SOLUTION: 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 3 times as old as her son is then. Find each of their ages
Algebra ->
Customizable Word Problem Solvers
-> Age
-> SOLUTION: 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 3 times as old as her son is then. Find each of their ages
Log On
Question 520060: 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 3 times as old as her son is then. Find each of their ages Found 2 solutions by mananth, MathTherapy:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! Son = x years
Mrs. Brooks 5x-7
after 6 years
son = x-6
Mrs. Brooks = 5x-7-6=>5x-13
5x-13=3x-3
5x-3x=13-3
2x=10
/2
x=5 son's age
Mrs. Brook's age =5x-7=> 5*5-7=18 years
m.ananth@hotmail.ca
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 3 times as old as her son is then. Find each of their ages
Let Mrs. Brooks be B, and son, S
B = 5S – 7 ------- eq (i)
In 6 years, Mrs. Brooks will be B + 6, and son will be S + 6
Therefore, we can say that: B + 6 = 3(S + 6) – 3
B + 6 = 3S + 18 - 3
B – 3S = 15 - 6
B - 3S = 9 ----- eq (ii)
5S – 7 – 3S = 9 ------- Substituting 5S – 7 for B in eq (ii)
