Question

Найдите корень уравнения: 8у-3(2у-3)=7у-2(5у+8) 5(2у-9)+6у=4(3у-2)-21 5у+7(3-у)=3(5-2у)-6 7у-4(3у+5)=2-3(4у-9)

Answer 1

8y-3(2y-3)=7y-2(5y+8)
8y-6y+9=7y-10y-16
8y-6y-7y+10y=-9-16
5y=-25
y=-25:5
y=-5
Жауабы:y=-5

2) 5(2y-9)+6y=4(3y-2)-21
10y-45+6y=12y-8-21
10y+6y-12y=45-8-21
4y=16
y=16:4
y=4
Жауабы:y=4

3) 5y+7(3-y)=3(5-2y)-6
5y+21-7y=15-6y-6
5y-7y+6y=-21+15-6
4y=-12
y=12:4
y=3
Жауабы:y=3

4)7y-4(3y+5)=2-3(4y-9)
7y-12y-20=2-12y+27
7y-12y+12y=20+2+27
7y=49
y=49:7
y=7
Жауабы:y=7