In order to help eighth graders and their parents prepare for the PARCC math test in Maryland and Illinois, Voxitatis will occasionally describe math problems that have been released into the public domain by the states that are part of the PARCC consortium. This is question 6 from the 2016 released item set.
Four systems of equations are shown. Indicate whether each system of equations has no solution, one solution, or infinitely many solutions.
The first system and third system have no solution. The second system has one solution, and the fourth system has infinitely many solutions.
According to the Common Core standards for eighth-grade math, under “expressions and equations,” students are required to be able to “analyze and solve pairs of simultaneous linear equations,” including the ability to:
- Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously.
- Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6.
As I say, there are multiple strategies for solving this type of problem in algebraic thinking that you might see on the PARCC test for eighth-grade math. Since I have four systems of two linear equations to figure out, I’ll do it four different ways.
System 1: Solve by graphing
If I plot the two lines (equations in the form y = mx + b are guaranteed to be lines when plotted on a coordinate grid), I can see if they are the same exact line, in which cases there are infinitely many solutions, they intersect, in which case there is exactly one solution, or they are parallel, in which case there are no solutions for the system.
As you can see, I got two parallel lines, which means the first system has no solutions. Note that the y intercept is given directly when the equations are in this form, as is the slope. With the same slope, the lines are parallel, and since the y intercept is different for the two lines, they will never meet.
System 2: Solve by slope-intercept
By inspecting the two equations, both in the standard slope-intercept format, y = mx + b, we can see that the slope of the top equation is positive 2 and the slope of the bottom equation is negative 2. This means if we were to plot the two lines on the same coordinate plane, they would make an X: one line starts at the top left somewhere and goes downward, while the other starts at the bottom left and goes upward.
Somewhere they will meet since lines keep going forever. The problem doesn’t ask for exactly where they will meet, so all I have to do is figure out how many times they will meet. That is, they will intersect each other at exactly one point, and X marks the spot.
System 3: Solve by inspection
Just looking at the two equations, I might notice that I have the same thing on the left side of both yet a different thing on the right. Is it possible that some expression—here, 3x + 2y—could equal 2 sometimes and 5 other times, given that the x and y have to be equal to the same thing in both equations (that’s what a system of equations means)?
No, that is not possible. These lines will never meet, because the system describes an impossible situation. The same expression with the same set of variables can only equal one thing at a time, and this system would require that expression to equal two things at the same time. That can’t happen; it’s a mathematical impossibility. So, no solution for this one.
System 4: Solve by solving the system
One way to solve a system of equations like this is to determine the actual point, if any, at which the two lines will intersect. That means, most often, that the best solution strategy is just to solve for y in the top equation and plug that value into the y in the bottom equation. Thus, y in the top equation is found like this:
Then we substitute that expression for y in the bottom equation as follows:
The final statement, 0x = 0 is true for all values of x. For example, if x is 3, the equation becomes 0×3=0; if x is 100, the equation becomes 0×100=0, which is also true.
No matter what x is, the system of equations will have a solution for y, and that means the two lines will intersect at infinitely many points. Any time you solve the system by substitution like this and you get a statement that is always true, there are infinitely many solutions.
Method 4 there requires a little more algebra than the others, but it has the benefit of always determining a solution. If the result for x is a single value, the system has one solution, and if the solving results in a statement that can’t possibly be true, such as 6=5, the system of equations would have no solution.
Also note, that while I picked only one solution strategy for each system of equations, any of the strategies (or a few others) would have worked on any of the systems. You can mix and match, and if you have enough time, try two on the same system of equations, just to check your work.
For example, it would have been easy to notice in the last system that the two equations are simply multiples of each other. PARCC set them up nicely like that: If you multiply every term in the top equation by 2, you’ll get the bottom equation. Any time you can do that, using any number as a multiple for one equation to get the other, the system has infinitely many solutions because both equations describe the same line.