Question

Five years ago, a mother was seven times as old as her daughter. Five years
hence, she will be three times as old as her daughter. Find their present ages.​

Answer 1

Answer:

Let the age of Daughter be x

And Age of Mother be y

Now according to question

Casr 1*= y-5= 7(x-5)

= y-5=7x-35

7x - y - 30 =0 ...... 1

Case 2*= y+5= 3(x+5)

= y+5=3x + 15

= 3x-y +10 = 0 ...... 2

Subtract both equations. We get

= 4x = 40

= x= 10 years

Now put x =10 , we get

= 7(10) - y - 30 = 0

= 70 - y = 30

= - y = - 40

= y =40*

So Daughter's present age is 10 years

And Mothers present age is 40 years

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Answer 2

Answer:

let the daughters age be x

hence mothers age= 7x

after 5 years

daughters age =x+5

mothers age = 7x+5

therefore according to the question

7x+5=3(x+5)

7x+5=3x+15

7x-3x = 15-5

4x= 10

x = 10/4

x= 2.5

daughters present age= 2.5

mothers present age = 11.5

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