Question

One year ago, a man was seven times as old as his daughter. Now, the sum of their ages is 42. Solve the simultaneous equations to find their current ages.

Answer 1

Answer:

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Step-by-step explanation:

Let the fathers age be x years and his son be y years.

Therefore, seven years ago, age of father = (x-7)

and age of son = (y - 7)

According to the given condition,

x-7 =7(y-7)

or x-7y = -42 ...(1)

After 3 years, fathers age = x+3 and

son's age = (y+3)

As per given condition,

x+3 = 3(y+3)

x-3y = 6 ...(2)

ON subtracting (2) from (1), we get

x - 7y - (x -3y) = -42 - 6

⇒ -4y = -48

⇒ y = 12

On putting value of y in (2), we get

x - 3(12) = 6

⇒ x = 6 + 36 = 42

Hence, present father's age = 42 years

and present son's age = 12 years.