Lesson 13

More Solutions to Linear Equations

13.1: Coordinate Pairs (5 minutes)


The purpose of this warm-up is for students to practice solving an equation for an unknown value while thinking about a coordinate pair, \((x,y)\), that makes the equation true. While the steps to solve the equation are the same regardless of which value of \(x\) students choose, there are strategic choices that make solving the resulting equation simpler. This should be highlighted in the discussion.


Encourage students to not pick 0 for \(x\) each time.

Student Facing

For each equation choose a value for \(x\) and then solve to find the corresponding \(y\) value that makes that equation true.

  1. \(6x=7y\)
  2. \(5x+3y=9\)
  3. \(y+5-\frac13 x=7\)

Student Response

For access, consult one of our IM Certified Partners.

Activity Synthesis

Collect the pairs of \(x\)’s and \(y\)’s students calculated and graph them on a set of axes. For each equation, they form a different line. Have students share how they picked their \(x\) values. For example:

  • For the first problem, choosing \(x\) to be a multiple of 7 makes \(y\) an integer.
  • For the last problem, picking \(x\) to be a multiple of 3 makes \(y\) an integer.

13.2: True or False: Solutions in the Coordinate Plane (15 minutes)


In the previous lesson, students studied the set of solutions to a linear equation, the set of all values of \(x\) and \(y\) that make the linear equation true. They identified that this was a line in the coordinate plane. In this activity, they are given graphs of lines and then are asked whether or not different \(x\)-\(y\) coordinate pairs are solutions to equations that define the lines. This helps students solidify their understanding of the relationship between a linear equation and its graph in the coordinate plane.

Consider asking students to work on the first 5 problems only if time is an issue. Since this activity largely reinforces the material of the previous lesson, it is not essential to do all 8 problems.


Arrange students in groups of 2. Students work through the eight statements individually and then compare and discuss their answers with their partner. Tell students that if they disagree, they should work to come to an agreement.

Action and Expression: Internalize Executive Functions. Check for understanding by inviting students to rephrase directions in their own words. Provide the following sentence frame to support student explanations: “Statement ___ is true/false because . . .”
Supports accessibility for: Organization; Attention

Student Facing

Here are graphs representing three linear relationships. These relationships could also be represented with equations.

Graph, origin O, no grid. Lines l, m, n. 

For each statement below, decide if it is true or false. Explain your reasoning.

  1. \((4,0)\) is a solution of the equation for line \(m\).

  2. The coordinates of the point \(G\) make both the equation for line \(m\) and the equation for line \(n\) true.

  3. \(x = 0\) is a solution of the equation for line \(n\).

  4. \((2,0)\) makes both the equation for line \(m\) and the equation for line \(n\) true.

  5. There is no solution for the equation for line \(\ell\) that has \(y = 0\).

  6. The coordinates of point \(H\) are solutions to the equation for line \(\ell\).

  7. There are exactly two solutions of the equation for line \(\ell\).

  8. There is a point whose coordinates make the equations of all three lines true.

After you finish discussing the eight statements, find another group and check your answers against theirs. Discuss any disagreements.

Student Response

For access, consult one of our IM Certified Partners.

Anticipated Misconceptions

Some students may try to find equations for the lines. There is not enough information to accurately find these equations, and it is not necessary since the questions only require understanding that a coordinate pair lies on a line when it gives a solution to the corresponding linear equation. Ask these students if they can answer the questions without finding equations for the lines.

Activity Synthesis

Display the correct answer to each question, and give students a few minutes to discuss any discrepancies with their partner. For the third question, some students might say yes because there is a solution to the equation for line \(n\), which has \(x = 0\), namely if \(y = 0\) as well. For the fifth question, make sure students understand that the line \(\ell\) meets the \(x\)-axis even if that point is not shown on the graph.

Some key points to highlight, reinforcing conclusions from the previous lesson as well as this activity:

  • A solution of an equation in two variables is an ordered pair of numbers.
  • Solutions of an equation lie on the graph of the equation.
Reading, Conversing: MLR3 Clarify, Critique, Correct. After students have had time to make decisions about whether the statements are true or false, offer an incorrect response such as: “The statement ‘\(x=0\) is a solution of the equation for line \(n\)’ is true because I see the line go through the \(x\) value at 0.” Invite students to work with a partner to clarify the meaning of this incorrect response and then critique it. Invite pairs to offer a correct response by asking, “What language might you add or change to make this statement more accurate?” Listen for key phrases such as “\(x=0\) is a line, not a point.” This will help students to solidify their understanding that the solution to an equation with two variables is both coordinates of the point on the line.
Design Principle(s): Cultivate conversation; Maximize meta-awareness

13.3: I’ll Take an X, Please (15 minutes)


Students are given linear equations—some of which represent proportional relationships—in various forms, and are also given solutions to their partner’s equations in the form of coordinates of a point. The student with the equation decides which quantity they would like to know, \(x\) or \(y\), and requests this information from their partner. They then solve for the other quantity. The activity reinforces the concept that solutions to equations with two variables are a pair of numbers, and that knowing one can give you the other by using the value you know and solving the equation. Students also have a chance to think about the most efficient way to find solutions for equations in different forms.

You will need the I’ll Take An \(x\) Please blackline master for this activity.


Consider demonstrating the first step with a student. Write the equation \(y = 5x - 11\) on one slip of paper and the point \((1, \text- 6)\) on another slip of paper. Give the slip with the coordinate pair to the student. You can ask for the \(x\) or \(y\) coordinate of a point on the graph and then need to find the other coordinate. Ask the student helper for either \(x\) or \(y\). (In this case, asking for the \(x\) coordinate is wise because then you can just plug it into the equation to find the corresponding \(y\)-coordinate for the point on the graph.) Display your equation for all to see and demonstrate substituting the value in and solving for the other variable. Alternatively, ask students how they might use the information given (one of the values for \(x\) or \(y\) to find the other given your equation.

Arrange students in groups of 2. One partner receives Cards A through F from the left side of the blackline master and the other receives Cards a through f from the right side. Students take turns asking for either \(x\) or \(y\) then solving their equation for the other, and giving their partner the information requested.

Students play three rounds, where each round consists of both partners having a turn to ask for a value and to solve their equation. Follow with a whole-class discussion.

Representation: Provide Access for Perception. Display or provide students with a physical copy of the written directions and read them aloud. Check for understanding by inviting students to rephrase directions in their own words. Display directions throughout the activity.
Supports accessibility for: Language; Memory

Student Facing

One partner has 6 cards labeled A through F and one partner has 6 cards labeled a through f. In each pair of cards (for example, Cards A and a), there is an equation on one card and a coordinate pair, \((x,y)\), that makes the equation true on the other card.

  1. The partner with the equation asks the partner with a solution for either the \(x\)-value or the \(y\)-value and explains why they chose the one they did.

  2. The partner with the equation uses this value to find the other value, explaining each step as they go.

  3. The partner with the coordinate pair then tells the partner with the equation if they are right or wrong. If they are wrong, both partners should look through the steps to find and correct any errors. If they are right, both partners move onto the next set of cards.

  4. Keep playing until you have finished Cards A through F.

Student Response

For access, consult one of our IM Certified Partners.

Student Facing

Are you ready for more?

Consider the equation \(ax + by = c\), where \(a, b,\) and \(c\) are positive numbers.

  1. Find the coordinates of the \(x\)- and \(y\)-intercepts of the graph of the equation.
  2. Find the slope of the graph.

Student Response

For access, consult one of our IM Certified Partners.

Activity Synthesis

The discussion should focus on using the given information to efficiently find a solution for the equation. Consider asking students:

  • “How did you decide whether you wanted the value of \(x\) or the value of \(y\)?” (One might be more efficient to solve for: for example, with card B asking for \(x\) makes sense while with card d the arithmetic to perform is similar whether asking for \(x\) or for \(y\).)
  • “Which equations represent proportional relationships? How do you know? Which do not?” (C and F are proportional because they can be written as \(y=kx\), although this is hidden at first glance in C.)
  • “Once you have identified one solution to your equation, what are some ways you could find others?” (Use the constant rate of change to add/subtract to the solution you know, solve the equation for \(x\) or \(y\), choose values for one variable and solve for the other, for proportional relationships you could find equivalent ratios.)

Point out that all of the equations in this activity are linear. They are given in many different forms, not just \(y = mx + b\) or \(Ax + By = C\).

Conversing: MLR8 Discussion Supports. Use this routine to support student discussions. Provide the following sentence frames, and invite the partner who had the equation to begin with: “I decided to ask for the value of \(x\) (or \(y\)) because…. The steps I took to find the other value were....” Encourage the listener to ask clarifying questions such as: “What would have happened if you chose the other variable?” or “Could you think of a more efficient way to solve?” This will help students justify their reasoning for choosing a certain value to start and explain the steps they took to obtain the other coordinate.
Design Principle(s): Support sense-making; Cultivate conversation (for explanation)

Lesson Synthesis

Lesson Synthesis

In order to highlight student thinking about different strategies for finding a solution to a linear equation, ask students:

  • “What are different ways to find a solution to the linear equation \(3y + x = 12\)?” (Substitute in a value for one variable and solve for the other; graph the equation and find points that lie on the line; rearrange the equation so that one variable is written in terms of the other variable.)
  • “How do you know when you have found a solution to the equation \(3y + x = 12\)?” (The coordinates of the point will make the statement true.)
  • “What are some easy values to substitute into the equation?” (In this case, a good strategic choice is \(x = 0\), which gives \(y = 4\), and \(y=0\), which gives \(x=12\). This says that the \(y\)-intercept of the graph of the equation is \((0,4)\). Similarly, the \(x\)-intercept of the equation’s graph is \((12,0)\)
  • “How can you find the slope of the line?” (Graphing the line shows that the slope is negative, and we can verify this by rewriting the equation as \(y = \text-\frac{1}{3}x + 4\).​)

13.4: Cool-down - Intercepted (5 minutes)


For access, consult one of our IM Certified Partners.

Student Lesson Summary

Student Facing

Let's think about the linear equation \(2x-4y=12\). If we know \((0,\text-3)\) is a solution to the equation, then we also know \((0,\text-3)\) is a point on the graph of the equation. Since this point is on the \(y\)-axis, we also know that it is the vertical intercept of the graph. But what about the coordinate of the horizontal intercept, when \(y=0\)? Well, we can use the equation to figure it out.

\(\begin{align}2x-4y&=12 \\2x-4(0)&=12 \\2x&=12\\x&=6 \end{align}\)

Since \(x=6\) when \(y=0\), we know the point \((6,0)\) is on the graph of the line. No matter the form a linear equation comes in, we can always find solutions to the equation by starting with one value and then solving for the other value.

Video Summary

Student Facing