Check sibling questions

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

Last updated at March 23, 2023 by

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

Get live Maths 1-on-1 Classs - Class 6 to 12

Book 30 minute class for β‚Ή 499 β‚Ή 299

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.

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.