The new SAT places a heavy emphasis on the “Heart of Algebra,” which is a bizarre and tortured euphemism for, mostly, working with linear equations. One of the kinds of questions you know you’re going to see, probably more than once, on your SAT is solving systems of linear equations. For example:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

There are a bunch of good ways to solve such a problem, and I think you should know all of them. So, I’m going to talk about all of them. That’s what you come here for, after all.

Substitution

To solve by substitution, first you’ll want to get one of the variables in one of the equations alone. It doesn’t matter which variable, and it doesn’t matter which equation, so pick the one that looks easiest to isolate. I’m gonna get the x from the first equation by itself. First, subtract the y-term from each side…

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\dfrac{1}{3}x=12-\dfrac{1}{6}y

Then multiply by 3 to get x alone…

3\left(\dfrac{1}{3}x\right)=3\left(12-\dfrac{1}{6}y\right)

x=36-\dfrac{3}{6}y

x=36-\dfrac{1}{2}y

Once I’ve got x alone, I’m going to substitute what I now know x equals into the other equation. This is important! You won’t get anywhere if you substitute back into the equation you just manipulated—you’ll eventually just end up at 0 = 0. Right, so:

x=36-\dfrac{1}{2}y

2x+5y=21

2\left(36-\dfrac{1}{2}y\right)+5y=21

Then I’m going to solve that equation for y:

72-y+5y=21

72+4y=21

4y=-51

y=-\dfrac{51}{4}

Once I’ve got my y I can actually look at the answer choices and see that I’m able to eliminate every choice but A. If I were in a hurry, I’d bubble that and move on. Since we’re practicing, though, I’m going to finish this process by putting the y I just found into one of the previous equations to get x. I like to use the equation I already solved for x to save me a step or two:

x=36-\dfrac{1}{2}y

x=36-\left(\dfrac{1}{2}\right)\left(-\dfrac{51}{4}\right)

x=36+\dfrac{51}{8}

x=\dfrac{288}{8}+\dfrac{51}{8}

x=\dfrac{339}{8}

So, there you have it. Choice A is for sure the answer: \left(\dfrac{339}{8},\dfrac{-51}{4}\right) is a solution for that system of equations.

Elimination

Substitution is the classic, the mainstay. Most people know how to do it and are comfortable with it. Elimination, to my constant amazement, seems less universally known. I say to my amazement because elimination is awesome.

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

To solve the same problem with elimination (I’ve pasted it in above so you don’t have to scroll to see it), what you’ll want to do is instead of trying to isolate any variables, find a way to multiply one of the equations by something so that either the x-coefficients in each equation are equal, or the y-coefficients are equal. I see an easy way to do that:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

6\left(\dfrac{1}{3}x+\dfrac{1}{6}y\right)=6(12)

\dfrac{6}{3}x+\dfrac{6}{6}y=72

2x+y=72

That’s what I’m talking about! I’ve got a 2x there, and a 2x in the other equation: 2x+5y=21. Now all I need to do is subtract one equation from the other, eliminating the 2x making it very easy to solve for y:

     \begin{align*} 2x+y &=72\\ -(2x+5y&=21)\\ -4y&=51\\ y&=-\dfrac{51}{4} \end{align*}

Easy, right? Now to find x, we just substitute -\dfrac{51}{4} in for y in either equation:

2x+5y=21

2x+5\left(-\dfrac{51}{4}\right)=21

2x-\dfrac{255}{4}=21

2x-\dfrac{255}{4}=\dfrac{84}{4}

2x=\dfrac{339}{4}

x=\dfrac{339}{8}

Cool, right? Unsurprisingly, because math works, we landed on the same answer with both methods. But we’re not done yet, folks. There are still more ways to solve a system of linear equations!

Graphing

Always remember, as you’re taking the SAT, that when two graphs intersect, they do so at a point (x,y) that is a solution set for the two equations that make those graphs. If you’re in the section where calculators are allowed (and with the numbers in this question we’re playing with, you would be), then you have still more weapons for solving systems of equations at your disposal.  Let’s look at that question again.

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

To solve a system of equations by graphing, first get each equation into y= form:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\dfrac{1}{6}y=12-\dfrac{1}{3}x

y=72-2x

2x+5y=21

5y=21-2x

y=\dfrac{21}{5}-\dfrac{2}{5}x

Pop those into your calculator and graph! If your window is currently set to standard zoom, you’ll probably only see one line. You should expect this to happen from time to time—the writers of the SAT would love to see you doing the algebra, not taking graphing shortcuts. Try zooming out a bunch by setting your window to go from –100 to 100 in both axes, instead of the standard –10 to 10.

Now just use your calculator’s intersect function! On a TI-83 or TI-84, you’re going to hit [2nd][TRACE] to open up the CALC menu, and then select “intersect.” Now just hit ENTER on the first line (your calculator asks for the “first curve,” but that’s only because this same function will also find intersections of non-linear functions), then hit ENTER on the second line. The calculator asks for a guess next—you don’t need to do anything here but hit ENTER. You should see something like this:

Of course, those are the decimal values of the answers we already know are correct:

\left(\dfrac{339}{8},-\dfrac{51}{4}\right)=(42.375,-12.75)

Backsolving

Finally, a multiple choice question like this can be solved by backsolving. All you have to do is try answer choices by making sure they work in both equations. Only the correct answer choice will result in true outcomes when substituted into the given equations.

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

Here’s what happens when you backsolve with a wrong answer choice (I’ll use choice B):

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\dfrac{1}{3}(18)+\dfrac{1}{6}(36)=12

6 + 6=12

12=12

That worked for the first equation. But let’s see what happens in the second equation:

2x+5y=21

2(18)+5(36)=21

36 + 180 =21

216=21

Nope—that’s not true at all!

The right answer, though, will work beautifully:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\left(\dfrac{1}{3}\right)\left(\dfrac{339}{8}\right)+\left(\dfrac{1}{6}\right)\left(-\dfrac{51}{4}\right)=12

\dfrac{113}{8}-\dfrac{17}{8}=12

\dfrac{96}{8}=12

12=12

Yep!

2x+5y=21

2\left(\dfrac{339}{8}\right)+5\left(-\dfrac{51}{4}\right)=21

\dfrac{339}{4} - \dfrac{255}{4} =21

\dfrac{84}{4}=21

21=21

Yep again! That’s a good answer.

Of course, I’m showing a lot of intermediate steps you won’t need to do yourself. All you’ll want to do is type the substituted left-side of the equation into your calculator and see what you get. For example, you might enter

2(339/8)+5(-51/4)

When you hit ENTER and get 21, which is what the question said you should get, then you know that ordered pair works for that equation.

Now you try

Think you’ve got all this? I bet you do. But just so that we can both sleep a little better tonight, why don’t you try a few more questions. Because this is practice and all, you might consider trying to solve all of these questions all 4 ways.

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