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Ex 2.6, 5 - Solve 7y + 4/ y+2 = -4/3 - Chapter 2 Class 8 NCERT Maths

Ex 2.6, 5 - Chapter 2 Class 8 Linear Equations in One Variable - Part 2

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Transcript

Ex 2.6, 5 Solve the following equations. (7𝑦 + 4)/(𝑦 + 2)=(βˆ’4)/3 (7𝑦 + 4)/(𝑦 + 2)=(βˆ’4)/3 3 (7y + 4) = βˆ’4 (y + 2) 21y + 12 = βˆ’4 (y + 2) 21y + 12 = βˆ’4y βˆ’ 8 21y + 4y + 12 = βˆ’8 25y + 12 = βˆ’8 25y = βˆ’8 βˆ’ 12 25y = βˆ’20 y = (βˆ’20)/25 y = (βˆ’πŸ’)/πŸ“ Check:- L.H.S = (7𝑦 + 4)/(𝑦 + 2) =(7 ((βˆ’4)/5) + 4)/((βˆ’4)/5 + 2)=((βˆ’28)/5 + 4)/((βˆ’4)/5 + 2) = ((βˆ’28 + 4(5))/5)/((βˆ’4 + 2(5))/5) = ((βˆ’28 + 20)/5)/((βˆ’4 + 10)/5)=((βˆ’8)/5)/(6/5)=(βˆ’8)/5Γ—5/6 =(βˆ’4)/3= R.H.S ∴ LHS = RHS Hence Verified.

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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.