8 will go into 14 once with a remainder of 6 (or 1 3/4 times)
First, to simplify and show you a easy way to do negative variable problems, multiply both sides, all terms, by a negative 1. -1(-X + 8 = 14) X - 8 = -14 add the 8 to both sides X = -6 check in original equation -(-6) + 8 = 14 that negative times a negative makes the 6 positive 6 + 8 = 14 14 + 14 checks
14 = 2*7 49 = 7*7 So 1, -1, 7, -7 all go into both 14 and 49.
Three times with eight remaining 14 + 14 + 14 + 8 = 50
As 8 is even, and 3 is odd, I think only 1 can go into both.
14 and 28 can both go into any multiple of 28 .
Assume that the expression is: x + 14 = 8 Then subtract both sides by 14 to get: x + 14 - 14 = 8 - 14 x = -6 If x = 14 + 8, then x = 22.
114 ÷ 8 = 14 2/8 = 14¼ = 14.25
They both go into 1440, for example.
1 2 4 7 14 28
8 will go into 14 once with a remainder of 6 (or 1 3/4 times)
If: 3x+8 = 14 Then: 3x = 14-8 So: 3x = 6 Divide both sides by 3: x = 2
14 = 2*7 49 = 7*7 So 1, -1, 7, -7 all go into both 14 and 49.
No. 104 is not evenly divisible by 14.
As 8 is even, and 3 is odd, I think only 1 can go into both.
The LCM of 12 and 14 is 84
Well, sweetheart, 4 and 8 can both go into numbers that are multiples of both 4 and 8. In other words, they can both go into numbers that are divisible by both 4 and 8. So, grab your calculator and start crunching those numbers!