Let the initial salary be ‘x’ and increment per year be ‘y’
Case I. Initial salary = x
Increment after 4 years = 4y
According to question
x + 4y = 4500
Case II. Initial salary = x
Increment after 10 years = 10y
According to question
x + 10y = 5400
Thus, we have x + 4y = 4500 ...(i)
x + 10y = 5400 ...(ii)
Since the coefficient of ‘x’ in (i) and (ii) are equal.
So we can simply eliminate the variable ‘x’ by subtracting.

Putting this value in (i), we get
x + 4y = 4500
⇒ x + 4(150) = 4500
⇒ x + 600 = 4500
⇒ x = 3900
Hence, initial salary = Rs. 3900 and annual Increment = Rs. 150.
Problems Based on Cross-Multiplication Method

We have
6 ax + 6by = 3a + 2b    ...(i)
6bx - 6ay = 3b - 2a    ...(ii)
For making the coefficient of ‘x’ in (i) and (ii) equal, we multiply eq. (i) by ‘b’ and (ii) by ‘a’ and then subtracting, we get

Let the number of 20 paise coins = x
and    number of 25 paise coins = y
Then,    value of 20 paise coins = 20x
and    value of 25 paise coins = 25y
Case I.    x + y = 50
Case II. 20x + 25y = 1125
[∵ 11.25 = 1125 Paise]
Thus, we have
x + y = 50    ...(i)
20x + 25y = 1125    ...(ii)
For making the coefficient of ‘x’ in (i) and (ii) equal, we multiply the (i) by 20 and then subtracting, we get

Putting the value of ‘y’ in (i), we get
x + y = 50
⇒    x + 25 = 50
⇒    x = 25

Hence, the number of 20 paise coins = 25 and the number of 25 paise coins = 25.

Let the digit at 10’s place be x
and digit at unit’s place be y.
∴ Number = 10x + y
Number obtained after reversing the order of the digits = 10y + x
Case I.
10x + y + 10y + x = 66
⇒ 11x + 11y = 66
⇒ 11(x + y) = 66
⇒ x + y = 6    ...(i)
Case II. x - y = ±2
⇒    x - y = + 2    ...(ii)
and    x - y = - 2    ...(iii)
Considering (i) and (ii), we get

Putting the value of ‘y’ in (i), we get
x + 2 = 6
⇒    x = 4
So,    Number = 10x + y
= 42
Considering (i) and (iii), we get

Putting the value of ‘y’ in (i), we get
x + y = 6
⇒    x + 4 = 6
⇒    x = 2
So,    Number = 10x + y = 24
Thus, there are two such numbers 42 and 24.