Seven years ago a women was four times old as her sons age at that time.In seven years years from now the mother will be twice as old as her son would be then.Find their present ages.
Answer
If the woman is w, and the son is s, then we know
w-7 = 4(s-7)
w+7 = 2(s+7)
woman is 35, son is 14
I don't know ple solve it
I CAN'T SOLVE IT
To solve this problem, let's assign variables to represent the present ages of the mother and her son.
Let's say the present age of the son is S, and the present age of the mother is M.
According to the problem, seven years ago, the son's age was S - 7, and the mother's age was M - 7.
1) Seven years ago, the mother was four times as old as her son at that time.
This can be written as M - 7 = 4(S - 7).
2) In seven years from now, the mother will be twice as old as her son would be then.
This can be written as M + 7 = 2(S + 7).
Now we have a system of two equations with two variables. Let's solve them to find the present ages of the mother and the son.
From equation 1) M - 7 = 4(S - 7), we can simplify it and rewrite it as:
M - 7 = 4S - 28
M = 4S - 21
Now substitute M = 4S - 21 into equation 2) M + 7 = 2(S + 7):
4S - 21 + 7 = 2S + 14
Simplify the equation:
4S - 14 = 2S + 14
Combine like terms:
4S - 2S = 14 + 14
2S = 28
Divide both sides by 2:
S = 14
Now substitute S = 14 back into equation M = 4S - 21:
M = 4(14) - 21
M = 56 - 21
M = 35
Therefore, the son's present age is 14, and the mother's present age is 35.