The new SAT has shifted the focus of the math questions away from geometry and more toward algebra and functions. One of the topics you’ll want to master is systems of linear equations. If the practice tests are any indication, you can expect to see four or five of these questions on each test.
Below you’ll find a link to 10 sample system of equations questions (and a second link to the solutions).
But first, let’s do a little review.
Most systems of linear equations have a single ordered pair as their solution. It’s the point where the two lines intersect, and you can find the coordinates of that point by graphing the lines on your calculator or using the substitution or elimination methods. But what about the special cases? Let’s take a look at systems that have either an infinite number of solutions or no solution at all.
A system of two linear equations will have an infinite number of solutions if the two equations represent the same line. That’s easy to recognize when the lines are in slope-intercept (y = mx + b) form, but what do we do when they are in standard form?
Let’s look at a sample system. Suppose that you are asked to find the solution to:
4x + 10y = 6
14x + 35y = 21
If you use the elimination method, you’ll try to get the coefficients of one of the variables to be the same and then you’ll subtract the two equations (or get the coefficients to be opposites and add the equations). For this problem, that might look something like this:
7(4x + 10y = 6)
2(14x + 35y = 21)
Multiplying these out gets us:
28x + 70y = 42
28x + 70y = 42
We can now clearly see that these two equations represent the exact same line! This system has an infinite number of solutions (all the points on that line).
A system of linear equations will have no solution if the two lines are parallel. If the equations are in slope-intercept form, you’ll be able to see that they have the same slope and different y-intercepts. Let’s look at what happens if the equations are in standard form. Suppose you want to solve this system:
6x – 10y = 8
9x – 15y = 10
Once again, we’ll multiply one or both of the equations so that we can get one set of coefficients to be the same (or opposites if you like to add the equations).
3(6x – 10y = 8)
2(9x – 15y = 10)
18x – 30y = 24
18x – 30y = 20
What do we notice about this system? The left side of both equations is the same, but the number on the right side of the equations is different. But that can’t be! The quantity on the left, 18x – 30y, can’t equal two different things. We can conclude that the system has no solution and the lines are parallel.
Below you’ll find links to 10 practice SAT problems that involve systems of equations. The first seven are “No Calculator” while the last three can be done with your calculator in hand. Give them a try!
If you have questions about these problems or anything to do with the New SAT, send us an email at firstname.lastname@example.org .