Question

№855 Найдите корень уравнения 1) 8у-3(2у-3)=7у-2(5у+8);2)5(2у-9)+6у=4(3у-2)-21;3)5у+7(3-у)=3(5-2у)-6;4)7у-4(3у+5)=2-3(4у-9)

Answer 1

1) 8у-3(2у-3)=7у-2(5у+8);
8y-6y+9=7y-10y-16
8y-6y -7y+10y=-16-9
5y=-25
y=-25/5 
y=-5 

2)5(2у-9)+6у=4(3у-2)-21;
10y-45+6y=12y-8-21
10y+6y-12y=45-29
4y=16
y=16/4
y=4

 

3)5у+7(3-у)=3(5-2у)-6;

5y+21-7y=15-6y-6
5y-7y+6y=9-21
4y=-12
y=-12/4

y=-3
 

4)7у-4(3у+5)=2-3(4у-9)

7y-12y-20=2-12y+27
7y-12y+12y=2+27+20
7y=49
y=49/7

y=7