Question

34. One year ago, a man 8 times as old his son. Now his age is equal to the square of his
son's age. Find their present ages.​

Answer 1

Answer:

Let the present age of the son be x years. It is given that one year ago; a man was 8 times as old as his son. If x = 1, then x2 = 1, which is not possible as father's age cannot be equal to son's age. Present age of man = x2 years = 49 years.

Step-by-step explanation:

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Answer 2

Answer:

Let , the man's age be 'x' and son's age by 'y'.

According to the question,

(x-1)=8(y-1)

x-1= 8y-8

x= 8y-7. (1)

and also,

$x = {y}^{2}$

(2)

From (1) and (2),

${y}^{2} - 8y + 7 = 0$

${y}^{2} - 7y - y + 7 = 0$

$y(y - 7) - 1(y - 7)$

$(y - 1)(y - 7)$

y= 7

x = 49

Hence , the present age of the man is 49 and son is 7.