – 8
-6x + 40 – 40 = -8 – 40
– 6x = – 48
\(\frac{-6 x}{-6}=\frac{-48}{-6}\)
x = 8
∴ The required number is 8.
Challenge Problems

Answer:

Let the required number be x. Six times the number = 6x.
Given 40 – 6x = – 8
-6x + 40 – 40 = -8 – 40
– 6x = – 48
\(\frac{-6 x}{-6}=\frac{-48}{-6}\)
x = 8
∴ The required number is 8.
Challenge Problems

Q.6From the sum of 5x + 7y -12 and 3x – 5y + 2, subtract the sum of 2x – 7y – 1 and – 6x + 3y + 9.v

Solution

Sum of 5x + 7y – 12 and 3x – 5y + 2 .
= 5x + 7y- 12 + 3x – 5y + 2 = (5 + 3) x + (7 – 5) y + ((- 12) + 2)
= 8x + 2y – 10.
Again Sum of 2x – 7y – 1 and – 6x + 3y + 9
= 2x – 7y – 1 + (- 6x + 3y + 9) = 2x – 7y – 1 – 6x + 3y + 9
= (2 – 6) x + (- 7 + 3) y + (- 1 + 9)
= – 4x – 4y + 8
Now 8x + 2y – 10 – (-4x – 4y + 8)
= 8x + 2y – 10 + (4x + 4y – 8)
= 8x + 2y – 10 + 4x + 4y – 8
= (8 + 4) x + (2 + 4) y + ((- 10) + (- 8))
= 12x + 6y – 18

Answer:

Sum of 5x + 7y – 12 and 3x – 5y + 2 .
= 5x + 7y- 12 + 3x – 5y + 2 = (5 + 3) x + (7 – 5) y + ((- 12) + 2)
= 8x + 2y – 10.
Again Sum of 2x – 7y – 1 and – 6x + 3y + 9
= 2x – 7y – 1 + (- 6x + 3y + 9) = 2x – 7y – 1 – 6x + 3y + 9
= (2 – 6) x + (- 7 + 3) y + (- 1 + 9)
= – 4x – 4y + 8
Now 8x + 2y – 10 – (-4x – 4y + 8)
= 8x + 2y – 10 + (4x + 4y – 8)
= 8x + 2y – 10 + 4x + 4y – 8
= (8 + 4) x + (2 + 4) y + ((- 10) + (- 8))
= 12x + 6y – 18

Q.7Find the expression to be added with 5a – 3b – 2c to get a – 4b – 2c?v

Solution

To get the required expression we must subtract 5a – 3b + 2c from a – 4b – 2c.
∴ a – 4b – 2c – (5a – 3b + 2c) = a – 4b – 2c + (- 5a + 3b – 2c)
= a – 4b – 2c – 5a + 3b -2c
= (1 – 5) a + (- 4 + 3) b + (- 2 – 2) c
= – 4a – b – 4c.
∴ -4a – b – 4c must be added.

Answer:

To get the required expression we must subtract 5a – 3b + 2c from a – 4b – 2c.
∴ a – 4b – 2c – (5a – 3b + 2c) = a – 4b – 2c + (- 5a + 3b – 2c)
= a – 4b – 2c – 5a + 3b -2c
= (1 – 5) a + (- 4 + 3) b + (- 2 – 2) c
= – 4a – b – 4c.
∴ -4a – b – 4c must be added.

Q.8What should be subtracted from 2m + 8n + 10 to get – 3m + 7n + 16?v

Solution

To get the expression we have to subtract – 3m + 7n + 16 from 2m + 8n + 10.
(2m + 8n + 10) – (-3m + 7n + 16) = 2m + 8n + 10 + 3m – 7n – 16
= (2 + 3) m + (8 – 7) n + (10 – 16)
= 5m + n – 6
SamacheerKalvi.Guru

Answer:

To get the expression we have to subtract – 3m + 7n + 16 from 2m + 8n + 10.
(2m + 8n + 10) – (-3m + 7n + 16) = 2m + 8n + 10 + 3m – 7n – 16
= (2 + 3) m + (8 – 7) n + (10 – 16)
= 5m + n – 6
SamacheerKalvi.Guru

Q.9Give an algebraic equation for the following statement: “The difference between the area and perimeter of a rectangle is 20”.v

Solution

Let the length of a rectangle = l and breadth = b then Area = lb; Perimeter = 2(1 + b)
Area – Perimeter = 20
∴ lb – 2(l + b)

Answer:

Let the length of a rectangle = l and breadth = b then Area = lb; Perimeter = 2(1 + b)
Area – Perimeter = 20
∴ lb – 2(l + b)

Q.10Add : 2a + b + 3c and a + \(\frac{1}{3}\)b + \(\frac{2}{5}\)cv

Solution

Answer: