Question 1063998
3 years ago a man is four times the age of the daughter. If the product of their ages is  430 what are their present ages? 


<pre>The original problem for this was:
<A HREF=https://www.algebra.com/algebra/homework/word/age/Age_Word_Problems.faq.question.776905.html>https://www.algebra.com/algebra/homework/word/age/Age_Word_Problems.faq.question.776905.html</A>
https://www.algebra.com/algebra/homework/word/age/Age_Word_Problems.faq.question.776905.html

The difference with the above problem (# 1063998) is that it failed to include the words: "IS NOW." 
As such, I agree with IKLEYN that it is DEFECTIVE, as age problems with ages that are NON-INTEGERS, are nonsense, in my opinion.
I think IKLEYN misunderstood me. When I referred to the 3 people who responded, I was talking about the 3 who responded to the ORIGINAL problem (# 776905),
not the CURRENT problem (# 1063998). I IMPLORE you to LOOK at the ORIGINAL problem (# 776905) and you'll see what I'm talking about.

This is UNBELIEVABLE! 3 other persons responded to the OLDER math problem (# 776905) and all 3 were UNABLE to solve this simple problem. Why so???
And, these people call themselves, TUTORS!

This is the ORIGINAL problem, solved a LONG time ago and ONLY yours truly's answer was correct.
<b>3 years ago a father was 4 times as old as his daughter <u>is now</u>. The product of their ages is 430. Calculate their present ages.</b>
Let daughter's, and father's current ages be D, and F, respectively
Then: F - 3 = 4D ----- F = 4D + 3 ----- eq (i)
Also FD = 430 ------ eq (ii)
D(4D + 3) = 430 ----- Substituting {{{4D + 3}}} for F in eq (ii)
{{{4D^2 + 3D - 430 = 0}}}
Solve this to get daughter's age. Then find father's: {{{430/D}}}
You should get:
Daughter's current age: {{{highlight_green(10)}}}
Father's current age: {{{highlight_green(43)}}}