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: Graph the solutions to a linear inequality in two variables as a half plane (excluding the boundary in the case of a strict inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the corresponding half-planes.
problem_activity: Activity
By now, students recognize that when a quadratic equation is in the form of
\(\text{expression} = 0\)
and the expression is in factored form, the equation can be solved using the zero product property. In this activity, they encounter equations in which one side of the equal sign is not 0. To make one side equal 0 requires rearrangement. For example, to solve
\(x(x+6)=8\)
, the equation needs to be rearranged to
\(x(x+6)-8=0\)
. Yet because the expression on the left is no longer in factored form, the zero product property won’t help after all and another strategy is needed.
Students recall that to solve a quadratic equation in the form of
\(\text{expression} = 0\)
is essentially to find the zeros of a quadratic function defined by that expression, and that the zeros of a function correspond to the horizontal intercepts of its graph. In the case of
\(x(x+6)-8=0\)
, the function whose zeros we want to find is defined by
\(x(x+6)-8\)
. Graphing
\(y=x(x+6)-8\)
and examining the
\(x\)
-intercepts of the graph allow us to see the number of solutions and what they are.
Launch
Arrange students in groups of 2 and provide access to graphing technology. Give students a moment to think quietly about the first question and then ask them to briefly discuss their response with their partner before continuing with the rest of the activity.
Action and Expression: Provide Access for Physical Action.
Support effective and efficient use of tools and assistive technologies. To use graphing technology, some students may benefit from a demonstration or access to step-by-step instructions.
Supports accessibility for: Organization; Memory; Attention
Student Facing
Han is solving three equations by graphing.
\((x-5)(x-3)=0\)
\(\displaystyle (x-5)(x-3)=\text-1\)
\(\displaystyle (x-5)(x-3)=\text-4\)
To solve the first equation,
\((x-5)(x-3)=0\)
, he graphed
\(y=(x-5)(x-3)\)
and then looked for the
\(x\)
-intercepts of the graph.
Explain why the
\(x\)
-intercepts can be used to solve
\((x-5)(x-3)=0\)
.
What are the solutions?
To solve the second equation, Han rewrote it as
\((x-5)(x-3)+1=0\)
. He then graphed
\(y=(x-5)(x-3)+1\)
.
Use graphing technology to graph
\(y=(x-5)(x-3)+1\)
. Then, use the graph to solve the equation. Be prepared to explain how you use the graph for solving.
Solve the third equation using Han’s strategy.
Think about the strategy you used and the solutions you found.
Why might it be helpful to rearrange each equation to equal 0 on one side and then graph the expression on the non-zero side?
How many solutions does each of the three equations have?
Student Response
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for free access to Student Response.
Student Facing
Are you ready for more?
The equations
\((x-3)(x-5)=\text-1\)
,
\((x-3)(x-5)=0\)
, and
\((x-3)(x-5)=3\)
all have whole-number solutions.
Use graphing technology to graph each of the following pairs of equations on the same coordinate plane. Analyze the graphs and explain how each pair helps to solve the related equation.
\(y=(x-3)(x-5)\)
and
\(y=\text-1\)
\(y=(x-3)(x-5)\)
and
\(y=0\)
\(y=(x-3)(x-5)\)
and
\(y=3\)
Use the graphs to help you find a few other equations of the form
\((x-3)(x-5)=z\)
that have whole-number solutions.
Find a pattern in the values of
\(z\)
that give whole-number solutions.
Without solving, determine if
\((x-5)(x-3)=120\)
and
\((x-5)(x-3)=399\)
have whole-number solutions. Explain your reasoning.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Extension Student Response.
Anticipated Misconceptions
If students enter the equation
\((x-5)(x-3)=0\)
into their graphing technology, they may see an error message, or they may see vertical lines. The lines will intersect the
\(x\)
-axis at the solutions, but they are clearly not graphs of a quadratic function. Emphasize that we want to graph the function defined by
\(y=(x-5)(x-3)\)
and use its
\(x\)
-intercepts to find the solution to the related equation.
All
the points on the two vertical lines do represent solutions to the equation (because the points along each vertical line satisfy the equation regardless of the value chosen for
\(y\)
), but understanding this is beyond the expectations for students in this course.
Activity Synthesis
Invite students to share their responses, graphs, and explanations on how they used the graphs to solve the equations. Discuss questions such as:
“
Are the original equation
\((x-5)(x-3)=\text-1\)
and the rewritten one
\((x-5)(x-3)+1=0\)
equivalent?” (Yes, each pair of equations are equivalent. In that example, 1 is added to both sides of the original equation.)
"Why might it be helpful to rearrange the equation so that one side is an expression and the other side is 0?" (It allows us to find the zeros of the function defined by that expression. The zeros correspond to the
\(x\)
-intercepts of the graph.)
“What equation would you graph to solve this equation:
\((x-4)(x-6)=15\)
?” (
\(y=(x-4)(x-6)-15\)
) “What about
\((x+3)^2-1=5\)
?" (
\(y=(x+3)^2-6\)
)
Make sure students understand that some quadratic functions have two zeros, some have one zero, and some have no zeros, so their respective graphs will have two, one, or no horizontal intercepts, respectively.
Likewise, some quadratic equations have two solutions, some have one solution, and some have no real solutions. (Because students won’t know about numbers that aren’t real until a future course, for now it is sufficient to say “no solutions.”)


aligns: yes