Provide each student with an Entrance Ticket (M-8-2-3_Entrance Ticket.doc). Students will model selections by writing three equations and three inequalities. Allow ample time for completion. Discuss the meaning of variables; ask students to define variable and explain the role of a variable in an equation/inequality. Ask students to explain the validity of their equations/inequalities.
Activity 1: Menu
Display the following example of a menu:
Cup of Soup
“How would we find out what one sandwich costs if we know that two sandwiches cost $7.30? Let’s use s to stand for the cost of one sandwich. We could write the following equation and solve for s:
2s = 7.30
s = 3.65
One sandwich costs $3.65.”
“What is the cost of one carton of milk if we know that four cartons of milk minus the cost of one tossed salad equal $4.25? Let m stand for the cost of one carton of milk and t for the cost of a tossed salad. We could write the following equation and solve for m:
4m – t = 4.25 m represents 1 carton of milk; t represents tossed salad
4m – 3.55 = 4.25 Substitute the value for t given in the menu.
4m = 4.25 + 3.55 Add 3.55 to both sides.
4m = 7.80 Perform the operation.
m = 1.95 Divide both sides by 4.
One carton of milk costs $1.95.”
“Sometimes we do not know the cost of either item, but are given the quantities of each and the total. For example, we might know that the cost of 5 rolls and 3 sandwiches equals $15.00. We would write the equation as 5r + 3s = 15.”
“What solution or ordered pair (r, s) would make this equation true? In other words, what values substituted for r and s would make the equation true?”
Have students discuss the possibilities. Some possibilities are
r = 1.05; s = 3.25 or (1.05, 3.25)
“Substituting back into the equation, we have
5(1.05) + 3(3.25) = 15
15 = 15
“Thus, the costs for a roll and sandwich could be $1.05 and $3.25, respectively.”
“If we determine a value for one variable, we can solve the equation for the other variable, so if we guessed that r = 1.05, then we can substitute for r and solve for s:
5r + 3s = 15
5(1.05) + 3s = 15 Substitute value of r.
5.25 + 3s = 15 Perform the operation.
3s = 15 – 5.25 Subtract 5.25 from each side.
3s = 9.75 Perform the operation.
s = 3.25 Divide each side by 3.”
This procedure can be used to find solutions of an equation with two variables. Lead students through the process of finding three more solutions and discuss the idea of infinitely many solutions.
Some other solutions include:
“Let’s substitute $2.25 for s and solve for r:
5r + 3s = 15
5r + 3(2.25) = 15 Substitute the value of s.
5r + 6.75 = 15 Perform the operation.
5r = 15 – 6.75 Subtract 6.75 from both sides.
5r = 8.25 Perform the operation.
r = 1.65 Divide each side by 5.
Thus . . .
r = 1.65; s = 2.25 the solution is (1.65, 2.25).”
“Let’s substitute $1.35 for r and solve for s:
5r + 3s = 15
5(1.35) + 3s = 15 Substitute the value of r.
6.75 + 3s = 15 Perform the operation.
3s = 15 – 6.75 Subtract 6.75 from both sides.
3s = 8.25 Perform the operation.
s = 2.75 Divide each side by 3.
Thus . . .
r = 1.35; s = 2.75 the solution is (1.35, 2.75).”
“Let’s substitute $3.85 for s and solve for r:
5r + 3s = 15
5r + 3(3.85) = 15 Substitute the value of s,
5r + 11.55 = 15 Perform the operation.
5r = 15 – 11.55 Subtract 11.55 from both sides.
5r = 3.45 Perform the operation.
r = 0.69 Divide each side by 5.
Thus . . .
r = 0.69; s = 3.85 the solution is (0.69, 3.86).”
“We can also examine solutions to an inequality using the menu.”
“Suppose we know the cost of four cartons of milk and three sandwiches is less than or equal to $15.50. How could we write this scenario using an inequality? How could we find possible solutions?”
“We can write the inequality as…”
“To find possible solutions, we will employ the same process as before. Let’s substitute a value for one of the variables and then solve for the other variable.”
“We will try a cost of $3.50 for a sandwich and substitute this value for s:”
“A possible solution is (1.25, 3.50). In other words, a carton of milk must cost $1.25 or less than $1.25 AND a sandwich must cost $3.50 or less than $3.50 for the total to be less than or equal to $15.50. Solving this type of problem occurs in everyday life as individuals determine whether or not they have enough money to purchase items from a menu.”
Have students find three more solutions to this inequality. Then ask them to write two more inequalities from the menu and find four solutions. Hold a class discussion reviewing equations, inequalities, variables, values substituted, and the relationship of each to one another.
Rewriting Equations and Inequalities in Slope-Intercept Form
“We have looked at several equations with different letter variables. Let’s look at some equations in two specific variables, x and y, and explore the slope-intercept form of an equation. The slope-intercept form of an equation is defined as a linear equation in the form of y = mx + b, where the value m represents the slope, or constant rate of change, and the value b represents the y-intercept, or starting point.” (Note: this brief definition assumes that students have already been exposed to slope-intercept form, and the terms “slope” and
“y-intercept.” If this is not the case, more discussion should take place here.)
“Consider the equation 4x + y = 15. Is this equation written in slope-intercept form?” (No.)
“This equation is not written in slope-intercept form. The slope-intercept form of an equation is solved for y. In other words, y is isolated. Why might we want to transform this equation so that it is written in slope-intercept form?” (Equations written in slope-intercept form easily show us valuable information, namely the slope and y-intercept of the graph or problem.)
“To determine the slope and y-intercept of the previous equation, we will transform it so that it is written in slope-intercept form. To do so, we must solve the equation for y. We can do this by subtracting 4x from each side. This process gives us:
y = -4x + 15 slope-intercept form.”
“Now that the equation is in slope-intercept form, we can easily determine the slope and y-intercept. What is the slope?” (−4) “What is the y-intercept?” (15)
“Let’s write another equation in slope-intercept form:
5x + 3y = 15
3y = -5x + 15 Subtract 5x from each side.
Divide each side by 3.”
“Sometimes, we have linear inequalities in two variables, x and y. In the same way that it is beneficial for us to transform such equations into slope-intercept form, we may also want to solve these types of inequalities for y. Consider the inequality -2x + y < 12.”
“What could we do to rearrange this inequality so that y is isolated?” (Add 2x to both sides.)
“By adding 2x to both sides, we obtain y < 2x + 12.”
“Let’s look at another one:
-3x + 18y ≥ 6
18y ≥ 3x + 6 Add 3x to both sides.
y ≥ 3/18x + 6/18 Divide both sides by 18.
y ≥ 1/6x + 1/3 Reduce all fractions.”
Activity 2: Memory Game
Prior to the lesson, cut out the Memory Cards (M-8-2-3_Memory Cards.docx) and place them in an envelope. Prepare enough envelopes for students to play in pairs or small groups and distribute accordingly. Instruct students to place each card face down and mix them around so the locations of specific cards are unknown. Students should take turns flipping two cards over at a time. If the cards flipped show equivalent equations or inequalities, the student wins this pair. If the cards do not match, the student flips them back over and play continues with the next student in the group. The game ends when all cards have been matched. The student with the most pairs wins.
Graphing Linear Equations and Inequalities in Two Variables
“We will now graph some linear equations and inequalities in two variables. You graphed linear equations and inequalities in one variable in Lesson 1 of this unit using a number line. In two variables, we graph equations and inequalities on a coordinate plane.”
“Consider the equation y – 5x = −10. Earlier, we discussed the benefit of rewriting equations in slope-intercept form. Doing this also makes these equations much easier to graph.”
“Let’s first solve for y:
y – 5x = -10 Solve for y if the equation is not already stated in slope-intercept form.
y = 5x – 10 Add 5x to each side.”
“Now that the equation is in slope-intercept form, how can we graph the corresponding line on a coordinate plane?” (First plot the y-intercept at (0, −10). Then count ‘up five, over one’ from point to point.) (Note: Here, it is assumed that students have graphed lines before. If not, more discussion will be needed.)
Demonstrate graph of line for students.
“Remember that the graph of a line shows all the sets of ordered pairs (x, y) that satisfy the equation.”
“Now let’s look at what happens to the similar inequality, y – 5x < −10. Use the same operations as tools in the same way to solve the inequality.
y – 5x < -10 Solve for y.
y < 5x – 10 Add 5x to each side.”
y – 5x ≤ -10 “Solve for y.
y ≤ 5x - 10 Add 5x to each side.”
y – 5x > -10 “Solve for y.
y > 5x – 10 Add 5x to each side.”
y – 5x ≥ -10 “Solve for y.
y ≥ 5x – 10 Add 5x to each side.”
“Like the graph of a linear equation, the graph of a linear inequality needs to show all the sets of ordered pairs (x, y) that satisfy the inequality.”
Instruct students to write down three ordered pairs that satisfy the inequality. When students have done this, ask them to write their ordered pairs on the board. Once all example solutions are written on the board, plot the corresponding points on a large coordinate grid for students to see.
“What do you notice about the graph of this inequality? Do all the ordered pairs line up like in the graph of an equation?” (No. An inequality has many more solutions than an equation.)
“As you can see, the graph of a linear inequality consists of more than just a line. Instead, when we graph a linear inequality, we are actually graphing a boundary line and a half plane of ordered pairs. The boundary line is the line we would graph if the inequality were actually an equation (or what happens when we replace the inequality symbol with an equal sign). The half plane refers to the half of the coordinate plane that holds the solutions to the inequality. This half will either be above or below the boundary line. We shade the half plane that holds the solutions to our inequality. Let’s use our inequality as an example.”
“First, we graph the boundary line.”
Graph the line y = 5x – 10 on a coordinate plane for students to see.
“Sometimes the boundary line will be solid, and sometimes it will be dotted or dashed. This depends on whether or not the ordered pairs of the boundary line are included in the solution set to the inequality. What are some ways we might be able to determine if the ordered pairs on the boundary line are included in the solution set to the inequality?” (Students may suggest testing an ordered pair from the boundary line by substituting these x and y values into the inequality. They should discover that in this case, the boundary line IS included in the graph. Lead them to discover that an easy way to know this is the simple fact that the inequality symbol used is a greater than OR EQUAL TO symbol.)
“Because our inequality uses a greater than or equal to symbol, the boundary line is included in the graph and should remain solid. Now, we need to determine the appropriate half plane to shade. This will either be ‘above’ or ‘below’ the line. How might we determine which half plane to shade?” (Students may suggest testing an ordered pair from either half plane and seeing which one has x and y values which satisfy the inequality. Lead them to discover that if a test point correctly satisfies the inequality, the half plane containing this test point is included in the graph of the inequality and should be shaded.)
Shade the half plane on the coordinate grid for students to see.
“So, to recap, to graph a linear inequality in two variables, we follow these three steps:
- 1. Graph it as an equation, and choose a dotted line for < or > and a solid line for ≤ or ≥.
- 2. Pick a test point (ordered pair) and substitute into the original inequality.
- 3. Shade the half plane that contains the point if the point is indeed a solution. If the point is not a solution, shade the other half plane.”
Look at the following additional examples with the class:
Two-Step Linear Inequalities in Two Variables
“Let’s look at a couple more examples:
2y + 6x ≤ -6
2y ≤ -6x – 6 Isolate the y term by subtracting 6x from each side.
y ≤ -3x – 3 Divide each term on both sides of the equation by 2.”
“To graph the inequality, we will need to first replace the inequality with an equal sign and graph it as an equation. We notice the inequality is ≤, so our line will be solid. We will use the test point (0, 0) to determine which half-plane to shade. Note that any test point would work. The point (0, 0) results in a false statement, so we will shade the half-plane that does not contain the point (0, 0).”
“Our next example is unique because it has a fraction for the slope:
−7x + 14y > 28
14y > 7x + 28 Isolate the y term by adding 7x to each side.
Divide each term on both sides of the equation by 14.”
“To graph the inequality, we will need to first replace the inequality with an equal sign and graph it as an equation. We notice the inequality is >, so our line will be dotted. We will use the same test point of (0, 0) to determine which half plane to shade. The point (0, 0) results in a false statement, so we will shade the half plane that does not contain the point (0, 0).”
Activity 3: Graphing Linear Inequalities
Distribute Graphing Inequalities (M-8-2-3_Graphing Inequalities.docx and M-8-2-3_Graphing Inequalities KEY.docx). Instruct students to complete it. Walk around the room as students work to answer questions and ensure that all students are on the right track.
- Briefly discuss graphing a system of linear inequalities. This involves graphing two inequalities on the same coordinate plane. The solution of the system is where the solution regions of each individual inequality overlap. Provide students with the Systems Extension worksheet (M-8-2-3_Systems Extension.docx and M-8-2-3_Systems Extension KEY.docx). This worksheet presents four examples of graphing a linear system of inequalities in two variables. It is a natural follow up to this lesson and can also be used as a review.