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

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
Answer by MathTherapy(10552) (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 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)

2S = 16

S, or son�s current age is , or

B, or Mrs. Brooks� age = 5(8) � 7, or

---------
Check
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5(8) - 7 = 40 � 7 = 33 ----- 33 = 33 (TRUE)

33 + 6 = 3(8 + 6) � 3 ---- 39 = 42 � 3 ----- 39 = 39 (TRUE)

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