The first ten positive integer multiples of 58 are: 1 x 58 = 58 2 x 58 = 116 3 x 58 = 174 4 x 58 = 232 5 x 58 = 290 6 x 58 = 348 7 x 58 = 406 8 x 58 = 464 9 x 58 = 522 10 x 58 = 580

There is no integral solution, because none of the factor pairs of 16 add to 6. x +y = -16 x * y = 6 x (-16 - x) = 6 - x2 - 16 x - 6 = 0 x2 + 16 x + 6 = 0 Quadratically, x and y are the two roots: (-8) + (sq rt 58) and (-8) - (sq rt 58) Check: (-8 +sq rt 58)*(-8 -sq rt 58)…

Well, the obvious answer is 1 x 58 = 58. Since 1 is the multiplicative identity, that is 1 x a = a for any number a. P.S. This is not Algebra, solving equations with variables, factoring etc. are.

I believe the converse is: if 2x equals 6 then x equals 3 inverse: if x doesn't equal 3 then 2x doesn't equal 6 contrapositive: if 2x doesn't equal 6 then x doesn't equal 3