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Question 29 (OR 2nd question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard

Last updated at Nov. 1, 2019 by

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

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

Check the answer here https://www.teachoo.com/1543/504/Example-17---Solve--2-x---3-y--13--5-x---4-y---2/category/Examples/

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  1. Class 10
  2. Solutions of Sample Papers for Class 10 Boards

    3. CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard

    CBSE Class 10 Sample Paper for 2019 Boards →

Transcript

Question 29 (OR 2nd question) Solve the following system of equations: 21/π‘₯ + 47/𝑦 = 110 47/π‘₯ + 21/𝑦 = 162, x, y β‰  0 Given equations 21/π‘₯ + 47/𝑦 = 110 47/π‘₯ + 21/𝑦 = 162 Let u = 1/π‘₯, v = 1/𝑦 So, our equations become 21u + 47v = 110 …(3) 47u + 21v = 162 …(4) From (3) 21u + 47v = 110 21u = 110 – 47v u = 110/21βˆ’47/21v Putting value of u in (4) 47u + 21v = 162 47(110/21βˆ’47/21 "v " ) + 21v = 162 47 Γ— 110/21βˆ’47"Γ—" 47/21 "v "+ 21v = 162 βˆ’47"Γ—" 47/21 "v "+ 21v = 162 – 47 Γ— 110/21 (βˆ’47 Γ— 47𝑣 + 21 Γ— 21𝑣)/21 " "= (162 Γ— 21 βˆ’ 47Γ—110)/21 (βˆ’γ€–47γ€—^2 𝑣 + γ€–21γ€—^2 𝑣)/21 " "= (162 Γ— 21 βˆ’ 47Γ—110)/21 (γ€–21γ€—^2 𝑣 βˆ’ γ€–47γ€—^2 𝑣 )/21 " "= (162 Γ— 21 βˆ’ 47Γ—110)/21 212v – 472v = 162 Γ— 21 – 47 Γ— 110 212v – 472v = 162 Γ— 21 – 47 Γ— 110 v(212 – 472) = 162 Γ— 21 – 47 Γ— 110 v(21 – 47) (21 + 47) = 162 Γ— 21 – 47 Γ— 110 v Γ— (–26) Γ— 68 = 162 Γ— 21 – 47 Γ— 110 v Γ— (–26) Γ— 68 = 3402 – 5170 v Γ— (–26) Γ— 68 = –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 – 47 21u = 63 u = 63/21 u = 3 Now, x = 1/𝑒 = 1/3 y = 1/𝑣 = 1/1 = 1 So, x = 𝟏/πŸ‘, y = 1

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