SOLUTION: Four members of one family have combined age of 145. Jeff is 3 times age of daughter Ann, while Tudy is 8years younger then her husband Jeff. Harry is a quarter of his mother Tru

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Question 1105692: Four members of one family have combined age of 145.
Jeff is 3 times age of daughter Ann, while Tudy is 8years younger then her husband Jeff. Harry is a quarter of his mother Trudys age.
How old are each of the family members.
So far my working I have:
Let the age of Ann be A
Jeff= 3A
Trudy= 3A-8
Harry= 1/4 of 3A-8
Ann= A
145= 3A + 3A-8+ (3A-8)/4+ a
The further I work out my equation it doesn't total 145. The best I've got is 114. So I'm just wonder where in my equation am I going wrong?
I feel it is how I'm writing 'Harry' into the equation but I don't know how else to do it.
Thank-you for your time.
Kirsty
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Your work and your equation are fine. You don't show us what you did to try to solve the equation, so we don't know why you couldn't solve the problem.

a%2B3a%2B3a-8%2B%283a-8%29%2F4+=+145
7a-8%2B%283%2F4%29a-2+=+145

Did you get that far with your work?
Can you finish from there?

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Added 1/4/2018...

a%2B3a%2B3a-8%2B%283a-8%29%2F4+=+145
7a-8%2B%283%2F4%29a-2+=+145

7a%2B%283%2F4%29a+=+155
%2831%2F4%29a+=+155
31a+=+620
a+=+20

The ages are...
Ann = a = 20
Jeff = 3a = 60
Trudy = 3a-8 = 52
Harry = (3a-8)/4 = 13