Question

a father says to his son 7 years ago I was seven times as old as you were and after 3 years I will be 3 times as you old as you will be find their present ages.

Answer 1

$\textbf {Hey there, here is the solution.}$

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STEP 1: Define the age 7 years ago:

Let the son be x years old 7 years ago.

7 years ago:

Son = x

Father =7x

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STEP 2: Define the age now:

Current age:

Son = x + 7

Father = 7x + 7

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STEP 3: Define the age 3 years later:

Son = x + 7 + 3 = x + 10

Father = 7x + 7 + 3 = 7x + 10

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STEP 4: Form the equation:

3 years later, the father is 3 times as old as son

⇒ 7x + 10 = 3(x + 10)

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STEP 5: Solve x:

7x + 10 = 3(x + 10)

7x + 10 = 3x + 30

7x - 3x = 30 - 10

4x = 20

x = 5

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STEP 6: Find their present age:

Son = x + 7 = 5 + 7 = 12 years old

Father = 7x + 7 = 7(5) + 7 = 42 years old

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Answer: The father is 42 years old and the son is 12 years old.

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⭐$\textbf {Cheers}$⭐

Answer 2

HEYA!!
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Let the age of Father be X years

Let the age of Son be Y years.

Ages seven years ago,
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( X - 7 ) = 7 ( Y - 7 )

X - 7 = 7 Y - 49

X - 7 Y + 42 =0------------------------------------------------------(1)

Ages three years later ,

( X + 3 ). = 3 ( Y + 3 )

X + 3 = 3Y + 9

X - 3Y -6 =0----------------------------------------------------(2)

Solving (1) and (2) by Elimination,

X - 7Y + 42 = 0
X - 3Y - 6 =0

-4Y + 48=0

-4Y = -48

Y = 12

Put Y = 9 in (1)

X -7 (12)+ 42 = 0

X - 84+42 = 0

X = 42

⏪Hence their Present Ages are 42 and 12 years respectively.

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