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

  1. Chapter 2 Class 8 Linear Equations in One Variable
  2. Serial order wise

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
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