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Solved Example

Let age of Cija = x

Age of Bob = Thrice as old as Cija = 3x

Age of Alice = Four years older than Bob = 3x + 4

The problem states:

x + 3x + (3x + 4) = 39

=> 4x + 3x + 4 = 39

=> 7x = 39 - 4 = 35

=> x = 5, age of Cija

Hence, Age of Bob = 3 x 5 = 15.

=>** Bob is of 15 years.**

Age of Bob = Thrice as old as Cija = 3x

Age of Alice = Four years older than Bob = 3x + 4

The problem states:

x + 3x + (3x + 4) = 39

=> 4x + 3x + 4 = 39

=> 7x = 39 - 4 = 35

=> x = 5, age of Cija

Hence, Age of Bob = 3 x 5 = 15.

=>

Let number = x

The problem states:

5 times a number plus 4 times its reciprocal is 21

=> 5x + $\frac{4}{x}$ = 21

=> $\frac{5x^2 + 4}{x}$ = 21

=> 5x^{2} + 4 = 21x

=> 5x^{2} - 21x + 4 = 0

=> 5x^{2} - 20x - x + 4 = 0

=> 5x(x - 4) - (x - 4) = 0

=> (5x - 1)(x - 4) = 0

either 5x - 1 = 0 or x - 4 = 0

=> x = $\frac{1}{5}$ or x = 4

The problem states:

5 times a number plus 4 times its reciprocal is 21

=> 5x + $\frac{4}{x}$ = 21

=> $\frac{5x^2 + 4}{x}$ = 21

=> 5x

=> 5x

=> 5x

=> 5x(x - 4) - (x - 4) = 0

=> (5x - 1)(x - 4) = 0

either 5x - 1 = 0 or x - 4 = 0

=> x = $\frac{1}{5}$ or x = 4

Let first integer = x (Smallest)

Second consecutive even integer = x + 2

Third consecutive even integer = x + 4 (Largest)

Since largest is twice the smallest integer

=> x + 4 = 2x

=> x = 4

Hence,

**First integer** = **4**,

**Second integer** = x + 2 = 4 + 2 =** 6**,

**Third integer** = x + 4 = 4 + 4 =** 8**.

Second consecutive even integer = x + 2

Third consecutive even integer = x + 4 (Largest)

Since largest is twice the smallest integer

=> x + 4 = 2x

=> x = 4

Hence,