Samacheer Class 7 Maths - Algebra Intext Questions

(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

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

(ii) 13
Hint: 7(3) – 4(2) = 21 – 8 = 13
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(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

(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

(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

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

(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

(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

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.

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

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.

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

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

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

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

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.

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

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

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.

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

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

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