<p>The answer is C. If the function is linear, as x increases at a constant rate, so does f(x). Basically the difference between C and 9 (9-C) has to be equal to the difference between 9 and D (D-9). C could be 3 (9-3=6) and D could be 15 (15-9=6), C could be 8 (9-8=1) and D could be 10 (10-9=1), etc. In both of those cases, the sum of C and D is 18 (3+15 or 8+10). Kind of sloppy, but that's how I did it.</p>
<p>Here's another way to solve this, which might look a little more familiar to you, depending on your algebra background:</p>
<p>Since the function f(x) is linear, f(x) = m x + b, where m is the slope and b is the y-intercept. So, list the values you are given (including letters), along with the result from the linear equation (with m and b still unknown). You might start the list with f(2), because a specific numerical value is given for it.
f(2) = 9 = 2 m + b
f(0) = c = b
f(4) = d = 4 m + b
Therefore, c + d = f(0) + f(4) = 4 m + 2 b.
But now notice that 4 m + 2 b is just twice (2 m + b), and (2m + b) is f(2).
So, c + d = 2 (2m + b) = 2 f(2) = 18.</p>
<p>shiomi has a given really elegant solution, which is preferable, if you see it. It's worthwhile to spend some time thinking about it. That approach does hinge on the x values being equally spaced, though.</p>
<p>As materr observed, 9 - c = d - 9. (This also hinges on the x values being equally spaced.) Once you have noticed this, you do not need to do any substitution or trial at all--just solve for c + d, and you have it.</p>
<p>For a faster solution, since 0, 2, 4 are an arithmetic sequence and the function is linear, c, 9 and d are also an arithmetic sequence. That means c+d=9*2=18</p>
<p>^cristina89 has also suggested an elegant method. To the OP: It's really worthwhile to understand how these methods work. They tend to be much faster than the "slog it out" method of my post #5--but it's worthwhile to know that you can also slog it out (usually), if you don't see one of the quicker methods.</p>
<p>Who doesn't use that method to solve it...
Does anyone actually use QuantMech's way? That is just a way to explain it, but its not really feasible to use it on the SAT.</p>
<p>personally... i think the easiest way is to find the 2 version of the slopes and put them equal to each other... once you have that just add c + d and you get your answer</p>
<p>@khoitrinh: I agree that the other ways are faster, and also more elegant. Someone who "gets" linear functions would almost certainly use one of them.</p>
<p>@arya202: The method suggested by lowriders715 is useful to remember, because it does not rely on the points being equally spaced, as they were in this particular problem. </p>
<p>My suggestion was really aimed at the OP who is probably more of a "verbal" person than a "math" person. In that case (and depending on the OP's algebra curriculum), a method that just uses the standard form of a linear function might be more appealing. I think there's plenty of time on the SAT math to use that method. </p>
<p>I'd say that if arya202 is taking the October SAT, then he/she might be better off using the definition-based method. If arya202 is taking any later SAT, then it would definitely be better to go for the level of understanding that makes the other approaches obvious.</p>
<p>Not to add yet another solution, but the following is pretty straightforward: the points are all on a line, so the slope (rise/run) between any pair of points is the same, so:</p>