Two years ago the age of father was three and a half times the age of his son then. Six years hence, the age of father will be ten years more than twice the age of his son then. Find their present age.
Answers
Answered by
8
Step-by-step explanation:
Present
Let the son's age be x.
Then, according to the condition given, father's age = 3x + 3
3 years later
Son's age= x + 3
Father's age = 3x + 3 + 3 = 3x + 6
According to the condition given,
3x + 6 = 10 + 2 ( x + 3 )
3x + 6 = 10 + 2x + 6
x = 10
Thus, the son's present age is 10 years and the father's age is 3(10) + 3 = 33 years.
Answered by
0
14 yrs. and 44 yrs
Step-by-step explanation:
let present age of father X years and the present age of Son be Y years
equation 1
[x _ 2] = 3½ * (y - 2)
[X- 2 ] = [ 7/ 2 ] [ y - 2]
2x - 4 = 7y - 14
2x - 7y+ 10= 0...... 1
equation 2
[ x+2] = 10 + 2 [y+ 6]
x+6 = 10+2y + 12
x - 2y - 16 =0...... 2
multiplying equation 1 with equation 2
2x - 4y - 32 = 0
2x - 7y + 10 = 0
- + -
3y = 4
y. = 14
x - [ 2* 14 ]- 16
x = 16 +28
=44 yrs.
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