Question 29 (OR 2nd question) Solve the following system of equations: 21/𝑥 + 47/𝑦 = 110 47/𝑥 + 21/𝑦 = 162, x, y ≠ 0
Given equations
21/𝑥 + 47/𝑦 = 110
47/𝑥 + 21/𝑦 = 162
Let u = 1/𝑥, v = 1/𝑦
So, our equations become
21u + 47v = 110 …(3)
47u + 21v = 162 …(4)
From (3)
21u + 47v = 110
21u = 110 – 47v
u = 110/21−47/21v
Putting value of u in (4)
47u + 21v = 162
47(110/21−47/21 "v " ) + 21v = 162
47 × 110/21−47"×" 47/21 "v "+ 21v = 162
−47"×" 47/21 "v "+ 21v = 162 – 47 × 110/21
(−47 × 47𝑣 + 21 × 21𝑣)/21 " "= (162 × 21 − 47×110)/21
(−〖47〗^2 𝑣 + 〖21〗^2 𝑣)/21 " "= (162 × 21 − 47×110)/21
(〖21〗^2 𝑣 − 〖47〗^2 𝑣 )/21 " "= (162 × 21 − 47×110)/21
212v – 472v = 162 × 21 – 47 × 110
212v – 472v = 162 × 21 – 47 × 110
v(212 – 472) = 162 × 21 – 47 × 110
v(21 – 47) (21 + 47) = 162 × 21 – 47 × 110
v × (–26) × 68 = 162 × 21 – 47 × 110
v × (–26) × 68 = 3402 – 5170
v × (–26) × 68 = –1768
v × 26 × 68 = 1768
v = 1768/(26 × 68)
v = 221/(13 × 17)
v = 17/17
v = 1
Putting v = 1 in equation (3)
21u + 47v = 110
21u + 47 × 1 = 110
21u + 47 = 110
21u = 110 – 47
21u = 63
u = 63/21
u = 3
Now,
x = 1/𝑢 = 1/3
y = 1/𝑣 = 1/1 = 1
So, x = 𝟏/𝟑, y = 1
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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