Question

A man is four times as older as his son. In four years time he will be three times as old. What are their ages now?

Answer 1

Answer:

The son is 8 years old

The father is 32 years old

Step-by-step explanation:

Let the man age be x

Let the son age = y

Right now the man is four times as older as his son = x=4y

In four years time he will be three times as old. ⇒x+4 =3(y+4)

Now substitute the value x=4y in x+4=3(y+4)

x+4=3(y+4)

4y+4=3(y+4)

4y+4=3y+12

Combine the like terms:

4y-3y =12-4

y=8

If the son is 8 years old than;

x=4y

x=4(8)

x=32

Father will be 32 years old....

Answer 2

Answer:

Present age of son: 8 years.

Present age of father: 32 years.

Step-by-step explanation:

Let x represent present age of the son and y represent present age of father.

We have been given that a man is four times as older as his son. 4 times of age of son would be $4x$.

We can represent this information in an equation as:

$y=4x$

We are also told that in four years time he will be three times as old.

Age of father in 4 years would be $y+4$.

We can represent this information in an equation as:

$y+4=3(x+4)$

Upon substituting $y=4x$ in 2nd equation, we will get:

$4x+4=3(x+4)$

$4x+4=3x+12$

$4x+4-4=3x+12-4$

$4x=3x+8$

$4x-3x=3x-3x+8$

$x=8$

Therefore, the present age of son is 8 years.

Upon substituting $x=8$ in equation $y=4x$, we will get:

$y=4x\Rightarrow 4(8)=32$

Therefore, the present age of father is 32 years.