LE 8 Traditional Math: Solving Systems of Equations by Graphing
Accelerated seventh grade and Math 8
In this lesson, the components of linear equations that have been presented previously come together in solving systems of linear equations. All the points on the graph of a linear equation represent all the solutions of that equation. This point, although stated on various occasions in this unit, becomes more obvious (or should become so) when we start to graph a system of two equations. The point where the lines intersect represents the x and y values common to the two equations.
Graphing the equations makes this point clear. I must confess, however, that I dislike this particular lesson because graphing to find the solution to a system of equations is time-consuming and frequently imprecise. That is the lesson that students discover and serves as an entre to subsequent lessons that provide instruction for more efficient techniques. I try not to prolong the agony, and assign very few problems, but enough so that some of the problems lend themselves to neat and easy solutions, while others do not. Students get the point quick enough.
As part of the lesson, I also focus on systems that have no solutions, (represented by parallel lines) and infinite solutions (represented by a single line).
Warm-Ups. The first problem is meant to allow for some instruction.
1. Solve for y. 2x + 4y = 8. (Hint; Divide ALL terms by the coefficient of y) Answer: 4y=-2x + 8; y = -1/2(x) + 2
2. Identify the slope and y-intercept of the equation. (Hint: divide ALL terms by the coefficient of y. 2x + 3y = -9 Answer: Solve for y:3y = -2x – 9; y=-2/3(x) – 3; m=-2/3, b = -3
3. Graph the equation -2x + y = -5 Answer: Solve for y: y = 2x+9; m=2, b =-5.
4. Simplify. 2/3(9x + 3) Answer: 2/3(9x) + 2/3(3) = 6x + 2
5. Find y if x = 1 for y = -3x + 4? Answer: y = -3(1) + 4 = 1
Students are used to solving equations for y when y has a coefficient of 1 (i.e., no coefficient). For Problems 1 and 2, y has a coefficient other than 1. For both problems, the y term is isolated. For Problem 1 we then have 4y = -2x + 8. Each term must then be divided by 4. For the x term this means it will be -2/4 (x) which reduces to ½(x). For Problem 2, it is similar. I walk them through Problem 1, and then have them work independently on Problem 2, but will provide guidance as necessary.
The Wind-Up: The Graph of a Linear Equation Represents its Solutions. Problem 5 of the Warm-Ups has students plugging in 1 for x to obtain the value for y. Referring to this problem, I point out that we have solutions for x and y: 1 and 1. “If we were to write this as an ordered pair, that is (x,y), how would we write it?”
If faced with blank stares I refrain from saying “It’s staring you in the face” or “It’s like asking who’s buried in Grant’s Tomb?” mainly because students may be overwhelmed despite the simplicity of the question, and secondly the response “Who’s Grant?” that I sometimes get is just as disturbing as their not seeing what the answer is.
“What did we say x and y equal?” Hearing 1 and 1, I continue: “So let’s plug this in to (x,y).” By now they’ll catch on and say it’s (1,1).
“Now suppose I ask how you can tell if the ordered pair (2,1) is on the line for y = -3x + 4?” If no response, I’ll prompt with “For (2,1) is 2 the x value or the y value?” Hearing “x”, I’ll ask what y equals when we plug in 2. The answer is -2. “Is (2, 1) on the line?”
No; for it to be on the line, we would have to get (2, -2).
“Let’s say we have the equation x+y=2; this is equivalent to y=-x+2. Now the graph of this line is made up of infinitely many points, each one an ordered pair of (x,y). That is there are many solutions to this equation; that is there are many ordered pairs that make the equation true. Is the point (1,1) on this line? Does it make the equation true? Work it out.” I’ll circulate around the room to see who is getting it. In fact, (1,1) is on the line since 1 = -1 + 2.
“Everyone graph the equation y = -x + 2. And plot the point (1,1) so you can see where it is on the line.”
Figure 1
“Now graph the equation x-y=0. Is (1,1) on that line?”
In fact, it is. “We can see that the point (1,1) is on both lines. Just to make sure, plug (1, 1) into x – y = 0 to see if it checks out.”
The Pitch: Systems of Equations. After graphing the equations I make the prounouncement that the point of intersection is the ordered pair that is on each of the lines. It is a common solution for both lines, or equations.
“Is it the only common solution for both lines?”
General consensus is always “yes”.
“This is a system of equations.” I write on the board:
System of Equations: A collection of two or more equations with the same set of variables.
“We can use graphing to find out the common solution for these equations. That is, we can s find values for x and y that make these equations true.” I wait and look over a sea of blank stares. “In other words, we can use graphing to solve for x and y.”
Examples. I have them find solutions to systems by graphing in their notebooks. The first one we do together, and the others they work independently while I go around to provide guidance. I announce answers as people progress, and show student work with a document camera.
1. x + y =4; 2x + y = 5 Answer: (1,3)
2. x+y = -1; y = 3x+7 Answer: (-2, 1)
3. y=-3x+2; y+3 = 2x Answer: (1, -1)
Sometimes the solution pair might have a fractional answer, or a pair of fractions instead of a pair of integers. I tell students that their answer does not have to be exact; just see how close they get to the actual answer. I will work with them to help them eyeball and estimate where the points are.
4. y= -x-1; y-4 = x; Answer: about (-2 ½ , 1 ½ )
5. y = ¼ x -3; y = -¼ x – 2; Answer: about (2, -2 ½ )
No Solution, Many Solutions. I have the students graph the following equations on the same set of axes:
“Do the graphs intersect?” In fact, they do not. They are parallel lines. “Sometimes a system of equations does not have a solution; if the equations have the same slope, which these do, they are parallel. Parallel lines do not intersect. So we say there is no solution.”
Next I write two equations on the board: y-4 = 2x and y-2x = 4. Some students, seeing that the slopes are the same will shout out “No solution; they’re parallel.” I reply “Take a closer look.”
Soon after someone else will say “They’re the same equation.” And in fact they are. When the equations are the same, then there is only one line. As discussed at the beginning of the lesson, a line represents all the possible solutions to the equation the line represents. “We say that there are infinitely many solutions.”
Homework. Problems will be similar to what was done in class. I tend to limit the number of problems to no more than ten.