A question like this on the SAT will *always* have answer choices, and in this case would be easy to backsolve. Don’t rob yourself of easy points—remember that answer choices are often a helpful tool! Use them to your advantage. In this case, you’d simply need to plug (2*x* – 1) in for *x* in each answer choice until one gave you *cx.*

Since you didn’t provide answer choices, though, we’ll have to do this the hard way.

The way I like to think about questions like this is pretty procedural. You’ve got some *f*(*x*) expression. You swap out the *x* for (2*x* – 1) and what happens? First, the 2*x* means there’s a doubling. We know we want to land on *cx*, so *f*(*x*) must have a *c*/2 in the *x* term that gets doubled. Make sense?

But what about those question marks? Well, we know that the end result we want is that *f*(2*x* – 1) = *cx*, so that question mark must cancel out the –*c*/2 we get from multiplying *c*/2 by (2*x* – 1). What cancels out –*c*/2? Easy: +*c*/2! So let’s see if making *f*(*x*) = (c/2)*x* + *c*/2 works:

Yep, that works. Again, it probably would have been easier to just start with the answer choices and see which one landed you on *cx*. Your work would end up looking just like that second bit of math above, but without so much abstract thinking required.

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