Question

five years ago a man was seven times old as his son.five year hence, the father will be three times as old as his son. Find their present ages.

solve with one variable x... best answer will be marked as brainiliest... spam will be reported​

Answer 1

Answer:

Age of son=10 years

Age of father=40 years

Step-by-step explanation:

Let the age of son = x

So, five years ago ,

the age of son was = x-5

the age of man was=7(x-5)=7x-35

Five years hence,

the age of son will be ,. = x+5

the age of father will be. = 3(x+5)=3x+15

So, {(x-5)+(7x-35)+10}-{(x+5)+(3x+15) -10}=0

(because adding 10 to the sum of the ages of father and son 5 years ago give the sum of their present ages and vice versa in case of 5 years later. And subtracting the sum of their present ages from itself gives 0)

=>(8x-30)-(4x+10)=0

=>8x-30-4x-10=0

=>4x=40

=>x=10

So age of , son = 10

Age of father

=(7x-35)+5

=70-35+5

=40