Pwn the sat 4th edition page 36 question 3

I am tutoring my sophomore on her SAT math and we are going through your book a page at a time.

This problem is under back solving.
Given the amount of time you need to spend multiplying the left hand side of the equation. Wouldn’t it be faster to solve for a than plugging in the values?

How would you approach this differently?
Thanks

The question:

I get the “Wouldn’t it be faster…” question a lot about both the plug in and backsolve techniques, and my stock answer is maybe. In this case, I agree with you that FOILing the left hand side until you get the right answer (as many as 3 times—if you fail 3 times then you know the 4th choice is correct) might take a minute. I also agree that if you know what you’re doing with complex numbers, solving for a is fairly straightforward. To make a stronger case for backsolving here, I maybe should have made the question a tiny bit tougher. Like this, perhaps:

(9+ai)(1-i)=12-2ai

As I say at the top of that drill, though, I think it’s worth practicing the techniques, even if they feel slower or actually are slower, because if you really internalize the technique, it can be an escape hatch for you on test day when you encounter a question you can’t figure out the math way. If you never practice backsolving because it always feels too slow, then it probably won’t be there for you in a pinch when you need it most.

Here’s how backsolving would look for me on this question. I’d start with C, since I pretty much always start with C:

    \begin{align*}(9+2i)(1-i)&=12-6i\\9-9i+2i-2i^2&=12-6i\\9-7i+2&=12-6i\\11-7i&=12-6i\end{align*}

That’s obviously false, but it’s CLOSE! Being 1 off would almost certainly clue me in to trying choice B next, and I’d follow the pattern of the FOILing I just did, so I’d be a little faster the 2nd time.

    \begin{align*}(9+3i)(1-i)&=12-6i\\9-9i+3i-3i^2&=12-6i\\9-6i+3&=12-6i\\12-6i&=12-6i\end{align*}