Question

Q.6. The difference between the ages of a father and son is 30 years. Five years ago, the father was
thrice as old as the son was, then find their present ages.
answers are sons age 20 years and fathers age is 50 years plz solve it​

Answer 1

Answer:

let the present age of father= x

so " " " " son. = 30-x

after 5 year father age = 5x

after 5 year son age. = 30-x

x+30+5= 3(x+5)

x+35 = 3x+15

35-15 = 3x-x

20 = 2x

20/2 = x

10 = x

the present age of father = 5x10 = 15

the present age of son. = 30-10 =20

that is your answer

plz mark as brainlist

Answer 2

Concept:

An equation is said to be linear if the maximum power of the variable is consistently 1. Another name for it is a one-degree equation. A linear equation with one variable has the conventional form Ax + B = 0. In this case, the variables x and A are variables, while B is a constant. A linear equation with two variables has the conventional form Ax + By = C. Here, the variables x and y, the coefficients A and B, and the constant C are all present.

Given:

The difference between the ages of a father and son is 30 years. Five years ago, the father was thrice as old as the son was

Find:

Find the age of father and son

Solution:

let the present age of father= x

so, the present age of son. = 30-x

after 5 year father age = 5x

after 5 year son age. = 30-x

⇒x+30+5= 3(x+5)

⇒x+35 = 3x+15

⇒35-15 = 3x-x

⇒20 = 2x

⇒20/2 = x

⇒10 = x

the present age of father = 5x10 = 50

the present age of son. = 30-10 =20

Therefore, the present age of son and Father is 20 and 50respectively

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