Найдите корень уравнения
(3y-5)/7=(2y+4)/5-2
5(3y-5)=7(2y+4)-70
5(3y-5)-7(2y+4)+70=0
15y-25-14y-28+70=0
y=17
(3у-5)/7= (2у+4)/5-(2*5)/5
(3у-5)/7=(2у+4)-10/5
5(3у-5)=7(2у+4)-7*10
15у-25-14у-28+70=0
у=17
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Answers & Comments
(3y-5)/7=(2y+4)/5-2
5(3y-5)=7(2y+4)-70
5(3y-5)-7(2y+4)+70=0
15y-25-14y-28+70=0
y=17
(3y-5)/7=(2y+4)/5-2
(3у-5)/7= (2у+4)/5-(2*5)/5
(3у-5)/7=(2у+4)-10/5
5(3у-5)=7(2у+4)-7*10
15у-25-14у-28+70=0
у=17