Question

Three years hence, a man’s age will be three times his son’s age. Seven years ago, he was seven

times as old as his son. Taking the man’s present age as ‘x’ and the son’s present age as ‘y’, find

their present age​

Answer 1

Answer:

Step-by-step explanation:

So let the present age of the father and the son be 'x' and 'y' respectively.

So, after 3 years their respective ages will be 'x+3' and 'y+3'.

So therefore according to the question the equation will be ,

x+3 = 3(y+3),as we know age of father is 3 times that of the son,

Similarly their ages 7 years ago will be 'x-7' and 'y-7' respectively. \

So therefore the equation will be ,

x-7 = 7(y-7)

Now as we know when there are 2 variables you need 2 equations, 3 variables and 3 equations and so on.....

S0,

x+3 = 3y+9 and x-7 = 7y-49.

If you subtract the two equations then on the R.H.S. and L.H.S. you get,

10 = -4y +58,transpose y to the other side and similarly for the +10.

You then get 4y = 48 and then u can divide and get y = 12.

Solve for 'x'.

Take the first equation for example .

We have already calculated y and therefore we can substitute.

So x+3 = 45.

So therefore x = 42.

So present age of father and son is 42 and 12.

Hope you found my answer helpful.