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Pair Of Linear Equations In Two Variables

Question
CBSEENMA10006965

A and B are friends and their ages differ by 2 years. A’s father D is twice as old as A and B is hvicc as old as his sister C. The age of D and C differ by 40 years. Find the ages of A and B.

Solution
Let the ages of A and B be x and y years respectively. Then,
space space space space space space straight x minus straight y minus plus-or-minus 2 space space space space space space space left square bracket Given right square bracket
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Clearly, D is older than C 
therefore space space space space space 2 straight x minus straight y over 2 equals 40 rightwards double arrow 4 straight x minus straight y equals 80

Thus, we have the following systems of linear equations
x - y = 2    ...(i)
4x - y = 80    ...(ii)
and    x - y = -2    ...(iii)
Case I : From (i), we have
[Considering (i) and (ii)]
x - y = 2
⇒    x = 2 + y    ...(iv)
Substituting the value of x in (ii), we get
4x - y - 80
⇒ 4(2 + y) - y = 80
⇒ 8 + 4y - y = 80
⇒ 8 + 3y = 80 ⇒ 3y = 72
⇒    y = 24
Substituting the value ofy in (iv), we get
x = 2 + y = 2 + 24=    26
Hence, Age of A = 26 years
and Age of B = 24 years
Case II : From (iii), we have
[Considering (ii) and (iii)]
x - y = -2 ⇒ x = y - 2 ...(v)
Putting the value of (v) in (ii), we get
4x - y = 80
rightwards double arrow 4 left parenthesis straight y minus 2 right parenthesis minus straight y space equals space 80
rightwards double arrow 4 straight y minus 8 minus straight y equals 80
rightwards double arrow space 3 straight y space equals space 88
rightwards double arrow space space space space straight y space equals space 88 over 3 equals 29 1 third
Substituting the value ofy in (v), we get


straight x equals 88 over 3 minus 2 equals fraction numerator 88 minus 6 over denominator 3 end fraction equals 82 over 3 equals 27 1 third
Hence comma space Age space of space straight A space equals space 29 1 third space space years
Age space of space straight B space equals space 27 1 third space years

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