3c+6=18 3c+6-6=18-6 3c=12 3c÷3=12÷3 c=4
17-3= 4c+3c c=2
2ab - 6ac + 3b - 9c = 2a(b - 3c) + 3(b - 3c) = (2a + 3)(b - 3c).
The solution (not sum) to the equation 21 = 3c + 6 is as follows:21 = 3c + 621 - 6 = 3c + 6 - 615 = 3c3c = 153c/3 = 21/3c = 7
4c+2 = 3c+9 4c-3c = 9-2 c = 7
3c+6=18 3c+6-6=18-6 3c=12 3c÷3=12÷3 c=4
17-3= 4c+3c c=2
2ab - 6ac + 3b - 9c = 2a(b - 3c) + 3(b - 3c) = (2a + 3)(b - 3c).
That factors to (b + 3)(a + c)
Multiplying out the brackets it is: 7a+21-ac-3c
(12c5 - 18c3 + 9c) / (-3c) = -4c4 + 6c2 - 3
The solution (not sum) to the equation 21 = 3c + 6 is as follows:21 = 3c + 621 - 6 = 3c + 6 - 615 = 3c3c = 153c/3 = 21/3c = 7
2ab - 6ac + 3b - 9c = b(2a + 3) - 3c(2a + 3) = (b - 3c)(2a + 3)
Without an equality sign the given expression remains as: 18-3c
a(3+b)+c(3+b) * * * * * This is easy to finish: . . . = (a + c)(3 + b).
4c+2 = 3c+9 4c-3c = 9-2 c = 7
Doing it step by step:-7 = -3c + 933c = 7 + 933c = 100c= 33 1/3