(i) \(\cfrac { 1 }{ 2 } \left( a+b \right) \)
(ii) p²+ q²

We have 6x, -5x, x, 16x are like terms
6y, y, 7y, are like terms
6, - 5, 1,3 are like terms

(i) 2x - 2y 
The co-efficient of x in 2x is 2
The co-efficient of y in - 2y is - 2
(ii) x +y + 3
The co-efficient of x is 1
The co-efficient of y is 1
The constant term is 3

(i) 2x³ - 3x² + 3xy + 8
(ii) 7x³ + 9y² - 2xy²
(iii) 9x² - 2x + 3xy - 1

(5a - 3b + 2c) + Expression = a - 4b - 2c 
Expression = (a - 4b - 2c)- (5a- 3b + 2c)
= (a - 4b - 2c) + (-5a - 3b +2c)
= (a - 5a) + (-4b + 3b) + (-2c - 2c)
= (1 - 5)a + (-4 + 3)b + (-2 - 2)c
= -4a - b - 4c

I. (3ab + 8) - (-3ab - 8)
= (3ab + 8) + (3ab + 8)
= (3ab + 3ab) + (8 + 8)
= (3 + 3)ab + (8 + 8)
= 6ab + 16
II. (-3ab - 8) - (3ab + 8)
= (-3ab - 8) + (-3ab - 8)
= (-3ab - 3ab) + (-8 - 8)
= (-3 - 3)ab + (-8 - 8)
=  - 6 ab - 16
Rule: Subtracting a term is the  same as adding its inverse

Let the whole nurnber be x.
Thrice of this number = 3x
Nine added to this gives 45.
3x + 9 = 45
3x + 9 - 9 = 45 - 9
[Subtract 9 on both sides]
3x = 36
\(\frac{3 x}{3}=\frac{36}{3}\)
[Divide by 3 on both sides]
x = 12
The number is 12.

(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

(i) 8x + 3x = (8 + 3) x = 11x
(ii) 7mn + 5mn = (7 + 5)mm = 12mm
(iii) -9y + 11y + 2y = (-9 + 11 + 2) y = (2 + 2) y = 4y

I step [addition]
(5x + 7y -12) + (3x - 5y + 2)
= (5x + 3x) + (7y - 5y) + (-12 + 2)
= (5 + 3)x + (7 - 5)y + (-12 + 2)
=8x + 2y - 10
II step [addition]
(2x - 7y - 1) + (- 6x + 3y + 9)
= (2x - 6x) + (-7y + 3y) + (-1 + 9)
= (2 - 6)x + (-7 + 3)y + (-l + 9)
= -4x - 4y + 8
III step [subtraction]
(8x + 2y - 10) - (- 4x - 4y + 8)
= (8x + 2y- 10) + (4x + 4y- 8)
= (8x + 4x) + (2y + 4y) + (-10- 8)
= (8 + 4)x + (2 + 4)y + (-18)
= 12x + 6y - 18

let the two consecutive odd numbers be 'x' and
x + 2
Given that their sum is 200
x + (x + 2) = 200
x + x + 2 = 200
(1 + 1) x + 2 = 200
2x + 2 = 20A
2x + 2 - 2 = 200 - 2
[Subtract 2 on both sides]
2x = 198
\(\frac{2 x}{2}=\frac{198}{2}\)
[Divide by 2 on both sides]
x = 99
x + 2 = 99 + 2
x + 2 = 101
The two consecutive odd numbers are 99 and 101.

(i) (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 + p - p - p + p) + (2 + 3 -4-5 + 10)
=(1 + 1 + 1 - 1 - 1 + 1)p  +(15 -9)
= 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 + n + n) + (m + m + m) + (1 + 2 + 3 + 4 + 5)
= (1 + 1 + 1) n + (1 + 1 + l)m + 15
= 3m + 3n + 15

i) 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
= 28p + 6q
(ii) 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 + (,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 + (-4)t = 6mn - 4t
(iv) 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)a+ (1 - 1 + 5 - 5)v
= 6u + 0v = 6u
(v) 5xyz - 3xy,3zxy - 5yx
(5xyz - 3xy) + (3zxy - 5yx)
= (5xyz + 3zxy) + (-3xy - 5yx)
= 8xyz + (-8xy) = 8xyz - 8xy

Let the number be x.
According to the condition,
Given 40 - 6x = -8
40 - 6x - 40 = - 8 - 4
[Subtract 40 on both sides]
-6x = - 48
\(\cfrac { -6x }{ -6 } =\cfrac { -48 }{ -6 } \)
[Divide by -6 on both sides]
x = 8
\(\therefore\) The number is 8

Let the required number be x
Thrice the number = 3x
Increased this by 5 = 3x + 5
According to the given condition,
3x + 5 = 44
3x + 5 - 5 = 44 - 5
[Subtract 5 on both sides]
3x = 39
\(\frac{3 x}{2}=\frac{39}{2}\)
[Divide by 3 on both sides]
x = 13
The number is 13.