Defining New Operations: Free Response

1. Reading the rule and working from the inside out . Foundational, 13 points. Question 1 of 5.

A new operation is defined by $a \clubsuit b = 3a - 2b + 4$ . Nothing about the symbol is special: it is a rule, evaluated by substituting into the two slots and simplifying with the order of operations you already use.

Part A.
Evaluate $2 \clubsuit 5$ and $(-3) \clubsuit 4$ .

Solve and show your work Write each step out, and end with the value and its units. 4 points
Part B.
Evaluate $(1 \clubsuit 2) \clubsuit 3$ . Evaluate the inner operation first, turn its result into a single number, and only then apply the outer operation.

Solve and show your work Write each step out, and end with the value and its units. 5 points
Part C.
Explain why $(1 \clubsuit 2) \clubsuit 3$ has to be evaluated by finding $1 \clubsuit 2$ first, rather than by substituting all three numbers into the rule at once or evaluating $2 \clubsuit 3$ first. Tie your explanation to why the multiplications inside the rule have to happen before the addition and subtraction.

Explain why it works A sentence or two. Reasons, not steps. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Every part of this question starts the same way: substitute the two given numbers into $a$ and $b$ in the rule $3a - 2b + 4$ , then simplify with multiplication before addition or subtraction.
Hint 2 of 3 · Part B
In $(1 \clubsuit 2) \clubsuit 3$ , the parentheses tell you which pair to combine first. Reduce $1 \clubsuit 2$ to a single number before that number ever meets the $3$ .
Hint 3 of 3 · Part C
Try evaluating $2 \clubsuit 3$ first instead of $1 \clubsuit 2$ first, then combine with the remaining number. Compare what you get with the value from part B.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Substitute into $3a - 2b + 4$ for each pair, multiplying before combining.

$2 \clubsuit 5 = 3(2) - 2(5) + 4 = 6 - 10 + 4 = 0.$

$(-3) \clubsuit 4 = 3(-3) - 2(4) + 4 = -9 - 8 + 4 = -13.$

Part B

Work from the inside out.

$1 \clubsuit 2 = 3(1) - 2(2) + 4 = 3 - 4 + 4 = 3.$

That result becomes the first input of the outer operation.

$3 \clubsuit 3 = 3(3) - 2(3) + 4 = 9 - 6 + 4 = 7.$

So $(1 \clubsuit 2) \clubsuit 3 = 7$ .

Part C

A custom operation takes exactly two inputs, so a three-number expression like $1 \clubsuit 2 \clubsuit 3$ is not itself a defined computation; only $(1 \clubsuit 2) \clubsuit 3$ or $1 \clubsuit (2 \clubsuit 3)$ is. The parentheses decide which pair the operation acts on first, exactly the way parentheses decide grouping in ordinary arithmetic. Reducing the inner pair to a single number before applying $\clubsuit$ again is not optional: it is the only way an operation of two inputs can ever accept a third number.

The two groupings need not even agree. Evaluating the other way,

$2 \clubsuit 3 = 3(2) - 2(3) + 4 = 6 - 6 + 4 = 4, \qquad 1 \clubsuit 4 = 3(1) - 2(4) + 4 = 3 - 8 + 4 = -1,$

so $1 \clubsuit (2 \clubsuit 3) = -1$ , not the $7$ from part B. The grouping is not a formality.

The same inside-out discipline protects the order of operations inside the rule itself: $3a$ and $2b$ are computed as products before the addition or subtraction touches them, the same way the inner operation is computed before the outer one is applied.

In one line

$2 \clubsuit 5 = 0$ and $(-3) \clubsuit 4 = -13$ ; working from the inside out, $(1 \clubsuit 2) \clubsuit 3 = 7$ ; and both facts rest on the same discipline, respect the grouping and multiply before combining, since ignoring either changes the answer.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Substitutes the given numbers into the correct slots of the rule for both computations. . Worth 1 point.

Carries out the multiplication before combining terms, and tracks the sign of the negative input correctly. . Worth 2 points.

Reports each result as a single simplified number. . Worth 1 point.

Part B 5 points

Evaluates the inner operation first and treats its result as a single number before evaluating the outer operation. . Worth 2 points.

Carries out both evaluations correctly, respecting the order of operations inside the rule each time. . Worth 2 points.

Reports the final value the whole nested expression simplifies to. . Worth 1 point.

Part C 4 points

Explains that the parentheses fix which pair of inputs combines first, and that the inner result must become a single number before the operation is applied again. . Worth 3 points. needs an explanation, not just an answer

Supports the explanation with a specific computation, from part B or a new one, showing that a different grouping or a different order changes the result. . Worth 1 point.

2. Pricing a package with two inputs . Application, 12 points. Question 2 of 5.

A shipping company prices a package with the rule $w \spadesuit d = 3w + 0.02d + 4$ , where $w$ is the package's weight in pounds, $d$ is the distance it travels in miles, and the output is the fee in dollars.

Part A.
A 5-pound package travels 250 miles. Write the fee as $w \spadesuit d$ with the correct inputs in place, then evaluate it.

Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
Part B.
A 12-pound package travels 150 miles. Find its fee.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part C.
The company considers combining the two packages above into a single 17-pound package that travels the longer of the two distances, rather than shipping them separately. Compute the fee for the combined package, and state which option costs less and by how much.

Carry your own answer forward Add your own two fees from parts A and B to get the total for shipping separately, even if they differ from the values you expected.

Compare the two methods Say what each one costs you, and when you would reach for it. 5 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Read the story first: figure out which number is the weight ( $w$ ) and which is the distance ( $d$ ) before you touch the rule. The rule only cares about the two numbers you feed it, in that order.
Hint 2 of 3
Keep the decimal coefficient's multiplication, $0.02$ times the distance, as its own separate step so it does not get merged with the whole-number multiplication by mistake.
Hint 3 of 3 · Part C
Work out the combined fee exactly the way you did in parts A and B, just with a different pair of inputs. Then compare that single number with a total you already have, rather than recomputing anything from scratch.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Matches the weight and the distance in the story to $w$ and $d$ , in that order, before substituting. . Worth 1 point.

Carries out the multiplication, including the decimal coefficient on $d$ , and adds the three terms correctly. . Worth 2 points.

Reports the fee with its units. . Worth 1 point.

Part B 3 points

Carries out the substitution and arithmetic correctly for this package's weight and distance. . Worth 2 points.

Reports the fee with its units. . Worth 1 point.

Part C 5 points

Uses the combined weight and the longer of the two distances as the inputs to the rule. . Worth 1 point.

Evaluates the combined fee correctly. . Worth 2 points.

Compares the combined fee to the sum of the two separate fees, and states which option is cheaper and by how much. . Worth 2 points.

3. Squaring the whole input, then solving backward . Foundational, 15 points. Question 3 of 5.

A new operation is defined by $a \heartsuit b = a^2 - 3b + 2$ . The first slot is squared, so whatever sits in it goes in whole.

Part A.
Evaluate $(-3) \heartsuit 5$ and $4 \heartsuit (-2)$ .

Solve and show your work Write each step out, and end with the value and its units. 4 points
Part B.
Simplify $(2k - 1) \heartsuit 7$ completely.

Write the expression An equation or an expression is enough here. Show how you built it. 4 points
Part C.
Solve $x \heartsuit 1 = 48$ for $x$ . Report every solution.

Solve and show your work Write each step out, and end with the value and its units. 4 points
Part D.
An equation built from $a \heartsuit b = a^2 - 3b + 2$ can put the unknown $x$ in either slot. Explain how that choice decides whether the resulting equation is quadratic or linear in $x$ , and how many solutions each of the two cases typically produces.

Explain why it works A sentence or two. Reasons, not steps. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 4
In every part, whatever sits in the first slot of $\heartsuit$ gets squared whole. If that slot holds an expression like $2k - 1$ , the entire expression is what gets squared, not just one piece of it.
Hint 2 of 4 · Part B
Before you square anything in part B, decide what single object occupies the first slot for this computation. It is the expression $2k - 1$ as a whole, not the letter $k$ by itself.
Hint 3 of 4 · Part C
Substitute $x$ and $1$ into the rule the same way you did with numbers in part A, then isolate $x^2$ before taking a square root of both sides. A square root taken across an equation keeps both signs.
Hint 4 of 4 · Part D
Ask what happens to the sign of a number once it is squared, and whether that loss of information can ever be recovered by working backward. Compare that to what happens when $x$ is only ever multiplied by a constant, never squared.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

$(-3) \heartsuit 5 = (-3)^2 - 3(5) + 2 = 9 - 15 + 2 = -4.$

$4 \heartsuit (-2) = 4^2 - 3(-2) + 2 = 16 + 6 + 2 = 24.$

Part B

The whole expression $2k - 1$ fills the first slot, so it is what gets squared, not just $k$ .

$(2k - 1) \heartsuit 7 = (2k - 1)^2 - 3(7) + 2.$

Expand the square first.

$(2k - 1)^2 = 4k^2 - 4k + 1.$

Then combine every term.

$4k^2 - 4k + 1 - 21 + 2 = 4k^2 - 4k - 18.$

Part C

Substitute $a = x$ and $b = 1$ into the rule to turn the equation into an ordinary one.

$x^2 - 3(1) + 2 = 48 \quad\Longrightarrow\quad x^2 - 1 = 48 \quad\Longrightarrow\quad x^2 = 49.$

Taking the square root of both sides keeps both signs.

$x = \pm\sqrt{49} = \pm 7.$

Both check: $7^2 - 1 = 48$ and $(-7)^2 - 1 = 48$ .

Part D

The two slots of $\heartsuit$ do different jobs to whatever fills them: the first slot is squared, and the second is only multiplied by $-3$ before the constant is added. Which job $x$ gets depends entirely on which slot it is placed in.

If $x$ fills the FIRST slot, substituting turns the left-hand side into $x^2 - 3b + 2$ for some fixed number $b$ , so once that fixed part is cleared away the equation reduces to $x^2$ equal to some number. An equation of that shape typically has two solutions, because both a positive and a negative number square to the same value, the way $7$ and $-7$ both satisfy the equation from part C:

$7^2 = 49, \qquad (-7)^2 = 49.$

If $x$ fills the SECOND slot instead, substituting gives $a^2 - 3x + 2$ for some fixed number $a$ , and $x$ appears only inside the term $-3x$ , to the first power. That equation is linear once solved, and because the coefficient of $x$ is $-3$ and not zero, it has exactly one solution: nothing squares $x$ , so there is no second sign to recover.

In one line

$(-3) \heartsuit 5 = -4$ and $4 \heartsuit (-2) = 24$ ; $(2k - 1) \heartsuit 7 = 4k^2 - 4k - 18$ , since the whole expression $2k - 1$ is squared; and solving $x \heartsuit 1 = 48$ gives $x = 7$ or $x = -7$ , two solutions because the unknown sat in the squared slot.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Substitutes the given numbers into the correct slots for both computations. . Worth 1 point.

Squares the first input and handles the sign of the $-3b$ term correctly in both computations, including when $b$ is negative. . Worth 2 points.

Reports each result as a single simplified number. . Worth 1 point.

Part B 4 points

Treats the expression $2k - 1$ as the whole content of the first slot, to be squared as a single unit. . Worth 2 points.

Expands $(2k - 1)^2$ correctly and combines the resulting terms with the rest of the rule. . Worth 2 points.

Part C 4 points

Substitutes the definition into the equation to turn it into an ordinary equation in $x$ . . Worth 1 point.

Isolates $x^2$ and solves it correctly. . Worth 2 points.

Reports BOTH solutions, not only the positive one. . Worth 1 point.

Part D 3 points

Explains that whether $x$ ends up squared depends on which slot it fills, and connects that to whether the resulting equation is quadratic or linear in $x$ . . Worth 2 points. needs an explanation, not just an answer

States how many solutions each of the two cases typically produces, and ties each count to whether squaring occurred. . Worth 1 point.

4. One pair is enough to break the claim . Reasoning, 13 points. Question 4 of 5.

A new operation is defined by $a \diamondsuit b = 2a - b^2$ . The two slots are treated differently: the first is doubled, and the second is squared.

Part A.
Evaluate $3 \diamondsuit 5$ and $5 \diamondsuit 3$ .

Solve and show your work Write each step out, and end with the value and its units. 4 points
Part B.
A claim is proposed: for every pair of numbers $a$ and $b$ , $a \diamondsuit b = b \diamondsuit a$ . Use your work from part A to refute this claim, and state precisely what a single counterexample does and does not establish about it.

Carry your own answer forward Use the two values you computed in part A, even if they differ from the ones you expected.

Construct a counterexample Give one specific case, and show it breaks the claim. 5 points
Part C.
Without computing $10 \diamondsuit 2$ or $2 \diamondsuit 10$ , state whether they are equal or different, and justify your prediction using the specific feature of the rule $2a - b^2$ that treats the two slots differently.

Justify your claim State the claim, then give the reason it has to be true. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Both parts of this question turn on the same fact: the rule $2a - b^2$ treats its two slots differently, since only the second one is squared. Keep track of which number lands in which slot.
Hint 2 of 3 · Part B
You do not need any new arithmetic for part B: the two numbers you already evaluated in part A either agree or they do not, and that is the whole content of a counterexample.
Hint 3 of 3 · Part C
Do not compute anything yet. Instead, ask which slot each of $10$ and $2$ would land in under each order, and whether swapping which one gets squared is likely to change the result.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

$3 \diamondsuit 5 = 2(3) - 5^2 = 6 - 25 = -19.$

$5 \diamondsuit 3 = 2(5) - 3^2 = 10 - 9 = 1.$

Part B

A claim quantified over every pair of numbers is destroyed by a single pair on which it fails, so part A's pair already does the work.

$3 \diamondsuit 5 = -19, \qquad 5 \diamondsuit 3 = 1.$

The two results are not equal, so the claim $a \diamondsuit b = b \diamondsuit a$ for every pair fails on $a = 3$ , $b = 5$ alone. That is all a counterexample is required to show. It does not establish that the two sides differ for every other pair too, since some other pair might happen to agree; it only establishes that they do not agree for ALL pairs, which is exactly what the claim asserted.

Part C

The two slots of $2a - b^2$ genuinely do different jobs: the first input is only doubled, and the second is squared. Swapping the two numbers moves each into the other job, so unless the arithmetic happens to cancel out, the two orders give different results. Nothing about $10$ and $2$ suggests such a cancellation, so the prediction is that they differ, matching the pattern already seen with $3$ and $5$ in part A. A direct check confirms it, though reaching the conclusion from the structure of the rule, without doing the arithmetic, is the point of this part:

$10 \diamondsuit 2 = 20 - 4 = 16, \qquad 2 \diamondsuit 10 = 4 - 100 = -96.$

In one line

$3 \diamondsuit 5 = -19$ and $5 \diamondsuit 3 = 1$ , which already refutes the claim that $a \diamondsuit b = b \diamondsuit a$ for every pair; and $10 \diamondsuit 2$ and $2 \diamondsuit 10$ are different too, because the rule squares only the second input, treating the two slots differently.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Substitutes the given numbers into the correct slots for both orders. . Worth 1 point.

Squares only the second input and doubles only the first, matching the rule, in both orders. . Worth 2 points.

Reports each result as a single simplified number. . Worth 1 point.

Part B 5 points

Identifies the specific pair from part A to use as the counterexample. . Worth 1 point.

States plainly that the two results found in part A are different, which is what defeats the claim. . Worth 2 points.

Explains precisely what a single counterexample establishes about the claim, and what it leaves open. . Worth 2 points.

Part C 4 points

Justifies the prediction using the fact that the rule squares only the second input, treating the two slots differently. . Worth 3 points. needs an explanation, not just an answer

States a definite prediction, equal or different, before justifying it. . Worth 1 point.

5. Searching for an identity element . Reasoning, 15 points. Question 5 of 5.

A new operation is defined by $a \dagger b = a + 2b$ . This question asks whether it has an identity element, a value that leaves every input unchanged from both sides.

Part A.
Evaluate $3 \dagger 4$ and $4 \dagger 3$ .

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part B.
Look for a value $e$ that works as a RIGHT input: search for $e$ such that $a \dagger e = a$ no matter which number $a$ is. Set up that condition using the definition, and solve for $e$ .

Justify your claim State the claim, then give the reason it has to be true. 4 points
Part C.
Test whether the value you just found for $e$ also works as a LEFT input, that is, whether $e \dagger a = a$ no matter which number $a$ is. Show your work using your own value of $e$ , and state whether it satisfies this second condition.

Carry your own answer forward Use the value of $e$ you found in part B, even if it is not the value you expected.

Justify your claim State the claim, then give the reason it has to be true. 4 points
Part D.
Using your own conclusions from parts B and C, decide whether $\dagger$ has an identity element, and explain your reasoning. If it does not, name exactly which of the two identity conditions could not be satisfied for every choice of $a$ at once.

Carry your own answer forward Use your own conclusions from parts B and C, even if they differ from what you expected.

Justify your claim State the claim, then give the reason it has to be true. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 4
An identity element has to satisfy TWO separate conditions, $a \dagger e = a$ and $e \dagger a = a$ , using the same value of $e$ in both, no matter which number $a$ is. Treat the two conditions as two separate searches.
Hint 2 of 4 · Part B
Write $a \dagger e = a$ using the definition, then subtract $a$ from both sides. Whatever is left has to equal zero no matter what $a$ is, which is what pins $e$ down to one value.
Hint 3 of 4 · Part C
Plug your value of $e$ into the OTHER order, $e \dagger a$ , and simplify. Then ask whether the result you get matches $a$ for every number, or only for one particular number.
Hint 4 of 4 · Part D
One of the two conditions in part B left no freedom to pick a different $e$ . Ask what that means for the operation's chances of having an identity if your part C result does not also come out satisfied for every $a$ .

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

$3 \dagger 4 = 3 + 2(4) = 3 + 8 = 11.$

$4 \dagger 3 = 4 + 2(3) = 4 + 6 = 10.$

Part B

Write the condition using the definition.

$a \dagger e = a + 2e = a.$

Subtract $a$ from both sides. The $a$ 's cancel completely, which is exactly what has to happen for a single value of $e$ to work no matter which number $a$ is.

$2e = 0 \quad\Longrightarrow\quad e = 0.$

Since $a$ dropped out on its own, $e = 0$ is forced, and it is the only value that could possibly work as a right input.

Part C

Substitute the value of $e$ found in part B into the OTHER order.

$e \dagger a = e + 2a.$

With $e = 0$ this becomes

$0 \dagger a = 0 + 2a = 2a.$

For this to equal $a$ no matter which number $a$ is, we would need $2a = a$ for every choice of $a$ , which simplifies to $a = 0$ . That is true only at the single number $a = 0$ , not for every $a$ : for example $0 \dagger 5 = 10$ , not $5$ . So $e = 0$ fails this second condition in general, even though it satisfied the first one.

Part D

An identity element has to satisfy BOTH $a \dagger e = a$ and $e \dagger a = a$ , for every choice of $a$ , using the SAME value of $e$ in both. Part B shows that the first condition, by itself, already pins $e$ down uniquely:

$a \dagger e = a \ \Longrightarrow\ e = 0,$

so there is no freedom left to choose a different value there. Part C then tests that forced value against the second condition and finds that it only works at the single number $a = 0$ , not for every $a$ .

Since the first condition allows no candidate for $e$ other than $0$ , and $e = 0$ fails the second condition for every $a$ except one, there is no value of $e$ left to try. The operation $\dagger$ has no identity element. The star operation in the lesson reached the same verdict by a different route: there the one-sided condition had no fixed solution at all, since it forced a value that still depended on $a$ , while here it has exactly one candidate and that candidate fails the other side.

In one line

Solving $a \dagger e = a$ forces $e = 0$ , but that same value fails $e \dagger a = a$ for every $a$ except $a = 0$ , so no single value satisfies both identity conditions for every choice of $a$ : the operation $\dagger$ has no identity element.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Evaluates both orders correctly, matching each number to its correct role in the rule. . Worth 2 points.

Reports both results, and states whether the two orders give the same value or different values. . Worth 1 point.

Part B 4 points

Writes the identity condition $a \dagger e = a$ using the definition. . Worth 1 point.

Isolates $e$ and explains why the $a$ terms must cancel, showing that a value of $e$ satisfying this condition cannot depend on $a$ . . Worth 2 points. needs an explanation, not just an answer

Solves the resulting equation for the single value of $e$ that works. . Worth 1 point.

Part C 4 points

Substitutes the value of $e$ into the other order and simplifies. . Worth 2 points.

Explains whether the simplified expression equals $a$ for every choice of $a$ or only for one special value, and identifies which. . Worth 2 points. needs an explanation, not just an answer

Part D 4 points

Connects the results of parts B and C to conclude that no single value of $e$ can satisfy both identity conditions for every choice of $a$ at once, and explains why the search is exhausted rather than merely incomplete. . Worth 3 points. needs an explanation, not just an answer

States the overall verdict about whether $\dagger$ has an identity element. . Worth 1 point.

1. Reading the rule and working from the inside out . Foundational, 13 points. Question 1 of 5.

2. Pricing a package with two inputs . Application, 12 points. Question 2 of 5.

3. Squaring the whole input, then solving backward . Foundational, 15 points. Question 3 of 5.

4. One pair is enough to break the claim . Reasoning, 13 points. Question 4 of 5.

5. Searching for an identity element . Reasoning, 15 points. Question 5 of 5.

Marking rubrics

1. Reading the rule and working from the inside out 13 points

2. Pricing a package with two inputs 12 points

3. Squaring the whole input, then solving backward 15 points

4. One pair is enough to break the claim 13 points

5. Searching for an identity element 15 points