Ex 3.3, 3
Form the pair of linear equations for the following problems and find their solution by substitution method.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Let Present age of Jacob = x years
& Present age of Jacob’s son = y years
Five years hence (later),
Jacob’s Age = x + 5
Jacob son’s Age = y + 5
Age of Jacob will be three times of his son.
x + 5 = 3(y + 5)
x + 5 = 3y + 15
x – 3y = 15 – 5
x – 3y = 10
Also,
Five years ago,
Jacob’s Age = x – 5
Jacob son’s Age = y – 5
Age of Jacob was seven times of his son.
x – 5 = 7(y – 5)
x – 5 = 7y – 7(5)
x – 5 = 7y – 35
x – 7y = −35 + 5
x – 7y = –30
Our equations are
x – 3y – 10 = 0 …(1)
x – 7y + 30 = 0 …(2)
From (1)
x – 3y = 10
x = 3y + 10
Putting x in (2)
x – 7y = −30
(3y + 10) – 7y = −30
3y + 10 – 7y = −30
– 4y = −30 − 10
– 4y = –40
y = (−40)/(−4)
y = 10
Putting y = 10 in (1)
x – 3y = 10
x – 3(10) = 10
x – 30 = 10
x = 10 + 30
x = 40
So, x = 40, y = 10
Thus
Present age of Jacob = x = 40 years
Present age of Jacob’s son = y = 10 years

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.