few_shots: 
Example 1:
Problem/activity:
You’re mixing ingredients for cookies. The recipe says to combine 6 tablespoons, or ¾ stick, of butter with 1 cup of sugar. You accidentally mix in a whole stick of butter (8 tablespoons) with the cup of sugar. How can you fix this?


Concept/skill: 
Recognize and represent proportional relationships between quantities.

Answer: 
No. Usually problems that teach recognizing and representing proportional relationships between quantities involve deciding whether two quantities are in a proportional relationship or identifying the constant of proportionality, and then representing proportional relationships by equations or explaining proportional relationship using coordinate graphs.

Example 2:
Problem/activity:
For each pair of numbers, determine if the values are equivalent. Explain your reasoning for each one.
a.
$$-0.25$$
and
$$\frac{1}{-4}$$
b.
$$-\left ( \frac{1}{4} \right )$$
and
$$\frac{-1}{-4}$$
c.
$$\frac{-1}{4}$$
and
$$-\frac{1}{4}$$


Concept/skill: 
Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.

Answer: 
Yes. This problem involves understanding fractions and division to divide rational numbers, such as 1, -4, 4, and -1. This problem also involves an understanding of multiplication: if p and q are integers, then -(p/q) = (-p)/q = p/(-q).

Example 3:
Problem/activity:
Problem 1
The table below contains single equations in a single variable, single equations in two variables, and pairs of simultaneous equations in two variables. Check the appropriate box to show whether there is no solution, exactly 1 solution, or if there is an infinite number of solutions.
|                                                          | **           No solution          ** | **           Exactly 1 solution          ** | **           Infinite number of solutions          ** |
|----------------------------------------------------------|--------------------------------------|---------------------------------------------|-------------------------------------------------------|
|                       $${9=4x+7}$$                       |                                      |                                             |                                                       |
|                       $${y=3y+5}$$                       |                                      |                                             |                                                       |
|                    $${y={x\over2}}$$                     |                                      |                                             |                                                       |
|                     $${3x+6y-9=0}$$                      |                                      |                                             |                                                       |
|    $${x=y}$$                        $${1.25x=1.25y}$$    |                                      |                                             |                                                       |
| $${9.9=6x+8y}$$                        $${9x=2.5y-8.8}$$ |                                      |                                             |                                                       |
| $${11x-2y=1.5}$$                        $${11x-2y=2.5}$$ |                                      |                                             |                                                       |
Problem 2
Kim has a small container and a large container as shown.
[IMG]
It takes 16 of the small containers to fill the large container. Three small containers leave 1.95 gallons of space in the big container.
What is the size of each of the two containers?


Concept/skill: 
Analyze and solve pairs of simultaneous linear equations.

Answer: 
Yes. The last three rows of the table in the first problem involves solving pairs of simultaneous linear equations. The second word problem can also be represented as two simultaneous linear equations.

Now, assess whether the following problem or activity is 'aligned' with the given Common Core standard in Mathematics.

shot_info: fl_problem_000972_neg_0

6

fl_problem_001097_pos_0

7

fl_problem_001392_pos_0

8
standard_description: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
problem_activity: Activity
In earlier activities, students gained some insights on the structure of equations in systems that have infinitely many solutions and those that have no solutions. In this activity, they apply those insights to sort systems of equations based on the number of solutions (one solution, many solutions, or no solutions).
Students could solve each system algebraically or graphically and sort afterwards, but given the number of systems to be solved, they will likely find this process to be time consuming. A more productive way would be to look for and make use of the structure in the equations in the systems (MP7), for example, by looking out for equivalent equations, equations with the same slope but different vertical intercepts, variable expressions with the same or opposite coefficients, and so on.
To effectively make use of the structure of the systems, students need to attend closely to all parts of each equation—the signs, variables, coefficients and constants—and to rearrange equations with care (MP6).
As students discuss their thinking in groups, make note of the different ways they use structure to complete the task. Encourage students who are solving individual systems to analyze the features of the equations and see if they could reason about the solutions or gain information about the graphs that way.
In this activity, students are analyzing the structure of equations in the systems, so technology is not an appropriate tool.
Here is an image of the cards for reference and planning:
[IMG]
Launch
Arrange students in groups of 2. Give one set of pre-cut slips or cards from the blackline master to each group.
Give students 7–8 minutes to sort the cards into groups. Emphasize to students that they should be prepared to explain how they place each system. Follow with a whole-class discussion.
Conversing: MLR8 Discussion Supports.
Use this routine to support small-group discussion during the card sort. Encourage students to take turns selecting a card and to explain to their partner whether the systems has no solutions, one solution, or infinitely many solutions. Display the following sentence frames for all to see: “This system has _____ solutions because….” Encourage students not only to challenge each other if they disagree, but also to press for clear explanations that use mathematical language.
Design Principle(s): Support sense-making; Maximize meta-awareness
Representation: Internalize Comprehension.
Chunk this task into more manageable parts to differentiate the degree of difficulty or complexity by beginning with fewer cards. For example, give students cards a subset of the cards to start with and introduce the remaining cards once students have completed their initial set of matches. Be sure to include at least one card from each category in the initial set.
Supports accessibility for: Conceptual processing; Organization
Student Facing
Your teacher will give you a set of cards. Each card contains a system of equations.
Sort the systems into three groups based on the number of solutions each system has. Be prepared to explain how you know where each system belongs.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Student Response.
Student Facing
Are you ready for more?
In the cards, for each system with no solution, change a single constant term so that there are infinitely many solutions to the system.
For each system with infinitely many solutions, change a single constant term so that there are no solutions to the system.
Explain why in these situations it is impossible to change a single constant term so that there is exactly one solution to the system.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Extension Student Response.
Anticipated Misconceptions
Some students may not know how to begin sorting the cards. Suggest that they try solving 2–3 systems. Ask them to notice if there's a point in the solving process when they realize how many solutions the system has or what the graphs of the two equations would look like. Encourage to look for similarities in the structure of the equations and see how the structure might be related to the number of solutions.
Activity Synthesis
Invite groups to share their sorting results and record them. Ask the class if they agree or disagree. If there are disagreements, ask students who disagree to share their reasoning.
Display all the systems—sorted into groups—for all to see and discuss the characteristics of the equations in each group. Ask students questions such as:
“How can we tell from looking at the equations in cards 2 and 5 that the systems have no solution?” (Possible reasoning:
Card 2: The equations are in slope-intercept form. The slope is 2 for both graphs, but the
\(y\)
-intercept is different, so the lines must be parallel.
Card 5: The second equation can be rearranged into
\(5x - 20y = \text-60\)
and multiplied by
\(\frac15\)
to give
\(x-4y=\text-12\)
. Both equations now have
\(x-4y\)
on one side, but that expression is equal to 4 in the first equation and equal to -4 in the second. There are no pair of
\(x\)
and
\(y\)
that can make both equations true.)
“What about the equations in cards 3, 7, and 8? What features might give us a clue that the systems have many solutions?” (Possible reasoning:
Card 3: The coefficients and constants in the second equation are 3 times those in the first, so they are equivalent equations.
Card 7: The first equation can be rearranged into the same form as the first:
\(x -4y = \text-4\)
. We can then see that the second equation,
\(4x-16y=\text-16\)
, is a multiple of the first equation, so they have all the same solutions.
Card 8: The first equation can be rearranged to
\(y=\text-25 x - \frac15\)
. We can then see that it is related to the second equation by a factor of 5, so they are equivalent equations.)
We can reason that all the other systems have one solution by a process of elimination—by noticing that they don’t have the features of systems with many solutions or systems with no solutions.


aligns: no