Final answer:
To show x is even if and only if 3x+5 is odd, one must assume x is even to prove 3x+5 is odd, and vice versa, using arithmetic and properties of even and odd numbers.
Step-by-step explanation:
To prove that x is even if and only if 3x+5 is odd, we need to work through the problem logically in two parts.
Part 1: Assume x is even. The definition of an even number is any number that can be expressed as 2k, where k is an integer. If x is even, then x = 2k for some integer k. Consequently, 3x + 5 = 3(2k) + 5, which simplifies to 6k + 5. Since 6k is clearly even, 6k+5 will always be odd because an even number plus 5 (an odd number) yields an odd number.
Part 2: Assume 3x + 5 is odd. If 3x + 5 is odd, it can be expressed as 2m + 1 where m is an integer. Therefore, 3x + 5 = 2m + 1, and subtracting 5 from both sides gives us 3x = 2m - 4. This further simplifies to 3x = 2(m - 2), showing that 3x is an even number as it is 2 times some integer (m - 2). However, since 3 is odd, x must be even for 3x to be even, since an odd times even is even.
This proves that x is even if and only if 3x + 5 is odd, as the two are intrinsically linked through a series of logical and arithmetic steps.