SOLUTION: 1. A man is 4 times as old as his son now. Six years ago, he was 7 times as ols as his son during that time . Find their present ages?

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Question 161974: 1. A man is 4 times as old as his son now. Six years ago, he was 7 times as ols as his son during that time . Find their present ages?
Found 2 solutions by vleith, KnightOwlTutor:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Let M be the age of the man and S be the age of his son.
Given M+=+4S and M-6+=+7%28S-6%29
M-6+=+7%28S-6%29
M-6+=+7S+-+42
M+=+7S+-+36
Now, substitute for M using the first equation M+=+4S
So
M+=+7S+-+36
4S+=+7S+-+36+
-3S+=+-36
S+=+12
So the son is 12 and the dad is 4S = 4%2A12 = 48
Make sure to check your answers

Answer by KnightOwlTutor(293) About Me  (Show Source):
You can put this solution on YOUR website!
x=boy's current age
4x=father's current age
The algebraic equation is as follows.We need to have the ages equal on both sides so we generate an equation that reflects their ages 6 years ago.
Using current age ratios to calculate their ages 6 years ago.
(4x-6)=father's age 6 years ago
x-6=age of son 6 years ago
4x-6+x-6=7(x-6)+(x-6)
Use the distributive property and simplify by combining like terms
5x-12=7x-42+x-6
5x-12=8x-48
Add 12 to both sides
Subtract 8x from both sides
-3x=-36
divide both sides by (-3)
x=12 is the son's current age
4x=48 is the father's current age

Let's double check our answer by calulation their ages 6 years ago
12-6=6
48-8=42
Six years ago the father was 7x older than his son
6y=42
y=7. The answer is correct.