Question

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.​

Answer 1

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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.

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Answer 2

Answer:

Marina is 34 yrs old and daughter is 14 yrs old.

Let x= present age of Marina

Let y= present age of daughter

x - 4 = 3( y - 4)

x - 3y +8 = 0

x + 6 = 2(y - 6)

x - 2y - 6= 0

Now, let x =2y +6

Substituting value of x,

2y + 6 -3y + 8 =0

-y +14 =0

y = 14

Now, x= 2y + 6 = 2×14 + 6= 34