SOLUTION: Two years ago, my age was 4 times that of my son. Eight years ago, my age was 10 times that of my son. Find age of my son now.

Question 530681: Two years ago, my age was 4 times that of my son. Eight years ago, my age was 10 times that of my son. Find age of my son now.
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
x = father's current age
y = son's current age
.
Two years ago...
x-2
y-2
.
Father's age was 4 times son's age:
(x-2) = 4(y-2)
.
Eight years ago...
x-8
y-8
Father's age was 10 times son's age:
(x-8) = 10(y-8)
.
So, we have two equations and two unknowns.
That suggests a system of linear equations can be used to solve them.
.
x-2 = 4y -8
x -4y = -6
.
x-8 = 10y -80
x -10y = -72
.
x -4y = -6
x -10y = -72
------------- subtract
6y = 66
divide both sides by 6
y = 11
.
So, the son is now 11 years old.
.
To check this answer...
substitute y=11 to find x
.
(x-2) = 4(11-2)
x -2 = 44 - 8
x = 38
.
check these values using the third equation
.
x-8 = 10(y-8)
x-8 = 30
y-8 = 3
10*3 = 30
Correct.
.
Answer: The son is now 11 years old.

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