The required expression can be obtained as follows
= 2x³ - 2x² + 3x - 5 - (2x³ + 7x²x  +7)
= 2x³ - 2x² + 3x + 5 + (-2x³ - 7x² + 2x - 7)
= 2x³ - 2x² + 3x + 5 - 2x³ - 7x² + 2x - 7
= (2 - 2)x³ + (-2 - 7)x² + (3 + 2)x + (5 - 7)
= 0x³ + (- 9x²) + 5x - 2
= - 9x² + 5x - 2
\(\therefore\) 2x³- 2x² + 3x + 5 is greater than 2x ³+ 7x² - 2x + 7 by -9x² + 5x - 2

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

Since we do not know the number of students who get down at the first stop, let us take the number as x. The number of students get down at the second bus stop is 2x.
x + 2x + 6 + 3 = 60
(1 + 2) x + 9 = 60
3x + 9 = 60
3x + 9 – 9 = 60 − 9 [Subtract 9 on both sides]
3x = 51
\(\frac { 3x }{ 3 } =\frac { 51 }{ 3 } \) [Divide by 3 on both sides]
Therefore, x = 17.
Thus, the number of students got down in the first bus stop is 17.

Let the number of notes in each denomination be x.
Then x + 5x + 10x = 960
(1 + 5 + 10) x = 960
16x = 960
Divide by 16 on both the sides,
\(\frac { 16x }{ 16 } =\frac { 960 }{ 16 } \) = 60
Therefore, x = 60
Thus, the number of notes in each denomination is 60.

(2m + 8n + 10)  Expression = -3 + 7n + 16
Expression = (2m + 8n + 10) - (-3m + 7n + 16)
= (2m+ 8n + 10) + (3m - 7n - 16)
= (2m + 3m) + (8n - 7n) + (10 - 16)
= 5m + n - 6

Total amount paid at the end of the trip = Rs. 740
Amount paid for 5 km = Rs. 100(-)
= Rs. 640
Let the distance travelled after completing 5 km = y
Charge for one additional km = Rs. 16
16 y = 640
\(\frac{16 y}{16}=\frac{640}{16}\) [ Divide by 16 on both sides]
y = 40
The total distance travelled = 5 km + 40 km
= 45 km

(3x + 6y) + (Expression) - 5x + 8y
Expression = (5x + 8y) - (3x + 6y)
(5x + 8y) + (3x - 6y)
= (5x- 3x) + (8y - 6y)
= ( 5 - 3)x + (8 - 6)y
= 2x + 2y
Rule : Subtracting a term is the same as adding its inverse

(i) 7x, -8x, -12x are like terms containing same algebraic variable 'x'.
(ii) 5y, 12 -9y are like terms containing same algebraic variable 'y'.
(ili) 6z, z and 11 z are like terms containing same algebraic variables 'z'.

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

The amount earned on the first day is Rs.200.
Let the amount spent on the first day be Rs. x. Amount with her at the end of the first day is (200 − x).
Amount earned on the second day is Rs.300 and the amount spent on the second day is Rs.2x. The amount left on the second day is 300 − 2x.
Similarly, the net amount that she would have on the third day is 400 − 4x. Therefore, the total amount that she would have at the end of three days is (200 − x) + (300 − 2x) + (400 − 4x).
That is, 200 + 300 + 400 + ( − 1 − 2 − 4)x
= 900 + ( −7)x
Thus the required algebraic expression is 900 − 7x.

In the given algebraic expression, x,y,z are the variables.
Let us group the like terms.
100x + 99y − 98z + 10x + 10y + 10z − x − y + z
= (100x + 10x − x) + (99y + 10y − y) + ( −98z + 10z + z)
= (100 + 10 − 1)x + (99 + 10 − 1)y + ( −98 + 10 + 1)z
= (110 − 1)x + (109 − 1)y + ( −98 + 11)z
= 109x + 108y + ( −87)z
= 109x + 108y − 87z.