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CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard

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

Last updated at March 23, 2023 by

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

Solve the equations: 21/x + 47/y = 110 and 47/x + 21/x = 162

Question 29 (OR 2nd question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 2
Question 29 (OR 2nd question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 3 Question 29 (OR 2nd question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 4 Question 29 (OR 2nd question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 5

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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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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.