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: Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
problem_activity: Activity
The purpose of this activity is for students to identify why equations are not equivalent. In an associated Algebra 1 lesson, students rewrite equations to express one variable in terms of the other. The work in this activity prepares students to be successful in the associated Algebra 1 lesson by allowing them to gain a deeper understanding of what makes two equations equivalent and to practice making appropriate moves to maintain the balance in an equation.
Student Facing
[IMG]
Tyler is practicing finding different equivalent equations that match the story. For each of the problems below, he gets one equation right but the other equation wrong. For each one, explain the error, give the correct equivalent equation, and explain your reasoning.
Situation: The yogurt at Sweet Delights costs \$0.65 per pound and \$0.10 per topping. The total cost of a purchase was \$1.70. Let
\(p\)
be the weight of the yogurt in pounds and
\(t\)
be the number of toppings bought.
Tyler’s first and correct equation:
\(0.65p+0.10t=1.70\)
Tyler’s second and
incorrect
equation:
\(t= (1.70 - .65p) \boldcdot 0.10\)
What is the error?
What is a correct second equation Tyler could have written?
What might Tyler have been thinking that led to his mistake?
Situation: The perimeter of a rectangle (twice the sum of the length and width) is 13.5 inches. Let
\(l\)
be the length of the rectangle and
\(w\)
be the width of the rectangle.
Tyler’s first and correct equation:
\(2(l + w) = 13.5​​​​​​​​​​​​​​\)
Tyler’s second and
incorrect
equation:
\(w = 13.5 - 2l\)
What is the error?
What is a correct second equation Tyler could have written?
What might Tyler have been thinking that led to his mistake?
Situation: For a fundraiser, a school is selling flavored waters for \$2.00 each and pretzels for \$1.50 each. The school has a fundraising goal of \$200. Let
\(w\)
be the number of waters sold and
\(p\)
be the number of pretzels sold.
Tyler’s first and correct equation:
\(2w + 1.5p = 200\)
Tyler’s second and
incorrect
equation:
\(1.5p = 198w\)
What is the error?
What is a correct second equation Tyler could have written?
What might Tyler have been thinking that led to his mistake?
Student Response
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Activity Synthesis
The purpose of this discussion is to reinforce the concept that different equations can represent the same relationship.
Display the school fundraiser situation from the practice activity for all to see.
Here are discussion questions:
What are some of the true equations that describe the school fundraiser situation? (
\(2w + 1.5p = 200, w = 100 - 0.75p, p = \frac{200 - 2w}{1.5}\)
)
What can be seen in each equation students share, and why might these equations be useful? (
\(2w + 1.5p = 200\)
shows how the water and pretzel revenue should equal
$
200. The other two equations help if you want to know how many waters you need to sell or how many pretzels you need to sell, if you know the other amount.)
Explain that because all these equations describe the same relationship, just in different ways, they are
equivalent
. Discuss how equivalent equations can be found by thinking about the situation in different ways, as well as by applying acceptable moves for solving equations, such as adding the same amount to each side or multiplying each side by the same amount.


aligns: no