Great question. Zac67 did a great job of answering it, but I'd like to give my own take on it.
Your question revolves around two other, but related questions:
- Why does UTP use a pair of wires?
- Why does UTP twist the wires around each other?
Both questions have to do with Electromagnetic Interference (EMI). The first mainly reduces outbound EMI, the second reduces inbound EMI.
Why does UTP use a pair of Wires?
Voltage applied to a wire emits outbound EMI. If there is only one cable in use, then that isn't much of a problem. But often cables are run alongside many other cables, and the idea of one cable's outbound EMI affecting the transmissions of other wires next to it is undesirable.
The solution devised is to transmit two signals inverse from each other.
For example, presume we need to send a voltage of +5v from one end of a cable to the other. Presume transferring +5v leaks +0.5v to neighboring cables. To offset this, use another wire to transfer the exact opposite: -5v, which will have the effect of also leaking -0.5v to neighboring cables. The combined effect of leaking +0.5v and -0.5v ends up being outbound emission of 0v.
(It isn't a perfect net 0 emission, but you can see the concept).
This is referred to in the electrical engineering world as a Balanced Pair, and is represented in twisted-pair wiring with the TX+ and TX- wire.
This allows you to use wiring schemes that don’t require heavy investments in shielding, and is a part of the reason for the prolific use of Unshielded Twisted Pair (UTP) cabling instead of Shielded Twisted Pair (STP)
So far we’ve only answered why we use a pair of wires, we will look into why they are twisted next.
Why does UTP twist the wires around each other?
Despite the strategy above of using a Balanced Pair, there is no avoiding the fact that any electrical wire will be exposed to some amount of inbound EMI. To offset this, another strategy was devised to reduce the absorbed EMI on a Balanced Pair.
The strategy revolves around the fact that the EMI interference will be greater the closer a wire is to the source. If two wires, who are sending the opposite voltage on a wire, "take turns" being closest to the source, they will both be exposed to the same amount of inbound EMI. Allowing the receiving end to extract the intended signal from the EMI.
Here is how that works:
The blue wire starts with +50v, and the green wire starts with the exact inverse, -50v. The source of the EMI is the red circle, and each wave that surrounds the EMI source impacts the wires progressively less and less. If you only add the EMI at each grey dot (the top and bottom of each twist), both wires end up receiving +22v of interference.
Even though the final voltage received on the right side of the wire is different, notice that the difference in voltage is consistent throughout the twisted pair of wires: it is always 100v apart. The EMI affected both wires identically. You could easily calculate the difference of the final values (100v), and display it on a number line to determine the starting voltages were +50v and -50v:
It should be said that the numbers used above were greatly simplified in order to communicate the concept. Typical EMI emission only affects signaling in the range of micro-volts (µV) — which is 1,000,000th of a Volt (V). But the concepts still remain true: because the original and inverse signals are being sent, the net outbound emission is canceled out, and because of the twists, both wires are equally exposed to the same amount of interference.
With all that said, we can speak directly to your questions:
However what I don't understand is why there's a seperate + and - transmission and receiver line, and what they do.
Are they each carrying their own signal?
Yes, but they are inverse of each other.
Or is one a reference voltage?
Yes, they are a mutual reference for each other.
The images and content above come from this section of an article on Ethernet Wiring on my blog. If you want to learn more about the subject, feel free to take a look.