Найдите корень уравнения 6(9-y)=7(4-y) 2)z(z-3)=(4+z)(z-5) 3) (a+8)(1-a)=8-a(a-11) 4) (2-b)(4-b)-17-(b+2)(7-b) 5) (11-2c)(5c-6)-80-10(c-3) ПОмогите прошу

1) 6(9-y)=7(4-y)

54 - 6y = 28 - 7y

-6y + 7y = 28 - 54

y = -26

2)z(z-3)=(4+z)(z-5)

z2 - 3z = 4z - 20 + z2 - 5z

z2 - 3z - 4z - z2 + 5z = -20

-2z = -20

z = -20 : (-2)

z = 10

3) (a+8)(1-a)=8-a(a-11)

a - a2 + 8 - 8a = 8 - a2 + 11a

a - a2 - 8a + a2 - 11a = -8 + 8

-18a = 0

a=0

4) (2-b)(4-b)-17-(b+2)(7-b) = 8 - 2b - 4b + b2 -17 - (7b -b2+ 14 -2b) =  8 - 2b - 4b + b2 -17 - 7b + b2 - 14 + 2b = -23 -11b

5) (11-2c)(5c-6)-80-10(c-3) = 55c - 66- 10c + 12c - 80 - 10c + 30 = 47c -116