Four years ago marina was three times old as her daughter .Six years from now the mother will be twice as old as her daughter.Find their present ages.

Let d = the present age of Marina’s daughter, and …
Let m = the present age of Marina.

Four years ago, Marina’s daughter was d – 4 years old and Marina was m – 4 years old, and Marina was three times older than her daughter, i.e.,   m – 4 = 3(d – 4).

Six years from now, Marina’s daughter will be d + 6 and Marina will be m + 6, and  Marina will be twice as old as her daughter, i.e., m + 6 = 2(d + 6).

So, we now have a system of two simultaneous linear equations in two unknowns, d and m:

1. m – 4 = 3(d – 4)
    
2. m + 6 = 2(d + 6)

We’ll solve this system of equations by the Substitution Method by arbitrarily starting with equation 1.) as follows:
m – 4 = 3(d – 4)
m – 4 = 3d – 12
m – 4 + 4 = 3d – 12 + 4

3.   m = 3d - 8

Now, substituting the expression on the right side of equation 3.) for the unknown quantity m into equation 2.), we have:

m + 6 = 2(d + 6)
(3d – 8) + 6 = 2d + 12
3d – 8 + 6 = 2d + 12
3d – 2 = 2d + 12
3d – 2 + 2 = 2d + 12 + 2
3d – 2d = 2d –2d + 14
d = 14 

Now, substituting this result, d = 14, into equation 3.), we get:

m = 3d – 8
    = 3(14) – 8
    = 42 – 8
m = 34
 
Checking our results for d and m in the two original equations 1.) and 2.):

m – 4 = 3(d – 4)                          m + 6 = 2(d + 6)
34 – 4 = 3(14 – 4)                       34 + 6 = 2(14 + 6)
30 = 42 – 12                                40 = 28 + 12
30 = 30                                        40 = 40

Therefore,  Marina’s daughter’s present age is d = 14 years old, and Marina’s present age is m = 34 years old.