Lesson 11
Connecting Equations to Graphs (Part 2)
11.1: Rewrite These! (5 minutes)
Warmup
In this warmup, students review how to apply the distributive property to rewrite expressions that involve division, preparing them to do so in the next activity in the lesson.
Launch
Arrange students in groups of 2. Give students a moment of quiet time to work on the first two questions and then time to discuss their responses with their partner before moving on the last two questions.
Student Facing
Rewrite each quotient as a sum or a difference.
 \(\dfrac {4x10}{2}\)
 \(\dfrac {1  50x}{\text2}\)
 \(\dfrac {5(x+10)}{25}\)
 \(\dfrac {\text \frac15 x + 5}{2}\)
Student Response
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Anticipated Misconceptions
Expect some students to give \(2x10\) or \(4x5\) as an answer to the first question. To illustrate why these are incorrect, take an example like \(\frac{4+6}{2}\). Explain that we know that 10 divided by 2 is 5, but if we divide only the 4 or only the 6 by 2 we won’t get 5. Alternatively, remind students that fraction bars can be interpreted as division, so each expression can be rewritten as, say, \((4x10) \div 2\), and we can apply the distributive property.
The signs of the numbers in the second expression might be a source of confusion. Students might be unsure if the expression should be \(\text  \frac12 25x\), \( \frac12 + 25x\), or another expression. Encourage students to substitute a number into the original expression, then try it in each potential answer. To explain why \(\text \frac12 + 25x\) is correct, appeal to the distributive property again.
Activity Synthesis
Invite students to share their equivalent expressions and how they reasoned about them. Display their expressions for all to see. Make sure students see that rewriting the expressions as sums or differences involves distributing the division (or applying the distributive property on division).
If not mentioned in students' explanations, point out each division could be thought of in terms of multiplication. For example, \(\frac {4x10}{2}\) is equivalent to \(\frac12 (4x10)\), because dividing by a number (in this case, 2) gives the same result as multiplying by the reciprocal of that number (in this case, \(\frac12\)). Applying the distributive property of multiplication to \(\frac12 (4x10)\) enables us to rewrite this product as a difference.
11.2: Graphs of Two Equations (15 minutes)
Activity
This activity reinforces the understandings that students began to develop in an earlier lesson about the connections between the structure of twovariable linear equations, their graphs, and the situations they represent.
Students first practice relating the parameters of an equation in slopeintercept form to the features of the graph and interpreting them in terms of the situation (MP2). Next, they practice making a case for how they know that a graph represents an equation given in standard form.
Some students may argue that substituting the \((x,y)\) pair of any point on the line gives a true statement, suggesting that the graph does match the equation. Or they may reason about the points on the graph in terms of almonds and figs and come to the same conclusion. For example, \((8,3)\) and \((11,1)\) are points on the line. If Clare buys 8 pounds of almonds and 3 pounds of figs, or 11 pounds of almonds and 1 pound of figs, the price is $75.
Ask these students how they would check whether the points with fractional \(x\) and \(y\)values (which are harder to identify precisely from the graph) would also produce true statements when those values are substituted. Use this difficulty to motivate rearranging the equation into slopeintercept form.
The work in this activity requires students to reason quantitatively and abstractly about the equation and the graph (MP2) and to construct a logical argument (MP3).
Launch
Display the two graphs in the task statement for all to see. Tell students the graphs represent two situations they have seen in earlier activities.
Arrange students in groups of 2. Give students a minute of quiet think time and ask them to be prepared to share at least one thing they notice and one thing they wonder about the graphs. Give them another minute to discuss their observations and questions with their partner before moving on to the task.
Design Principle(s): Support sensemaking
Student Facing
Here are two graphs that represent situations you have seen in earlier activities.

The first graph represents \(a=45020t\), which describes the relationship between gallons of water in a tank and time in minutes.
 Where on the graph can we see the 450? Where can we see the 20?
 What do these numbers mean in this situation?

The second graph represents \(6x + 9y = 75\). It describes the relationship between pounds of almonds and figs and the dollar amount Clare spent on them.
Suppose a classmate says, “I am not sure the graph represents \(6x + 9y = 75\) because I don’t see the 6, 9, or 75 on the graph.” How would you show your classmate that the graph indeed represents this equation?
Student Response
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Activity Synthesis
Focus the discussion on students' explanations for the last question. If no one mentions that \(6x+9y=75\) can be rearranged into an equivalent equation, \(y=8\frac13 \frac23 x\), point this out. (Demonstrate the rearrangement process, if needed.)
Ask students if we can now see the \(8\frac13\) and the \(\text \frac23\) on the graph and if so, where they are visible. To help students connect these values back to the quantities in the situation, ask what each value tells us about almonds and figs. Make sure students see that the \(8 \frac13\) tells us that if Clare bought no almonds, she can buy \(8\frac13\) pounds of figs. For every pound of almonds she buys, she can buy less figs—\(\frac23\) pound less, to be exact.
11.3: Slope Match (15 minutes)
Activity
Previously, students have studied the structure of equations concretely and contextually. In this activity, they shift to reasoning symbolically and abstractly about linear equations in two variables.
Students are prompted to match equations in standard forms to pairs of slopes and \(y\)intercepts. They could go about making the matches in various ways. Here are a few likely strategies, from less reliant on structure to more reliant on structure:
 Substituting the coordinates of the given \(y\)intercepts into each equation and see which one produces true equations.
 Calculating the \(y\)values when \(x\) is 0 and \(x\) is 1. The former gives the \(y\)intercept. Subtracting the latter from the former gives the slope.
 Graphing each equation (either by hand or using technology) and analyzing the graph.
 Rearranging the equation into slopeintercept form, \(y=mx+b\), and identifying the \(m\) and \(b\).
Some students may notice patterns from manipulating and graphing equations in the past few lessons. For instance, they may observe that when \(A\), \(B\), and \(C\) in \(Ax + By = C\) are positive, the graph always slants down from left to right and therefore has a negative slope.
Others may notice that when isolating \(y\) in an equation in standard form, the constant term in the resulting equation is \(\frac{C}{B}\) and the coefficient of \(x\) is \(\frac{\text A}{B}\), and that these values tell them the vertical intercept and the slope. In addition to making use of structure (MP7), students who make and apply these observations also practice expressing regularity through repeated reasoning (MP8).
Making graphing technology available gives students an opportunity to choose appropriate tools strategically (MP5).
As students work, monitor the strategies they use. Encourage those who approach the task by substituting values into the equations to consider rearranging equations or otherwise using the structure of the equations.
Launch
Keep students in groups of 2. Consider asking students to take turns finding a match and explaining their strategy to their partner.
Design Principle(s): Support sensemaking; Maximize metaawareness
Supports accessibility for: Organization; Attention
Student Facing
Match each of the equations with the slope \(m\) and \(y\)intercept of its graph.
 \(\text4x+3y=3\)
 \(12x4y=8\)
 \(8x+2y=16\)
 \(\textx+\frac 13 y=\frac 13\)
 \(\text4x + 3y = \text6\)
A: \(m=3\), \(y\text{int}=(0,1)\)
B: \(m=\frac 43\), \(y\text{int}=(0,1)\)
C: \(m=\frac 43\), \(y\text{int}=(0,\text2)\)
D: \(m=\text4\), \(y\text{int}=(0,8)\)
E: \(m=3\), \(y\text{int}=(0,\text2)\)
Student Response
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Student Facing
Are you ready for more?
Each equation in the statement is in the form \(Ax+By=C\).
 For each equation, graph the equation and on the same coordinate plane graph the line passing through \((0,0)\) and \((A,B)\). What is true about each pair of lines?
 What are the coordinates of the \(x\)intercept and \(y\)intercept in terms of \(A\), \(B\), and \(C\)?
Student Response
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Anticipated Misconceptions
Students will likely use the strategy of rewriting the equations in slopeintercept form. Common mistakes here include isolating \(x\) rather than \(y\), changing the sign of only one term when dividing by a negative number, and dividing only one of two terms by the coefficient of \(y\). (For these last two mistakes, remind students of the work in the warmup).
Students who recognize that the slope of a line with equation \(Ax+By=C\) is \(\frac{\text A}{B}\) and that the \(y\)intercept is \(\frac{C}{B}\)may also write the wrong signs, or get a ratio reversed. Students who use this strategy are likely shortcutting the process of isolating \(y\). Asking them to isolate \(y\) for one equation can help them to identify errors.
Activity Synthesis
Select students to present their strategies. Sequence the order of presentation as listed in the Activity Narrative. After each strategy is presented, ask if other students approached it the same way and consider discussing questions such as:
 "How would you describe the matching process or the strategy? Was it fairly efficient or laborious? Was it prone to errors?"
 "If the list of slopes and \(y\)intercepts were not available for you to choose from, would you still be able to determine the slope and \(y\)intercept?" (Yes, except those who rely on the first strategy—using the coordinate values of each \(y\)intercept on the list to test the equations.)
 (For students who use graphing technology:) "Would you still use graphs to make the matches if the graphing needed to be done by hand?"
Highlight that it is helpful and efficient to use the structure of an equation to get insights about the properties of its graph. At this stage, it is not essential that students recognize that the slope of an equation of the form \(Ax+By=C\) is \(\frac{\text A}{B}\) and that it crosses the \(y\)axis at \(\frac{C}{B}\). Students should, however, recognize that solving for \(y\) involves a predictable process and the resulting equation makes the slope and \(y\)intercept visible.
Lesson Synthesis
Lesson Synthesis
Display the following description and graphs for all to see.
Suppose Clare went back to the store to get more almonds and dried figs and spent \$108 this time. Almonds cost \$6 a pound and dried figs cost \$9 a pound. Clare's purchase can be represented by the equation \(6x+9y=108\).
Here are two graphs that represent the relationship between pounds of almonds, \(x\), and pounds of figs, \(y\).
Discuss with students:
 "Without calculating anything, can you tell which graph could represent the equation \(6x+9y=108\) and which graph could not?" (Yes, graph B could represent the equation. The graph of the equation would have a negative slope because the more almonds Clare bought, the fewer dried figs she could have afforded. Graph A shows that as more almonds are bought, more figs are also bought, which couldn't be true if Clare spent a fixed amount of money.)
 "What does the vertical intercept mean in this situation?" (It shows the pounds of dried figs Clare could buy if she bought no almonds.)
 "What is the vertical intercept of the graph? How can we find it?" (The vertical intercept is \((0,12)\). One way to find it is to substitute 0 for \(x\) and solve for \(y\), which gives \(y=12\). Another way is to rearrange the equation into slopeintercept form.)
 "What is the slope of the graph? How can we find it?" (The slope is \(\text\frac23\). We can rearrange the equation into slopeintercept form, or we can find the coordinates of another point and calculate the slope.)
 "What does the slope tell us about the almonds and dried figs?" (It tells us that for every additional pound of almonds that Clare buys, she could buy \(\frac23\) fewer pound of dried figs.)
11.4: Cooldown  Features of a Graph (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
Here are two situations and two equations that represent them.
Situation 1: Mai receives a \$40 bus pass. Each school day, she spends \$2.50 to travel to and from school.
Let \(d\) be the number of school days since Mai receives a pass and \(b\) the balance or dollar amount remaining on the pass.
Situation 2: A student club is raising money by selling popcorn and iced tea. The club is charging \$3 per bag of popcorn and \$1.50 per cup of iced tea, and plans to make \$60.
Let \(p\) be the bags of popcorn sold and \(t\) the cups of iced tea sold.
\(b = 402.50d\)
\(3p + 1.50t = 60\)
Here are graphs of the equations. On each graph, the coordinates of some points are shown.
The 40 in the first equation can be observed on the graph and the 2.50 can be found with a quick calculation. The graph intersects the vertical axis at 40 and the 2.50 is the slope of the line. Every time \(d\) increases by 1, \(b\) decreases by 2.50. In other words, with each passing school day, the dollar amount in Mai's bus pass drops by 2.50.
The numbers in the second equation are not as apparent on the graph. The values where the line intersects the vertical and horizontal axes, 40 and 20, are not in the equation. We can, however, reason about where they come from.
 If \(p\) is 0 (no popcorn is sold), the club would need to sell 40 cups of iced tea to make \$60 because \(40 (1.50) = 60\).
 If \(t\) is 0 (no iced tea is sold), the club would need to sell 20 bags of popcorn to make \$60 because \(20 (3) = 60\).
What about the slope of the second graph? We can compute it from the graph, but it is not shown in the equation \(3p+1.50t = 60\).
Notice that in the first equation, the variable \(b\) was isolated. Let’s rewrite the second equation and isolate \(t\):
\(\begin {align} 3p + 1.50t &= 60\\ 1.50t &= 60  3p\\ t &= \dfrac{603p}{1.50}\\ t &=40  2p \end{align}\)
Now the numbers in the equation can be more easily related to the graph: The 40 is where the graph intersects the vertical axis and the 2 is the slope. The slope tells us that as \(p\) increases by 1, \(t\) falls by 2. In other words, for every additional bag of popcorn sold, the club can sell 2 fewer cups of iced tea.