CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard

Class 10
Solutions of Sample Papers for Class 10 Boards

Solve the following system of equations: 21/x + 47/y = 110 47/x + 21/x = 162, x, y ≠ 0

Note : This is similar to Example 17 of NCERT – Chapter 3 Class 10

Transcript

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 = 110/21β47/21v Putting value of u in (4) 47u + 21v = 162 47(110/21β47/21 "v " ) + 21v = 162 47 Γ 110/21β47"Γ" 47/21 "v "+ 21v = 162 β47"Γ" 47/21 "v "+ 21v = 162 β 47 Γ 110/21 (β47 Γ 47π£ + 21 Γ 21π£)/21 " "= (162 Γ 21 β 47Γ110)/21 (βγ47γ^2 π£ + γ21γ^2 π£)/21 " "= (162 Γ 21 β 47Γ110)/21 (γ21γ^2 π£ β γ47γ^2 π£ )/21 " "= (162 Γ 21 β 47Γ110)/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 (β26) Γ 68 = 162 Γ 21 β 47 Γ 110 v Γ (β26) Γ 68 = 3402 β 5170 v Γ (β26) Γ 68 = β1768 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