Ex 3.4, 2
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
Let Numerator be x
and Denominator be y
So, Fraction is 𝒙/𝒚
Given that,
If 1 is added to numerator and 1 is subtracted from the denominator, fraction becomes 1.
(𝑁𝑢𝑛𝑒𝑟𝑎𝑡𝑜𝑟 + 1)/(𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 −1)= 1
(𝑥 + 1)/(𝑦 − 1)= 1
(x + 1) = (y – 1)
x – y = –1 – 1
x – y = –2
Also,
If we add 1 to the denominator, fraction becomes 1/2.
𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟/(𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 + 1)=1/2
𝑥/(𝑦 +1)=1/2
2x = y + 1
2x – y = 1
Hence, our equations are
x – y = –2 …(1)
2x – y = 1 …(2)
We use elimination method with equation (1) & (2)
–x = –3
x = 3
Putting x = 3 in equation (1)
x – y = –2
3 – y = –2
–y = –2 – 3
–y = –5
y = 5
So, x = 3, y = 5 is the solution of our equation
∴ Numerator = x = 3
& Denominator = y = 5
Hence, Original Fraction = 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟/𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 "= " 𝑥/𝑦=𝟑/𝟓

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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