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.
{"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?


Description: 
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 aligns with the provided description.

shot_info: fl_problem_001392_pos_0

8
standard_description: Solve quadratic equations in one variable.
problem_activity: Optional activity
This activity is optional. It shows three different methods for solving an equation: by rewriting it in factored form, by transforming it into an expression in which the squared term has a coefficient 1 and then rewriting in factored form, and by completing the square. The second method involves temporarily substituting a part of the expression with another variable so that the squared term has a coefficient of 1, which makes it easier to rewrite in factored form. Students encountered this strategy in an optional activity (“Making It Simpler”) in an earlier lesson. If students did not do that optional activity, consider doing it first before using this activity.
The numbers in the equations here do not lend themselves to be easily rewritten in factored form (without guessing and checking or substitution), or to be easily rewritten into perfect squares (because the leading coefficient and the constant terms are not perfect squares), but students see that these equations can be solved with these strategies. The activity further illustrates that any of these methods can be rather tedious, providing further motivation to solve using the quadratic formula, which students will learn in a future lesson.
Launch
Display the perfect squares students saw in the preceding activity. Ask them what they notice about the coefficient of the square term in each expression.
\(\displaystyle 100x^2 + 80x + 16\)
\(\displaystyle 36x^2 - 60x + 25\)
\(\displaystyle 25x^2 + 40x + 16\)
\(\displaystyle 0.25x^2- 14x + 196\)
Students are likely to notice that the coefficient of every
\(x^2\)
is a perfect square, which made it easier to rewrite the completed square as squared factors. Solicit some ideas about whether we could still complete the square if a quadratic expression does not have a perfect square for the coefficient of
\(x^2\)
. Then, tell students that they will explore a few methods for solving such equations.
Arrange students in groups of 2. Ask one partner to study the first two methods and the other partner to study the third method, and then take turns explaining their understanding to each other. Discuss the methods, especially the third one, before students begin using them to solve equations.
Make sure students see that, in both the second and third methods, the first step involves multiplying both sides of the equation by 3 to make the coefficient of
\(x^2\)
a perfect square. Each term on the left side of the equation changes by a factor of 3, but the right side of the equation remains 0 because multiplying 0 by any number results in 0.
If time is limited, consider asking students to solve only three equations, using each method once.
Action and Expression: Internalize Executive Functions.
To support development of organizational skills in problem-solving, chunk this task into more manageable parts. For example, cut the three methods apart into separate pieces. Begin by presenting one method at a time and invite students to annotate the examples as they analyze the steps. After students annotate the first method, invite students who have great clarity in their annotations to share their work, and display for all to see. Encourage students to refer to their annotations as they approach the last problems themselves.
Supports accessibility for: Memory; Organization
Student Facing
Here are three methods for solving
\(3x^2 + 8x + 5 = 0\)
.
Try to make sense of each method.
Method 1:
\(\displaystyle \begin {align}3x^2 + 8x + 5 &= 0\\ (3x + 5)(x + 1) &= 0 \end{align}\)
\(\displaystyle \begin {align} x = \text- \frac53 \quad \text{or} \quad x = \text-1\end {align}\)
Method 2:
\(\displaystyle \begin {align} 3x^2 + 8x + 5 &= 0\\ 9x^2 + 24x + 15 &= 0\\ (3x)^2 + 8(3x) + 15 &= 0\\ U^2 + 8U + 15 &= 0\\ (U+5)(U+3) &= 0 \end{align}\)
\(\displaystyle \begin {align} U = \text-5 \quad &\text{or} \quad U = \text-3\\3x = \text-5 \quad &\text{or} \quad 3x = \text-3\\ x = \text- \frac53 \quad &\text{or} \quad x = \text-1 \end{align}\)
Method 3:
\(\displaystyle \begin {align} 3x^2 + 8x + 5 &= 0\\ 9x^2 + 24x + 15 &= 0\\9x^2 + 24x + 16 &= 1\\(3x + 4)^2 &= 1 \end{align}\)
\(\displaystyle \begin {align} 3x+4 = 1 \quad & \text{or} \quad 3x+4 = \text-1\\x = \text-1 \quad & \text{or} \quad x = \text- \frac53 \end {align}\)
Once you understand the methods, use each method at least one time to solve these equations.
\(5x^2 + 17x + 6 = 0\)
\(6x^2 + 19x = \text-10\)
\(8x^2 - 33x + 4 = 0\)
\(8x^2 - 26x = \text-21\)
\(10x^2 + 37x = 36\)
\(12x^2 + 20x - 77=0\)
Student Response
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Student Facing
Are you ready for more?
Find the solutions to
\(3x^2 -6x + \frac{9}{4} = 0\)
. Explain your reasoning.
Student Response
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Activity Synthesis
Consider asking students who solve the same equation using different methods to compare and contrast their solution strategies. Then, invite them to reflect on the solving process:
“Did the numbers in certain equations make it easier to solve by one method versus the others?”
“What are the advantages of each method? What about its drawbacks?”
“Is there a strategy you prefer or find reliable? Which one and why?”
Tell students that in upcoming lessons, they will look at a more efficient method for solving quadratic equations.
Speaking, Representing: MLR8 Discussion Supports.
Use this routine to support whole-class discussion. Before addressing the two questions, “What are the advantages of each method? What about its drawbacks?”, give students 2–3 minutes to plan what they will say when they present their responses to the class. Encourage students to consider what details are important to share and to think about how they will explain their reasoning using mathematical language. If needed, revoice student ideas to demonstrate mathematical language use by restating a statement as a question in order to clarify, apply appropriate language, and involve more students.
Design Principle(s): Support sense-making


aligns: yes