Question 119992: Hey, I need help with this age problem, because I seriously don't get it. A detailed explation will be greatly appreciated! :) Thanks in advanced.
Mrs. Brooks lacks 7 years from being five times as old as her son. Six years from now she will lack 3 years from being three times as old as her son is then. Find each of their ages.
I tried to solve it for an hour before I gave up and looked online for help. Here's what I came up with:
Mrs. Brooks= x; Son (Brooks Junior)= y
5x-7=y
x+6=3(y+6)-3
x+6=3y+18-3
x+6=3(5x-7)+18-3
x+6=15x-21+18-3
x+6=15x-6
-6 -6
-------------
x=15x-12
-15x -15x
----------------
14x=-12
--- ----
14 14
x= -6
--
7
My answer makes no sense, because it's not possible to have a negative age. Help me, please!!!!!
Thanks.--- Rae.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Mrs. Brooks lacks 7 years from being five times as old as her son. Six years from now she will lack 3 years from being three times as old as her son is then. Find each of their ages.
:
B = Brooks present age; S = Son's present age
:
Mrs. Brooks lacks 7 years from being five times as old as her son.
(B+7) = 5S
B = 5S - 7
:
Six years from now she will lack 3 years from being three times as old as her son is then.
That translates to:
B + 6 + 3 = 3(S + 6)
B + 9 = 3S + 18
B = 3S + 18 - 9
B = 3s + 9
:
Find each of their ages.
Substitute (5S-7) for B and find S
5S - 7 = 3S + 9
5S = 3S + 9 + 7
5S - 3S = 16
S = 16/2
S = 8 yr is the son
:
B = 5S - 7
B = 40 - 7
B = 33 yrs is the Mom (Brooks)
:
Check solutions the statement:
Six years from now she will lack 3 years from being three times as old as her son is then.
33 + 6 + 3 = 3(8 + 6)
42 = 3 * 14; confirms our solution
:
:
I'm not sure what you did, You seemed to start out right
Hope this helps you.
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