Question

three types the age of my father added to the 7 times the age of my sister 183 year 6 times the difference between their age added to 9 years three times the sum of their ages determine their present ages

Answer 1

Given :

Three times the age of  father added to the 7 times the age of  sister =  183 years

Six times the difference between their ages added to 9 years = three times the sum of their ages

To Find :

The present ages of father and sister = ?

Solution :

Let the age of father be 'x' years  and age of sister be 'y' years .

From the 1st case we have :

3x +7y = 183    -(1)

From the 2nd case we have :

6 (x-y) +9 = 3 (x+ y )

6x - 6y + 9 = 3x + 3y

3x - 9y + 9 = 0      - (2)

Subtracting eq (2) from eq (1) :

3x + 7y - 3x + 9y -9 = 183

Or , 16 y = 192

Or , $y = \frac{192}{16}$

         = 12

Putting y = 12 in eq (2) :

$3x - 9 \times 12 +9 = 0$

Or, 3x = 108 - 9

Or , 3x = 99

So , x = 33

Hence age of father = x = 33 years , and

Age of sister = y = 12 years .