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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Question 5 Solve the following equations. (7š‘¦ + 4)/(š‘¦ + 2)=(āˆ’4)/3 (7š‘¦ + 4)/(š‘¦ + 2)=(āˆ’4)/3 3 (7y + 4) = āˆ’4 (y + 2) 21y + 12 = āˆ’4 (y + 2) 21y + 12 = āˆ’4y āˆ’ 8 21y + 4y + 12 = āˆ’8 25y + 12 = āˆ’8 25y = āˆ’8 āˆ’ 12 25y = āˆ’20 y = (āˆ’20)/25 y = (āˆ’šŸ’)/šŸ“ Check:- L.H.S = (7š‘¦ + 4)/(š‘¦ + 2) =(7 ((āˆ’4)/5) + 4)/((āˆ’4)/5 + 2)=((āˆ’28)/5 + 4)/((āˆ’4)/5 + 2) = ((āˆ’28 + 4(5))/5)/((āˆ’4 + 2(5))/5) = ((āˆ’28 + 20)/5)/((āˆ’4 + 10)/5)=((āˆ’8)/5)/(6/5)=(āˆ’8)/5Ɨ5/6 =(āˆ’4)/3= R.H.S ∓ LHS = RHS Hence Verified.

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