Question

a father is 7 times as old as his son . two years ago the father was 13times as old as his son . how old are they now​

Answer 1

Step-by-step explanation:

DAD=7X WHEN THE SON'S AGE IS X NOW.

2 YEARS AGO THEY WERE:

DAD=(7X-2) & THE SON=(X-2)

SO WE HAVE:

7X-2=13(X-2)

7X-2=13X-26

7X-13X=-26+2

-6X=-24

X=-24/-6

X=4 THE AGE OF THE SON NOW.

THUS THE DAD=7*4=28 YEARS OLD NOW.

PROOF:

7*4-2=13(4-2)

28-2=13*2

26=26

Answer 2

Step-by-step explanation:

Let the age of the son be x years. Then is fathers age will be 7x.

2 years ago :

Father's age = 7x-2

Son's age = x-2

Given that father's age was 13 times older than his son.

7x-2 =13(x-2)

7x-2 =13x-2

6x=24

x=4

If the age of son is 4 then the age of the father will be 7×4=28

Their present ages are son=4 and father=28

Hope

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