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 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.