(Q15).Mary told her daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." Find the present age of Mary and her daughter.

Let Mary's present age be x years and her daughter's age be y years.

Then, seven years ago Mary's age was x - 7 and daughter's age was y - 7.

x - = (y - )

x - = y -

x - y + = 0 ....................(1)

Three years hence, Mary's age will be x + 3 and daughter's age will be y + 3.

x + = (y + )

x + = y +

x - y - = 0 ......................(2)

Elimination method

Equation 1 ⇒ x - y = -
Equation 2 ⇒ x - y =
(-) (+) (-) same sign for x, so subtract.

-y = -

y =
-
-
 =

Substitute the value of y in equation (2)

x -3() - 6 = 0

x = + 6 =

Therefore, Mary's present age is years and her daughter's age is years.