seven years ago, Mary was 6 times as old as Cora, In 1 year, she will be only twice as old as Cora. Find their present ages
Let:
M=Mary's age now
C=Cora's present age
M-7=6(C-7)
M+1=2(C+1)
M=2C+1
(2C+1)-7=6C-42
2C-6=6C-42
-4C=-36
C=9
therefore Cora's age now is 9 yrs. old
M=2C+1
M=2(9)+C
M=19
therefore Mary's age now is 19 yrs. old.
Using only one variable:
seven years ago:
Cora --- x
Mary --- 6x
present ages:
Cora = x+7
Mary = 6x+7
one year from now
Cora = x+8
Mary = 6x+8
6x+8 = 2(x+8)
6x +8 = 2x+16
4x=8
x =2
present ages:
Cora = x+7 = 9
Mary = 6x+7 = 19
To find their present ages, let's first assign variables to represent their ages. Let's assume Mary's present age is M, and Cora's present age is C.
According to the given information, seven years ago, Mary was 6 times as old as Cora. This can be written as:
M - 7 = 6(C - 7)
In one year, Mary will be only twice as old as Cora. This can be written as:
M + 1 = 2(C + 1)
We now have a system of two equations with two variables. We can solve this system of equations to find the values of M and C.
First, let's solve the equation M - 7 = 6(C - 7):
M - 7 = 6C - 42
M = 6C - 42 + 7
M = 6C - 35
Next, let's substitute this value of M in the second equation:
6C - 35 + 1 = 2(C + 1)
6C - 34 = 2C + 2
6C - 2C = 2 + 34
4C = 36
C = 36 / 4
C = 9
Substituting the value of C = 9 back into the equation M = 6C - 35:
M = 6(9) - 35
M = 54 - 35
M = 19
Therefore, Mary's present age (M) is 19 years old, and Cora's present age (C) is 9 years old.