To find present ages of mother and son in a Simultaneous equations Solutions

Q. Twenty years ago, a mother was 4 times as old as her son and 4 years hence, she will be twice as old her son. What are their present ages?

Solution:

Let M be the mother's present age and S be the son's present age. 

Now solving the problem according to the given conditions in the problem:

1. Twenty years ago:

M - 20 = 4(S - 20) 

or (simplifying)

M - 20 = 4s - 80

or

M = 4S - 60 _____________ eqn (1)

2. Four years after from now:

M + 4 = 2(S + 4) 

or (simplifying)

M + 4 = 2s + 8

or

M = 2S + 4 _____________ eqn (2)

Equating (comparing) the two expressions for M:

4S - 60 = 2S + 4

or (simplifying)

⇒    4S - 2S =  4 + 60

⇒    2S =  64

⇒    S = 32 Years

Then, by putting the value of S=32 in any expressions (1) 0r (2) we get the mother age M:

Let in eqn (2)

M = 2S +4 = 2(32) + 4 = 68 years

The present ages are

⇒ Mother: 68 years.

⇒ Son: 32 years.

<Note: You Check (verify) both expressions putting the value of M and S. >