Question

1. The age of a person is double the age of his son. Ten years ago, it was four times the age of his son. Use the concept of algebra and find the present age of the son. ​

Answer 1

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Let the age of a man = x years And the age of his son = y years

Ten years hence,

Man's age =(x+10) years

Son's age =(y+10) years

According to the question, (x+10)=2(y+10)

⇒x+10=2y+20

⇒x=2y+20−10

⇒x=2y+10…(i)

Ten years ago,

Father's age =(x−10) years

Son's age =(y−10) years

According to the question, (x−10)=4(y−10)

⇒x−10=4y−40

⇒x=4y−30…(ii)

From Eq. (i) and (ii), we get 2y+10=4y−30

⇒2y−4y=−30−10

⇒−2y=−40

⇒y=20

On putting the value of y=20 in Eq. (i), we get x=2y+10

⇒x=2(20)+10

⇒x=50

Hence the age of the man is 50 yrs and the age of his son is 20 yrs.

Answer 2

Answer:

The man age is 50

and his son age is 20

Step-by-step explanation:

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