# 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 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:

- m – 4 = 3(d – 4)

- m + 6 = 2(d + 6)

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.

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