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

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Solution

Let his age be x,an dthe son's age be y one year ago.
$x = 8 y; x + 1 = (y + 1)^{2} ⟹ y = 6; x = 48$
Their present ages are 7,49

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

Let his age be x,an dthe son's age be y one year ago.x=8y;x+1=(y+1)2⟹y=6;x=48Their present ages are 7,49

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

Let his age be x,an dthe son's age be y one year ago.
$x = 8 y; x + 1 = (y + 1)^{2} ⟹ y = 6; x = 48$
Their present ages are 7,49