few_shots: 
Example:
Concept/skill: 
Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.

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.
{"table": [{"_attributes": {"align": "center", "class": ["table", "table-bordered"], "style": "table-layout:fixed; width:70%"}, "tbody": [{"tr": [{"td": [{"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "No solution"}]}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "Exactly 1 solution"}]}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "Infinite number of solutions"}]}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9=4x+7}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${y=3y+5}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${y={x\\over2}}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${3x+6y-9=0}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${x=y}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${1.25x=1.25y}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9.9=6x+8y}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9x=2.5y-8.8}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${11x-2y=1.5}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${11x-2y=2.5}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {}]}]}]}]}
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?


Answer: 
No. Completing neither of these problems requires replacing one equation by the sum of that equation and a multiple of the other equation.

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

shot_info: fl_problem_001392_neg_0

8
standard_description: Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent.
problem_activity: Activity
This activity prompts students to solve a problem involving division of fractions in a less-scaffolded way. Students can see two relevant numbers to work with, but they need to interpret the context, the visual information, and the written question to decide whether the missing value is the size of one group, the number of groups, or the given amount (MP1). By carefully making sense of the context, students see that part of a group is known and they have to find the size of the whole group.
As students work, monitor the strategies they use and the way they reason about the situation. Students may draw directly on the photo to make sense of the quantities. Some may draw a tape diagram or another type of diagram to represent the partially-filled water container. Others may bypass diagramming and instead reason verbally or by writing equations. Select a few students who use different but equally effective strategies to share later.
Launch
Video
Water in Containers
available at
https://player.vimeo.com/video/304137827
.
Arrange students in groups of 3–4. Show the short video. Ask students what questions we could ask about the amount of water in this situation that would require working with fractions to determine the answers. Give groups a moment to think about their questions. If needed, show the video again, or refer to the photos to identify the fractions.
Possible questions:
How much water fits into the whole water dispenser?
How much water is indicated by each tick mark of the water dispenser?
How far would 1 liter of water fill the water dispenser?
After hearing students’ ideas, give students 4–5 minutes of quiet work time and then another 2 minutes to share their responses with their group. Ask students to discuss any disagreements they might have about their interpretation of the problem, the solving process, or the answer. Remind students they can check their solution using the multiplication equation.
Representation: Internalize Comprehension.
Guide information processing and visualization. To support working memory, play the video multiple times. Students may also benefit from explicit guidance on which aspects of the video to focus on each time.
Supports accessibility for: Memory; Organization
Student Facing
[IMG]
After looking at these pictures, Lin says, “I see the fraction
\(\frac 25\)
.” Jada says, “I see the fraction
\(\frac 34\)
.” What quantities are Lin and Jada referring to?
Consider the problem: How many liters of water fit in the water dispenser?
Write a multiplication equation and a division equation for the question.
Find the answer and explain your reasoning. If you get stuck, consider drawing a diagram.
Check your answer using the multiplication equation.
Student Response
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Anticipated Misconceptions
For the second question, students may not immediately see that, to answer the question “how many liters of water fit in the dispenser?” they need to relate the amount in liters (as shown in the measuring cup) to the fraction of the dispenser that is filled with water. Suggest that they think about the problem in terms of equal-sized groups and try to identify the groups, number of groups, etc.
Activity Synthesis
Invite previously identified students to share their solutions and reasoning. Sequence their presentations so that the more concrete strategies are presented before the more abstract (e.g., a student who used only equations should present last). Display their diagrams and record their reasoning for all to see.
Connect the reasoning done using the diagram to the mathematical operations. Regardless of the path students took, finding the amount of water that fills the entire dispenser requires dividing
\(\frac34\)
liters by 2 (to find the amount in
\(\frac15\)
of a dispenser), and then multiplying that amount by 5. Consider revisiting this activity later, after students have learned an algorithm for dividing a fraction, and using this problem to reinforce that multiplying by the denominator and dividing by the numerator makes sense as a way to divide by a fraction.


aligns: no