SAT Math Prep: Solving Systems of Linear Equations

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The SAT is one of the most important tests you’ll take in your academic career, and folks often dread the math section in particular. If solving systems of linear equations is your idea of a nightmare and finding a best-fit equation for a scatter plot makes you feel scatter-brained, this is the guide for you. The SAT math sections are a challenge, but they’re easy enough to master if you handle your preparation right.

Get to Grips with the SAT Math Test

The math SAT questions are broken up into a 25-minute section that you can’t use a calculator for and a 55 minute section that you can use a calculator for. There are 58 questions in total and 80 minutes to complete them in, and most are multiple choice. The questions are loosely ordered by least difficult to most difficult. It’s best to familiarize yourself with the structure and format of the question paper and the answer sheets (see Resources) before you take the test.

On a larger scale, the SAT Math Test is divided into three separate content areas: Heart of Algebra, Problem Solving and Data Analysis, and Passport to Advanced Math.

Today we'll look at the first component: Heart of Algebra.

Heart of Algebra: Practice Problem

For the Heart of Algebra section, the SAT covers key topics in algebra and generally relate to simple linear functions or inequalities. One of the more challenging aspects of this section is solving systems of linear equations.

Here is an example system of equations. You need to find values for x and y:

\begin{alignedat}{2} 3&x+ &\;&y = 6 \\ 4&x-&3&y = -5 \end{alignedat}

And potential answers are:

a) (1, −3)
b) (4, 6)
c) (1, 3)
d) (−2, 5)

Try to solve this problem before reading on for the solution. Remember, you can solve systems of linear equations using the substitution method or the elimination method. You could also test each potential answer in the equations and see which one works.

The solution can be found using either method, but this example uses elimination. Looking at the equations:

\begin{alignedat}{2} 3&x+ &\;&y = 6 \\ 4&x-&3&y = -5 \end{alignedat}

Note that y appears in the first and −3_y_ appears in the second. Multiplying the first equation by 3 gives:

9x+3y=18

This can now be added to the second equation to eliminate the 3_y_ terms and leave:

(4x + 9x) + (3y-3y) = (– 5 + 18)

So...

13x=13

This is easy to solve. Dividing both sides by 13 leaves:

x=1

This value for x can be substituted into either equation to solve. Using the first gives:

(3 × 1) + y = 6

So

3 + y = 6

Or

y = 6 – 3 = 3

So the solution is (1, 3), which is option c).

Some Useful Tips

In math, the best way to learn is often by doing. The best advice is to use practice papers, and if you make a mistake on any questions, work out exactly where you went wrong and what you should have done instead, rather than simply looking up the answer.

It also helps to work out what your main issue is: Do you struggle with the content, or do you know the math but struggle to answer the questions in time? You can do a practice SAT and give yourself extra time if needed to work this out.

If you get the answers right but only with extra time, focus your revision on practicing solving problems quickly. If you struggle with getting answers right, identify areas where you’re struggling and go over the material again.

Check Out for Part II

Ready to tackle some practice problems for Passport to Advanced Math and Problem Solving and Data Analysis? Check out Part II of our SAT Math Prep series.

References

About the Author

Lee Johnson is a freelance writer and science enthusiast, with a passion for distilling complex concepts into simple, digestible language. He's written about science for several websites including eHow UK and WiseGeek, mainly covering physics and astronomy. He was also a science blogger for Elements Behavioral Health's blog network for five years. He studied physics at the Open University and graduated in 2018.

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