Why does time dilation cause you to age slower? And is time considered relative to the observer? I understand that the higher your velocity the slower light will move. But how does time itself slow down while you are moving faster?
 A: There are two notions of time you'll find in special relativity, "coordinate time" relative to an inertial frame of reference, and the "proper time" measured by a clock which may or may not be moving on an inertial path through spacetime. If you pick two events--like the event of two twins departing from one another, and the later event of them reuniting, then different inertial frames can disagree about the amount of coordinate time between these two events, so that version of time is "relative to the observer". However, the amount of proper time that each twin will measure along their own path through spacetime (their own world line) is an objective fact which doesn't depend on the choice of reference frame. And different inertial frames can predict how much proper time $\tau$ each twin will measure, if they know the twin's coordinate velocity as a function of coordinate time $v(t)$, as measured in their own reference frame. In that case, the elapsed proper time the twin measures between two events which happen at coordinate times $t_0$ and $t_1$ is given by the formula $\int_{t_0}^{t_1} \sqrt{1 - (v(t)/c)^2} \, dt$
You probably won't understand that formula since you indicated you know algebra but not calculus, but there is a special case you might be able to follow better. Suppose the twin is moving at a constant coordinate velocity $v$ (i.e., the twin was moving inertially) during the time interval where we want to calculate their proper time. In that case, in a coordinate time interval of $\Delta t$, the twin will age by a proper time $\Delta \tau$ according to the formula $\Delta \tau = \sqrt{1 - (v/c)^2} \Delta t$, or equivalently $\Delta t = \frac{\Delta \tau}{\sqrt{1 - (v/c)^2}}$. This is what's usually given for the "time dilation" formula in special relativity, showing how a clock which is moving relative to a given frame of reference will also be running slow relative to that frame, in terms of the rate it ticks (the rate its proper time increases) compared to coordinate time in the frame. Just as an example, if a clock is moving at $v=0.8c$ for a time interval $\Delta t$ of 10 years in the coordinates of my reference frame, I can predict that the clock will elapse a proper time $\Delta \tau$ of just $\sqrt{1 - 0.8^2} * 10$ = 6 years.
Since time dilation is comparing the proper time to the coordinate time, and coordinate time is observer-dependent (the coordinate time between a given pair of events depends on what inertial reference frame you use), the amount of "time dilation" a given clock experiences is observer-dependent too. But as I said, the actual total amount of proper time a given clock experiences between two events that happen next to it is not observer-dependent. This means you can have a situation like the twin paradox, where two twins carrying clocks depart from one another, move apart at constant velocity for a while, and then one accelerates to turns around and they approach each other, and when they eventually reunite they discover that the twin that accelerated to turn around has experienced less total proper time in total than the twin that remained at constant velocity throughout the trip. Different frames can disagree on which twin was aging faster during any particular constant-velocity "leg" of the trip--for example, even though the twin who accelerates ages less in total, there is a frame where that twin was aging faster than the non-accelerating twin during the first leg of the trip while they were moving apart. But that frame will find that after the twin accelerates and changes velocity, the twin will now be moving at a faster velocity during the second leg of the trip where the distance between the two twins is shrinking, and so the twin who accelerated will be aging slower during this leg, and when you add up the aging on both legs this frame agrees with every other frame that the twin who accelerated ages less in total.
As for "why" this is true, in physics the only way to address "why" questions about particular formulas like time dilation is to derive them from other assumptions. As Moonraker said, time dilation can be derived from the two fundamental postulates of special relativity, which demand that the speed of light be constant in all inertial reference frames and the equations of the laws of physics look the same when expressed in the coordinates of different inertial reference frames. And there are also analogies which may be helpful to intuition--in the case of coordinate time vs. proper time, there is a close analogy to ordinary plane geometry (the geometry of a 2D surface), where one can measure the length along a particular path (such as a straight line) between two endpoints, and one can also measure the difference in some coordinate like the y-coordinate for each point, in the context of a Cartesian coordinate system defined on the plane. 
Note for example that if you know the difference in x-coordinate $\Delta x$ and the difference in y-coordinate $\Delta y$ for a given pair of points in Cartesian coordinates, then according to the Pythagorean formula the distance along a straight-line path between those points is $\sqrt{\Delta x^2 + \Delta y^2}$, which is equivalent to $\sqrt{1 + (\Delta x / \Delta y)^2} * \Delta y$. This looks very similar to the formula for time dilation for a clock whose velocity is in the x-direction of an inertial frame, since in that case clock's coordinate velocity $v$ can be expressed as $\Delta x / \Delta t$, the distance the clock travels in a given time in this coordinate system, meaning the proper time elapsed on the clock is $\sqrt{1 - (\Delta x / \Delta t)^2} * \Delta t$ according to the time dilation formula. Aside from changing a positive sign to a negative sign, this is just like the earlier formula for distance between points on a 2D place, except with a t-coordinate in place of a y-coordinate. The negative sign tells you that the geometry of spacetime is not exactly like the geometry of 2D space (it's more like the geometry of the complex plane, if you're familiar with that idea), but it's conceptually very similar. 
One important difference introduced by the minus sign is that while in 2D spatial geometry a straight line is always the shortest path between points, in spacetime a "straight" path through spacetime between two events (i.e. the path of a clock moving inertially, at constant velocity) is always the one with the longest proper time, and this is why in the twin paradox, the twin who moves inertially between the departure and the reunion has aged more than the twin who accelerated to turn around (meaning his path through spacetime is non-inertial).
A: You ask for the reason of time dilation, so I suppose that you know what time dilation is (there are plenty of questions in this forum and also Wikipedia to get information about time dilation).
Time dilation is a phenomenon of spacetime. Spacetime is 4D, but following the principles of special relativity, the fourth time dimension is different from space dimensions. 
The second postulate of special relativity tells us that light speed is constant for any observer. This implies (for mathematical and geometrical reasons) a spacetime concept were the spacetime interval of light and particles moving at light speed is always 0 (represented in form of the light cone in the Minkowski spacetime diagram) which is in contradiction with the 4D geometry of former times. Accordingly, the spacetime interval of particles moving near light speed is strongly reduced. Their spacetime interval is their proper time, which is smaller than the time measured by observers who are not moving at light speed.
This spacetime geometry applies also to you: if you are approaching light speed during a space travel, you will age less than your twin on Earth.
