
Get live Maths 1-on-1 Classs - Class 6 to 12
Last updated at March 23, 2023 by Teachoo
Ex 2.6, 5 Solve the following equations. (7π¦ + 4)/(π¦ + 2)=(β4)/3 (7π¦ + 4)/(π¦ + 2)=(β4)/3 3 (7y + 4) = β4 (y + 2) 21y + 12 = β4 (y + 2) 21y + 12 = β4y β 8 21y + 4y + 12 = β8 25y + 12 = β8 25y = β8 β 12 25y = β20 y = (β20)/25 y = (βπ)/π Check:- L.H.S = (7π¦ + 4)/(π¦ + 2) =(7 ((β4)/5) + 4)/((β4)/5 + 2)=((β28)/5 + 4)/((β4)/5 + 2) = ((β28 + 4(5))/5)/((β4 + 2(5))/5) = ((β28 + 20)/5)/((β4 + 10)/5)=((β8)/5)/(6/5)=(β8)/5Γ5/6 =(β4)/3= R.H.S β΄ LHS = RHS Hence Verified.