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?


Description: 
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.

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 aligns with the provided description.

shot_info: fl_problem_001392_neg_0

8
standard_description: Fluently divide multi-digit numbers using the standard algorithm.
problem_activity: Activity
In this activity, students use long division to divide whole numbers whose quotient is not a whole number. Previously, students found the quotient of
\(62 \div 5\)
using base-ten diagrams and the partial quotients method. Because the long division is a particular version of the partial quotients method, and because students have been introduced to long division, they have the tools to divide whole numbers that result in a (terminating) decimal. In this activity, students evaluate and critique the reasoning of others (MP3).
Teacher Notes for IM 6–8 Accelerated
Adjust this activity to 15 minutes. Consider skipping the third and fourth sets of questions in this activity to save time for the next activity, where students will continue practicing these skills with problems where the dividend is a decimal.
Launch
Arrange students in groups of 2. Provide access to graph paper. Give students 7–8 minutes to analyze and discuss Lin’s work with a partner and then complete the second set of questions.
Pause to discuss students’ analyses and at least one of the division problems. Students should understand that, up until reaching the decimal point, long division works the same for
\(62 \div 5 = 12.4\)
as it does for
\(657 \div 3 = 219\)
. In
\(62 \div 5\)
, however, there is a remainder of 2 ones, and we need to convert to the next smaller place value (tenths), change the 2 ones into 20 tenths, and then divide these into 5 equal groups of 4 tenths.
Prepare students to do the last set of questions by setting up the long division of
\(5 \div 4\)
for all to see. Discuss the placement of the decimal point. Reiterate that we can write extra zeros at the end of the dividend, following the decimal point, so that remainders can be worked with. Have students work through this problem and the others. Follow with a whole-class discussion.
Representation: Access for Perception.
Display and read Lin’s method for calculating the quotient of
\(62 \div 5\)
aloud. Students who both listen to and read the information will benefit from extra processing time.
Supports accessibility for: Language; Conceptual processing
Student Facing
Here is how Lin calculated
\(62 \div 5\)
.
[IMG]
Discuss with your partner:
Lin put a 0 after the remainder of 2. Why? Why does this 0 not change the value of the quotient?
Lin subtracted 5 groups of 4 from 20. What value does the 4 in the quotient represent?
What value did Lin find for
\(62 \div 5\)
?
Use long division to find the value of each expression. Then pause so your teacher can review your work.
\(126 \div 8\)
\(90 \div 12\)
Use long division to show that:
\(5 \div 4\)
, or
\(\frac 54\)
, is 1.25.
\(4 \div 5\)
, or
\(\frac 45\)
, is 0.8.
\(1 \div 8\)
, or
\(\frac 18\)
, is 0.125.
\(1 \div 25\)
, or
\(\frac {1}{25}\)
, is 0.04.
Noah said we cannot use long division to calculate
\(10 \div 3\)
because there will always be a remainder.
What do you think Noah meant by “there will always be a remainder”?
Do you agree with him? Explain your reasoning.
Student Response
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for free access to Student Response.
Anticipated Misconceptions
Students may be perplexed by the repeating decimals in the last question and think that they have made a mistake. Ask them to compare their work with a partner’s, and then clarify during discussion that some decimals do repeat. Because this work comes into focus in grade 7, the goal here is simply for students to observe that not all decimals terminate.
Activity Synthesis
Focus the whole-class discussion on the third and fourth sets of questions. Ask a few students to show their long division for all to see and to explain their steps. Some ideas to bring to uncover:
Problems like
\(1 \div 25\)
are challenging because the first step is 0: there are
zero
groups of 25 in 1. This means that we need to introduce a decimal and put a 0 to the right of the decimal. But one 0 is not enough. It is not until we add the second 0 to the right of the decimal that we can find 4 groups of 25 in 100. Because we moved two places to the right of the decimal, these 4 groups are really
\(0.04\)
, which is the quotient of 1 by 25.
Problems like
\(1 \div 3\)
are not fully treated until grade 7. At this point, we can observe that the long division process will go on and on because there is always a remainder of 1.
Some questions to ask students that highlight these points include:
When you found
\(1 \div 8\)
, what was your first step? (I put a 0 above the dividend 1 and added a decimal because I cannot take any 8’s from 1.)
When you found
\(1 \div 25\)
, what were your first few steps? (I put a 0 above the dividend 1 and added a decimal because I cannot take any 25’s from 1. I needed to add another 0 to the right of the decimal because I still cannot take any 25’s from 10.)
Why do we not write 0’s in advance to the right of the decimal? (We do not know in advance if we will need these 0’s or how many of them we might need, so we usually write them as needed.)
Another valuable link to make is to connect the values for the decimals in Problem 4 to percentages. For example, the fact that
\(\frac{4}{5} = 0.8\)
means that
\(\frac{4}{5}\)
of a quantity is 80% of that quantity. Similarly
\(\frac{1}{25}\)
of a quantity is the same as 4% of that quantity.
Writing, Listening, Conversing: MLR1 Stronger and Clearer Each Time.
Use this routine to help students improve their written responses to the question, “Noah said we cannot use long division to calculate
\(10÷3\)
because there will always be a remainder. Do you agree or disagree with this statement?” Give students time to meet with 2–3 partners to share and get feedback on their responses. Provide listeners with prompts that will help their partners strengthen their ideas and clarify their language (e.g., “Can you explain why there is always a remainder?”, “What number always remains?”, and “Could you try to explain this using a different example?”). Give students 1–2 minutes to revise their writing based on the feedback they received.​ Communicating their reasoning with a partner will help students understand situations in which long division does not work
Design Principle(s): Optimize output (for explanation); Cultivate conversation


aligns: yes