I’m helping a relative prep for the SAT later this year. One recent practice problem involved solving a system of two linear equations, but had lots of fractions and negative signs, which made it fiendishly difficult to solve via the ordinary algebra she’s been taught. I ended up solving it with Gaussian elimination, which seemed much better suited to problems like this under the conditions of the SAT.
I haven’t taken the SAT in over forty years, so I’m seeking advice: should I advise her to use Gaussian elimination for problems like this? I also worked it with Cramer’s Rule, but that method seemed more prone to errors if you forget the formula.
Just two linear equations? Gaussian elimination may be overkill. It would be good to know, if taking linear algebra, which isn’t usually tackled until sophomore year for engineers. But I didn’t learn it until grad school, myself.
I’d just do the usual back-substitution, i.e. isolate one variable in equation 1, plug it into equation 2, solve for the remaining variable. Then use value that back in the first equation to get the value of that first variable.
Elimination is indeed a method taught in high school algebra classes, and it is often an easier method to use than substitution (especially, I find, for real-world problems). Whether it’s “Gaussian” elimination specifically is an almost moot point, for a system as simple as two equations.
Of course, if (like most SAT questions) this is multiple choice, then an even easier method to use is the method of checking.
That’s what we did, but neither of us got it right the first few times. Which made me wonder if the SAT tries to test the ability to spot the right technique for the right problem. In which case, she should practice several methods, and also practice deciding which one to use for the specific problem.
I should hope that it tries to test for that, at least.
Oh, and not relevant to this particular problem, but since you’re prepping for the SAT math test, one simple fact that will be enough to solve a vast number of problems: In any triangle, the longest side will be opposite to the largest angle, the middle-length side will be opposite to the middle-sized angle, and the smallest side will be opposite to the smallest angle. That will usually be enough to determine whether any given angle is greater than or less than 45º, which is often enough to narrow down to a single answer choice, without even needing to pick up a calculator.
When solving linear equations (and it may be some good 25 years since I last did it), I usually check if the terms in the equations neatly eliminate each other, and then take the elimination route. If not, I just solve for one variable and plug it into the other equation. Cramer’s rule for a two variable problem surely is overkill, it only really becomes useful for third or higher grade linear systems.
[my bold]
Sorry, but I have to ask: if your relative isn’t comfortable with negative signs and fractions, wouldn’t it be best to teach them about that first? I don’t know how hard this test is, but how could you do it without a command of negative numbers and fractions?
So, kill the fractions, by multiplying both sides by a common denominator? If it would reduce the number of negative signs, you could multiply both sides by a negative.
She has a very good command of negative numbers, and fractions. She worked about twenty problems, and the only one she had trouble with was this one. (It took me a few tries to get the right answer). And it didn’t help that the answer in the back of the book was wrong.
As I said upthread, I assume that fluency in multiple methods, and the ability to spot which method to use, are what they are trying to select for.
It sounds to me like the problem wasn’t conceptually difficult; just that it had a lot of steps, and thus a lot of opportunities for a mistake to creep in.
Cramer’s Rule is overkill for any system. It’s an interesting mathematical curiosity that it works, but finding even one of the determinants is already more work than just solving the system using any of the other methods.
Yeah, that’s the point, you must have practice to get a feel for the best method for different cases. I’m an electrical engineer, so solving linear equations was my bread and butter during college, and I developed an eye for it.
In retrospect, you’re right, and I really didn’t use it if I had the choice. But I learned it in college, and have a special fondness for Cramer’s rule, because that’s also my last name .
Christ, it’s been about 30 years since I’ve done this, so forgive me if this is a stupid question, but would using a matrix equation be helpful? I seem to recall doing that for solving systems of linear equations, but I may be conflating the concept with something else.
Gaussian elimination, mentioned by the OP, involves matrices. So does Cramer’s rule, and also one or two other possible methods.
Like other responders, I doubt that any of these would be the easiest way to solve a system with only two equations. That is, unless you have a calculator that can do the matrix work for you.
Ah. I don’t remember either of those names. The way I was taught, I think, was this one. At least this looks familiar to me, with the inverse matrix and all that:
If you have a TI-80 series calculator, you can solve a system by plugging the coefficients into a matrix, then running a function in the matrix menu called rref [reduced row-echelon format].
3x + 4y = 9
4x - 3y = 12
[3 4 9]
[4 -3 12]
Run rref, you’ll see this:
[1 0 3]
[0 1 0]
In other words, x = 3, and y = 0.
There are lots of little shortcuts built into the things, to the extent that students who have one, and know how to use them, have a big advantage over those who just have a simple 5 function jobber.
Yeah, that was one of the “one or two other possible methods” I had in mind. (The other was Gauss-Jordan elimination, if you consider that distinct from Gaussian elimination.)
And that was what I had in mind when I said “unless you have a calculator that can do the matrix work for you.” If you know what you’re doing and aren’t prone to typos when entering numbers into the calculator, that may well be the easiest approach.
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