Any of its multiples! Simply calculate any of the following: 57 x 0 57 x 1 57 x 2 57 x 3 57 x (-1) 57 x (-2) etc., and you get a number that is a multiple of 57, and therefore, divisible by 57.
The six multiples are: 57 x 1 = 57 57 x 2 = 114 57 x 3 = 171 57 x 4 = 228 57 x 5 = 285 57 x 6 = 342
60
1 x 114, 2 x 57, 3 x 38, 6 x 19, 19 x 6, 38 x 3, 57 x 2, 114 x 1
Algebraically.X + ( X + 2) = - 1162X + 2 = - 1162X = - 118X = - 59X + 2 = - 57- 59 (+) - 57 = - 116( - 57, - 59 )------------------solution set
Any of its multiples! Simply calculate any of the following: 57 x 0 57 x 1 57 x 2 57 x 3 57 x (-1) 57 x (-2) etc., and you get a number that is a multiple of 57, and therefore, divisible by 57.
The six multiples are: 57 x 1 = 57 57 x 2 = 114 57 x 3 = 171 57 x 4 = 228 57 x 5 = 285 57 x 6 = 342
The first 5 positive integer multiples of 57 are: 1 x 57 = 57 2 x 57 = 114 3 x 57 = 171 4 x 57 = 228 5 x 57 = 285
57 = 3 x 19, 76 = 2 x 2 x 19 so LCM = 3 x 19 x 2 x 2 = 228
2(x - 176)(x + 57) x = 176, -57
1 x 57, 3 x 19
57 = 19 x 3, 60 = 5 x 2 x 2 x 3. They have 3 as a common factor, so 5 x 2 x 2 x 3 x 19 = 1140
60
1, 2, 3, 6, 19, 38, 57, 114Prime Factors of 114=2x3x19.1,2,3,6,19,38,57,1141 x 114, 2 x 57, 3 x 38, 6 x 19.
1 x 114, 2 x 57, 3 x 38, 6 x 19, 19 x 6, 38 x 3, 57 x 2, 114 x 1
The way you actually wrote it: x+2(57). This is just x+114 and does not help you find x.The way I think you meantto write it: x+2x=57. 3x=57. x=19
x--57 is an algebraic expression equivalent to x+57