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

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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.