Question 29 (OR 2nd question) Solve the following system of equations: 21/π₯ + 47/π¦ = 110 47/π₯ + 21/π¦ = 162, x, y β 0
Given equations
21/π₯ + 47/π¦ = 110
47/π₯ + 21/π¦ = 162
Let u = 1/π₯, v = 1/π¦
So, our equations become
21u + 47v = 110 β¦(3)
47u + 21v = 162 β¦(4)
From (3)
21u + 47v = 110
21u = 110 β 47v
u = 110/21β47/21v
Putting value of u in (4)
47u + 21v = 162
47(110/21β47/21 "v " ) + 21v = 162
47 Γ 110/21β47"Γ" 47/21 "v "+ 21v = 162
β47"Γ" 47/21 "v "+ 21v = 162 β 47 Γ 110/21
(β47 Γ 47π£ + 21 Γ 21π£)/21 " "= (162 Γ 21 β 47Γ110)/21
(βγ47γ^2 π£ + γ21γ^2 π£)/21 " "= (162 Γ 21 β 47Γ110)/21
(γ21γ^2 π£ β γ47γ^2 π£ )/21 " "= (162 Γ 21 β 47Γ110)/21
212v β 472v = 162 Γ 21 β 47 Γ 110
212v β 472v = 162 Γ 21 β 47 Γ 110
v(212 β 472) = 162 Γ 21 β 47 Γ 110
v(21 β 47) (21 + 47) = 162 Γ 21 β 47 Γ 110
v Γ (β26) Γ 68 = 162 Γ 21 β 47 Γ 110
v Γ (β26) Γ 68 = 3402 β 5170
v Γ (β26) Γ 68 = β1768
v Γ 26 Γ 68 = 1768
v = 1768/(26 Γ 68)
v = 221/(13 Γ 17)
v = 17/17
v = 1
Putting v = 1 in equation (3)
21u + 47v = 110
21u + 47 Γ 1 = 110
21u + 47 = 110
21u = 110 β 47
21u = 63
u = 63/21
u = 3
Now,
x = 1/π’ = 1/3
y = 1/π£ = 1/1 = 1
So, x = π/π, 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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