A man has two sons, one twice as old as the other. The man is four times as old as the older boy. In three years he will be five times as old as the younger boy. Find their present ages.

Man's present age: 32 years

Older son's present age: 8 years

Younger sons present age: 4 years

Step-by-step explanation:

Let their present ages be represented by:

Man = a

Older boy = b

Younger boy = c

A man has two sons, one twice as old as the other:

b = 2 × c

b = 2c Equation 1

The man is four times as old as the older boy:

a = 4 × b

a = 4bEquation 2

In three years he will be five times as old as the younger boy:

a + 3 = 5 (c + 3)

a + 3 = 5c + 15Equation 3

Since b = 2c and a = 4b

Subtitute 2c for b in Equation 2

a = 4b

a = 4 × 2c

a = 8c

Subtitute 8c for a in Equation 3

a + 3 = 5cEquation 3

8c + 3 = 5c + 15

Collect like terms

8c - 5c = 15 - 3

3c = 12

c = 4

Therefore since c, represents the present age of the younger son, the younger son is 4 years old

b = 2c

b = 2 × 4

b = 8

Since b is the present age of the older son, the older son is 8 years old

a = 4b

b = 8

a = 4 × 8

a = 32

Since a is the present age of the man, the man is 32 years old

Therefore,

Man's present age: 32 years

Older son's present age: 8 years

Younger sons present age: 4 years

center= (-2,0)

radius= sqrt2/2

step-by-step explanation: