few_shots: 
Example:
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, does the following problem or activity enable students to learn the full intent of the following concept or skill?

shot_info: fl_problem_001392_pos_0

8
standard_description: Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.
problem_activity: Activity
In general,
\(P\%\)
of something is
\(\frac{P}{100}\)
times that thing. The purpose of this activity is to make this explicit. In this activity, students are asked to find 80% of several different values, generalize their process, and express their generalization in different ways. As they make repeated calculations, students look for and express regularity in their work (MP8) and see more explicitly that
\(P\%\)
of a number is
\(\frac{P}{100}\)
times that number. Once they arrive at one or more generalizations, students practice articulating why each generalization always works (MP3).
Launch
Give students quiet think time to complete the activity and then time to share their explanation with a partner.
Representation: Internalize Comprehension.
Activate or supply background knowledge about equivalent fractions. Incorporate explicit opportunities for review and practice if necessary. Allow students to use calculators to ensure inclusive participation in the activity.
Supports accessibility for: Memory; Conceptual processing
Writing, Speaking, Listening: MLR1 Stronger and Clearer Each Time.
Use this with successive pair shares to give students a structured opportunity to revise and refine their own strategies to determine 80% of any value to find a sale price. Make sure students include a table as well as an expression in their explanations. Ask each student to meet with 2–3 other partners in a row for feedback. Provide students with prompts for feedback that will help them strengthen their ideas and clarify their language (e.g., “Can you explain how…”, “Will this expression always work? How do you know?”, etc.). Students can borrow ideas and language from each partner to refine and clarify their original explanation. This will help students refine their own strategies and learn about other ways to find the sale price of an item.
Design Principles(s): Optimize output (for explanation); Maximize meta-awareness
Student Facing
During a sale, every item in a store is 80% of its regular price.
If the regular price of a T-shirt is
$
10, what is its sale price?
The regular prices of five items are shown here. Find the sale price of each item.
|               | item 1 | item 2 | item 3 | item 4 | item 5 |
|---------------|--------|--------|--------|--------|--------|
| regular price |        |        |        |        |        |
|  sale price   |        |        |        |        |        |
You found 80% of many values. Was there a process you repeated over and over to find the sale prices? If so, describe it.
[IMG]
Select
all
of the expressions that could be used to find 80% of
\(x\)
. Be prepared to explain your reasoning.
\(\frac{8}{100} \boldcdot x\)
\(\frac{80}{100} \boldcdot x\)
\(\frac{8}{10} \boldcdot x\)
\(\frac{4}{10} \boldcdot x\)
\(\frac85 \boldcdot x\)
\(\frac45 \boldcdot x\)
\(80 \boldcdot x\)
\(8 \boldcdot x\)
\((0.8) \boldcdot x\)
\((0.08) \boldcdot x\)
Student Response
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Activity Synthesis
Make sure students have the correct values in the table for the second question. Then, discuss the expressions, starting with
\(\displaystyle \frac{80}{100} \boldcdot x.\)
Guide students to see that if we know the value of 100%, dividing that value by 100 (or equivalently, multiplying it by
\(\frac{1}{100}\)
) tells us the corresponding value for 1%. We can then multiply that value by the desired percentage. Consider using a table to organize this argument, as shown below:
| percentage |         regular price          |
|------------|--------------------------------|
|    100     |             \(x\)              |
|     1      | \(\frac{1}{100} \boldcdot x\)  |
|     80     | \(\frac{80}{100} \boldcdot x\) |
After everyone understands where this expression comes from, ask students to discuss the remaining expressions with a partner. This is a good opportunity for just-in-time review on equivalent fractions, if needed.


aligns: no