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Introduction to Algebra: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 150 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. The trolley and the shelves . 15 points. Question 1 of 10.

A trolley of books is being reshelved and nobody counted what it started with. After $28$ books are lifted off it, $47$ are still on it.
1. Part A.
  Write a one-step equation in $t$ for the number of books the trolley started with, solve it, and confirm your value by putting it back into your equation.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  All of those books are then placed evenly on $5$ shelves. Write a one-step equation in $s$ for the number on one shelf, solve it, and name the operation you applied to both sides.
  
  Carry your own answer forward Build this equation from the starting count you reached in part A, whatever number that was, and work honestly from there.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A helper says the trolley must have started with $47 - 28 = 19$ books. Explain what quantity that subtraction actually produces, say what the equation in part A calls for instead, and give the reason the equals sign permits that move.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer
Part A

$t - 28 = 47$ , so $t = 75$ books. The check gives $75 - 28 = 47$ .
- $47 = t - 28$ is the same equation with its two sides written the other way round, and $t = 75$ either way
Part B

$5s = 75$ , so $s = 15$ books on one shelf, by dividing both sides by $5$ .
- $75 = 5s$ is the same equation with its two sides written the other way round, and $s = 15$ either way
Part C

It produces the gap between the two given numbers, not the starting count, which was the larger of the two since books were removed from it. The equation calls for the inverse of the subtraction, adding $28$ to both sides, which is permitted because two equal amounts changed equally stay equal.
Worked solution

Part A

The trolley started with $t$ books and $28$ were taken off, leaving $47$ . That is a subtraction attached to the letter.

$t - 28 = 47$

The inverse of subtracting $28$ is adding $28$ , applied to both sides so the equation stays level.

$t - 28 + 28 = 47 + 28 \;\Rightarrow\; t = 75$

Check by substituting into the original equation: $75 - 28 = 47$ , and the two sides agree.

Part B

Five equal shelves of $s$ books hold the whole trolley between them.

$5s = 75$

The letter is multiplied by $5$ , and the inverse of multiplying by $5$ is dividing by $5$ .

$\frac{5s}{5} = \frac{75}{5} \;\Rightarrow\; s = 15$

Check it: five shelves of $15$ books is $75$ books, which is what came off the trolley.

Part C

The helper has repeated the operation the situation already performed instead of undoing it. Books were taken off the trolley, so the starting count is the larger number, and $47 - 28$ can only produce something smaller than $47$ .

$47 - 28 = 19 \text{ is the gap between the two given numbers, not the starting count}$

The equation says what is attached to the letter: $28$ has been subtracted from it. To free the letter you apply the inverse of that, and you apply it to both sides.

$t - 28 + 28 = 47 + 28$

The reason both sides may be changed is the balance principle. If the two sides are the same number, then adding the same amount to each cannot make them different, so every solution the equation had, it still has.

Watch out. A quick check kills the helper's answer in one line: $19 - 28 = -9$ , not $47$ .

In one line

The trolley started with $t = 75$ books, from $t - 28 = 47$ solved by adding $28$ to both sides, and $75 - 28 = 47$ confirms it. Spread over $5$ shelves, $5s = 75$ gives $s = 15$ books on one shelf, by dividing both sides by $5$ . The helper's $47 - 28 = 19$ is the gap between the two given numbers, not the starting count, because books were removed from the larger amount; the equation calls for adding $28$ to both sides instead, and the equals sign permits that because two equal amounts changed equally stay equal.

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 5 points

Writes an equation in which the number removed is subtracted from the starting count, not from the amount left. . Worth 2 points.

Applies the inverse of the attached operation to both sides and carries the arithmetic out. . Worth 2 points.

Substitutes the value back and reports the result as a number of books. . Worth 1 point.

Part B 5 points

Writes an equation in which one shelf's count is multiplied by the number of shelves to give the whole trolley. . Worth 2 points.

Divides both sides by the coefficient rather than subtracting it, and names that operation. . Worth 2 points.

States the result as a number of books on one shelf. . Worth 1 point.

Part C 5 points

Says what the helper's subtraction computes and why it cannot be the starting count, referring to which of the two numbers is the larger. . Worth 2 points. needs an explanation, not just an answer

Names the operation the equation calls for as the inverse of the one attached to the letter, applied to both sides. . Worth 2 points. needs an explanation, not just an answer

Gives the balance reason that doing the same thing to both sides keeps the equation true. . Worth 1 point.
2. One notice board, two kinds of claim . 12 points. Question 2 of 10.

A community garden is being planned. Registering one plot costs $14$ dollars, and $n$ stands for the number of plots one family registers. Several claims about the same scheme are pinned to the notice board, and they are not all the same kind of statement.
1. Part A.
  Write an expression in $n$ for what a family pays to register its plots. Then write an expression for what a family would have paid had it registered seven fewer plots than it did.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  The board also states that one family may register at most $9$ plots, and that the garden goes ahead only if at least $40$ plots are registered altogether. Write each of those as an inequality, using $n$ for one family's plots and $T$ for the total registered, and name the symbol you chose in each.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  A neighbour writes $14n - 7$ for what a family would have paid with seven fewer plots. Decide whether that agrees with your second expression from part A, support the decision with a value of $n$ of your own choosing, and say what each of the two expressions describes.
  
  Carry your own answer forward Compare the neighbour's expression against whichever second expression you wrote in part A, and argue from that one.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer
Part A

$14n$ dollars to register, and $14(n - 7)$ dollars for seven fewer plots.
- $14 \times n$ and $14 \times (n - 7)$ say the same thing; so does $(n - 7)14$
Part B

$n \le 9$ , using "less than or equal to", and $T \ge 40$ , using "greater than or equal to".
- $9 \ge n$ and $40 \le T$ are the same two claims read from the other end
Part C

They do not agree. At $n = 10$ the correct expression gives $14(3) = 42$ dollars while the neighbour's gives $140 - 7 = 133$ dollars. The neighbour's expression describes the full registration with seven dollars knocked off the bill; the correct one describes registering seven fewer plots, each of which costs $14$ dollars.
Worked solution

Part A

Each plot costs the same, so the charge is that cost repeated once per plot.

$14n$

"Seven fewer plots than it did" starts at $n$ and takes seven away, so the count becomes $n - 7$ . The whole reduced count is then charged at $14$ dollars each, which is what the parentheses hold together.

$14(n - 7)$

Part B

"At most $9$ " means no more than $9$ , and $9$ itself is still allowed, so the symbol is the non-strict one that points toward the smaller side.

$n \le 9$

"At least $40$ " means no fewer than $40$ , and $40$ itself is enough, so the symbol is the non-strict one pointing the other way.

$T \ge 40$

Both wordings include their boundary, which is what the bar under each symbol records.

Part C

Test both at a value that makes the reduced count easy to read, say $n = 10$ .

$14(10 - 7) = 14(3) = 42, \qquad 14(10) - 7 = 140 - 7 = 133$

One value where they differ is enough to settle that they are different expressions, and the two numbers are nowhere near each other.

The difference is what the seven is taken from. In $14(n - 7)$ the seven is taken from the plot count, so seven whole plots at $14$ dollars each are removed, which is $98$ dollars less. In $14n - 7$ the seven is taken from the money, so the bill drops by seven dollars and the plot count never changes.

Watch out. Both expressions are perfectly good algebra. What decides between them is which quantity the phrase says the seven belongs to.

In one line

Registering costs $14n$ dollars, and seven fewer plots would have cost $14(n - 7)$ dollars. The two board rules are $n \le 9$ and $T \ge 40$ , both non-strict because "at most" and "at least" allow their boundary numbers. The neighbour's $14n - 7$ does not agree with $14(n - 7)$ : at $n = 10$ they give $133$ dollars and $42$ dollars. The neighbour has taken seven dollars off the bill, while the phrase takes seven plots off the count, and each of those plots was worth $14$ dollars.

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

Writes the registration charge as the cost of one plot multiplied by the number of plots. . Worth 1 point.

Subtracts the seven from the plot count in the order the phrase means rather than the order it is spoken. . Worth 2 points.

Groups the reduced count so that the whole of it is charged, not just part of it. . Worth 1 point.

Part B 4 points

Chooses a symbol pointing toward the smaller side for the cap and toward the larger side for the requirement. . Worth 2 points.

Uses non-strict symbols in both, so each boundary number is allowed. . Worth 2 points.

Part C 4 points

Rejects the agreement and supports it by evaluating both expressions at one stated value of the letter. . Worth 2 points. needs an explanation, not just an answer

Says which quantity the seven is subtracted from in each expression, the count in one and the money in the other. . Worth 2 points.
3. Two expressions from one worksheet . 14 points. Question 3 of 10.

Two lines are copied from the same worksheet. $7(3c + 2) - 5(c - 9) \qquad \text{and} \qquad 9c - (4c - 11)$ Neither has been given a value for its letter, and neither is yet as short as it will go.
1. Part A.
  Shorten the first line to as few terms as it will go. Write out the line with both sets of parentheses cleared before you gather anything.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Shorten the second line, and state what the minus sign written in front of the group does to each term inside it.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Explain what decides whether a pair of terms may be merged into one. Name the property that licenses the merges you made in part A, and say why the terms you were left with could go no further.
  
  Carry your own answer forward Argue from whichever part A result you reached, naming the terms you merged in your own working.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer
Part A

Cleared, the line reads $21c + 14 - 5c + 45$ , and gathered it is $16c + 59$ .
- $59 + 16c$ is the same expression with its two terms written the other way round
Part B

$9c - (4c - 11) = 5c + 11$ . The minus in front of the group is a factor of $-1$ , so it reverses the sign of every term inside, turning $4c$ into $-4c$ and $-11$ into $+11$ .

Part C

Two terms merge when the whole variable part is shared, so pulling it out leaves a plain number behind. The two cleared terms carrying $c$ give $21c - 5c = c(21 - 5) = 16c$ , the distributive property run backward. The two terms left share no variable part, and a shared number alone would not be enough.
Worked solution

Part A

Multiply each outside factor across every term inside its own parentheses. The second factor is $-5$ , a negative, so it flips the sign of what it hits.

$7(3c + 2) = 21c + 14, \qquad -5(c - 9) = -5c + 45$

Write the line with no parentheses left in it, then gather the two kinds of term separately.

$21c + 14 - 5c + 45 = 16c + 59$

The negative times the negative $9$ became $+45$ , which is the step this line was built to test.

Part B

A minus sign standing in front of a bracket is a factor of $-1$ waiting to be distributed.

$-(4c - 11) = (-1)(4c) + (-1)(-11) = -4c + 11$

So the line becomes an expression with no parentheses, and the two multiples of $c$ gather.

$9c - 4c + 11 = 5c + 11$

Watch out. Writing $9c - 4c - 11$ leaves the second sign unflipped, which is the single commonest slip on a bracket with a minus in front of it.

Part C

The rule that lets like terms combine is not a rule of its own. It is the distributive property, $a(b + c) = ab + ac$ , read from right to left, which says a common factor can be pulled out of a sum.

$21c - 5c = c(21 - 5) = c(16) = 16c$

What makes that a merge rather than a mere factoring is that the factor pulled out was the whole variable part, $c$ . Once it is outside, only numbers are left inside the parentheses, so they collapse to a single number and the letter goes straight back on.

Be precise about the test, because it is not that some factor is shared. Pull the $3$ out of $6c - 21$ and you get $3(2c - 7)$ , a perfectly good factoring, and yet the expression is still two terms: what came out was a number rather than the letter, so nothing collapsed.

$16c + 59 \text{ has no shared variable part, so no merge is available}$

The second term carries no $c$ at all, so there is nothing to pull out that would leave numbers behind, and the expression stops at two terms.

In one line

The first line clears to $21c + 14 - 5c + 45$ and gathers to $16c + 59$ , the $+45$ coming from the negative factor meeting the $-9$ . The second line clears to $9c - 4c + 11$ and gathers to $5c + 11$ , because the minus in front of the group is a factor of $-1$ that reverses the sign of every term inside. Terms merge exactly when the whole variable part is shared, since only then does pulling it out leave numbers behind: $21c - 5c = c(21 - 5) = 16c$ is the distributive property run backward. A shared number is not the test, as $6c - 21 = 3(2c - 7)$ shows, and $16c$ and $59$ share no variable part at all, so they stay apart.

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 5 points

Multiplies each outside factor by every term inside its parentheses, not only by the first. . Worth 2 points.

Carries the minus sign of the second factor into both of its products, so the second constant arrives positive. . Worth 2 points.

Shows the cleared line before gathering, and gathers the multiples of the letter separately from the constants. . Worth 1 point.

Part B 4 points

Treats the leading minus as a factor of $-1$ and reverses the sign of both terms inside the group. . Worth 2 points.

Gathers the two multiples of the letter and reports a result of two terms. . Worth 1 point.

States the effect of the leading minus on each term rather than only on the first. . Worth 1 point.

Part C 5 points

States the test for merging as the whole variable part being shared, so that pulling it out leaves only numbers behind. . Worth 2 points. needs an explanation, not just an answer

Names the distributive property run backward and writes the line of algebra that performs the merge. . Worth 2 points. needs an explanation, not just an answer

Says why the remaining pair has no shared variable part, so no merge is available and the expression stops there. . Worth 1 point.
4. Two letters and one order of operations . 15 points. Question 4 of 10.

A rule is written with two letters in it. $4(a - b)^2 - ab$ Its value depends on both, and the order in which its operations run is fixed before either letter is given a number.
1. Part A.
  Work out the value of the rule at $a = 5$ and $b = 8$ . Show the substitution with each value inside its own parentheses, and show each stage of the order of operations.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Now work out the value of the same rule at $a = -3$ and $b = -7$ , again writing each substituted value inside its own parentheses.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A classmate replaces $(a - b)^2$ with $a^2 - b^2$ , saying the exponent reaches each letter separately. Using the values from part A, work out both versions and explain what the first one squares that the second one does not.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer
Part A

The rule is worth $-4$ at these values.
- $-4$ , whether the grouped difference is written $(5) - (8)$ or collapsed to $-3$ before squaring
Part B

The rule is worth $43$ at these values.
- $43$ , reached from $4(4)^2 - 21$
Part C

At $a = 5$ and $b = 8$ the first gives $(5 - 8)^2 = 9$ and the second gives $25 - 64 = -39$ , so the replacement is not allowed. The exponent sits outside the parentheses, so it squares the single number the difference comes to; the classmate's version squares each letter first and subtracts afterwards, which is a different order.
Worked solution

Part A

Substitute both values, each inside its own parentheses, and copy every operation across unchanged.

$4\big((5) - (8)\big)^2 - (5)(8)$

The grouped piece is finished first, then the exponent, then the two multiplications, then the subtraction.

$(5) - (8) = -3, \qquad (-3)^2 = 9, \qquad 4(9) = 36, \qquad (5)(8) = 40$

The line is now a single subtraction.

$36 - 40 = -4$

Part B

Substitute both negative values in parentheses so their signs cannot drift.

$4\big((-3) - (-7)\big)^2 - (-3)(-7)$

Subtracting a negative adds its opposite, so the grouped piece comes out positive.

$(-3) - (-7) = -3 + 7 = 4, \qquad 4^2 = 16, \qquad 4(16) = 64$

The last term is a negative times a negative, which is positive, and it is being subtracted.

$(-3)(-7) = 21, \qquad 64 - 21 = 43$

Part C

Test both at the part A values, since one disagreement is enough to settle it.

$(5 - 8)^2 = (-3)^2 = 9 \qquad \text{against} \qquad 5^2 - 8^2 = 25 - 64 = -39$

The two are not even the same sign, so they are not the same expression.

What separates them is what the exponent is attached to. In $(a - b)^2$ the exponent sits outside the parentheses, so the subtraction runs first and the exponent squares the one number that comes out of it. In $a^2 - b^2$ each exponent is attached to a single letter, so both squarings run first and the subtraction happens afterwards. Reordering operations changes an answer, which is exactly what the order of operations exists to prevent.

Watch out. An exponent never distributes over a sum or a difference. It applies to whatever it is written against, and here that is the whole grouped piece.

In one line

At $a = 5$ and $b = 8$ the rule gives $4(-3)^2 - 40 = 36 - 40 = -4$ . At $a = -3$ and $b = -7$ it gives $4(4)^2 - 21 = 64 - 21 = 43$ , the group coming out positive because subtracting a negative adds its opposite, and the last product coming out positive because both factors are negative. The classmate's replacement fails: at the part A values $(5 - 8)^2 = 9$ while $5^2 - 8^2 = -39$ . The exponent written outside the parentheses squares the single number the difference comes to, whereas squaring each letter first and subtracting afterwards runs the operations in a different order.

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 5 points

Finishes the grouped difference before applying the exponent to it. . Worth 2 points.

Squares the difference to a positive value and multiplies by the outside factor afterwards, not before. . Worth 2 points.

Reports a single number as the value of the rule at these inputs. . Worth 1 point.

Part B 5 points

Handles the subtraction of a negative inside the group and reaches a positive difference. . Worth 2 points.

Gets a positive product from the two negative values and then subtracts it rather than adding it. . Worth 2 points.

Reports a single number as the value of the rule at these inputs. . Worth 1 point.

Part C 5 points

Evaluates both versions at a stated pair of values and reports two numbers that differ. . Worth 2 points.

Explains that the exponent outside the parentheses acts on the result of the subtraction, while the classmate's version squares each letter before subtracting. . Worth 3 points. needs an explanation, not just an answer
5. Two tank records . 14 points. Question 5 of 10.

A pumping station's log records two different tanks. For tank A, the whole day's loss $D$ was shared evenly over the four hours the pump ran, and each of those hours lost $53$ litres. For tank B, the reading fell steadily from $260$ litres to $62$ litres across six hours, losing $L$ litres in each hour. $\dfrac{D}{4} = 53 \qquad \text{and} \qquad 260 - 6L = 62$
1. Part A.
  Solve tank A's equation for $D$ . Name the operation you apply to both sides, and confirm your value by substituting it back.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Solve tank B's equation for $L$ . Set down each inverse move as you apply it, name the number you divide by including its sign, and check the value against the equation as printed.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Explain what feature of each equation's left side decides how many inverse moves are needed to free its letter. Then say whether clearing the constant before the coefficient is the only order that works on the equation that needs two, and what taking that order is worth.
  
  Carry your own answer forward Argue from the two solutions you produced in parts A and B, describing the moves you actually made.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer
Part A

$D = 212$ litres, by multiplying both sides by $4$ . The check gives $\tfrac{212}{4} = 53$ .
- $212$ litres, however the multiplication is written, for example $4 \times 53$
Part B

$L = 33$ litres an hour. Subtracting $260$ from both sides gives $-6L = -198$ , and dividing by the coefficient $-6$ gives $33$ . The check gives $260 - 198 = 62$ .
- $33$ litres per hour, whether the second move is written as dividing by $-6$ or as dividing $-198$ by $-6$
Part C

How many moves are needed is how many operations are attached to the letter: tank A's is only divided, tank B's is multiplied and that product then subtracted from a constant. Clearing the constant first is not the only order, since dividing by $-6$ first also gives $L = 33$ , but it keeps the numbers whole.
Worked solution

Part A

The letter is divided by $4$ and nothing else is attached to it, so a single move frees it. The inverse of dividing by $4$ is multiplying by $4$ .

$4 \cdot \frac{D}{4} = 4 \cdot 53 \;\Rightarrow\; D = 212$

Check it in the original equation: $\tfrac{212}{4} = 53$ , and the two sides agree, so the day's loss was $212$ litres.

Part B

Read the left side as $260 + (-6L)$ . The constant is the outer layer, so clear it first.

$260 - 6L - 260 = 62 - 260 \;\Rightarrow\; -6L = -198$

Now divide both sides by the whole coefficient, sign included. A negative divided by a negative is positive.

$\frac{-6L}{-6} = \frac{-198}{-6} \;\Rightarrow\; L = 33$

Check in the original: $260 - 6(33) = 260 - 198 = 62$ , so the tank lost $33$ litres in each hour.

Part C

Look at how each left side is built up from its letter. Tank A's letter has one thing done to it, a division by $4$ , so one inverse operation undoes it and the letter stands alone.

$\frac{D}{4} \;\longrightarrow\; \text{one operation, one inverse}$

Tank B's letter is first multiplied by $6$ and then that product is subtracted from $260$ . Two operations went on, so two must come off, and the standard order takes the outer one first: the subtraction of $6L$ from $260$ wraps around the multiplication, so it is the outer layer.

$260 - 6L = 62 \;\Rightarrow\; -6L = -198 \;\Rightarrow\; L = 33$

That order is a preference, not a requirement. Dividing every term by $-6$ first is a perfectly good balance move and lands on the same value.

$\frac{260 - 6L}{-6} = \frac{62}{-6} \;\Rightarrow\; L - \frac{130}{3} = -\frac{31}{3} \;\Rightarrow\; L = \frac{99}{3} = 33$

So what clearing the constant first buys is arithmetic rather than correctness. It keeps every number whole, where dividing first splits both the $260$ and the $62$ into thirds. Prefer it for that reason, and never on the grounds that the other route is closed.

Watch out. Counting the moves is not the same as counting the numbers on the page. What matters is how many operations sit between the letter and its value.

In one line

Tank A gives $D = 212$ litres, by multiplying both sides by $4$ , and $\tfrac{212}{4} = 53$ confirms it. Tank B gives $L = 33$ litres an hour: subtracting $260$ from both sides leaves $-6L = -198$ , and dividing by the whole coefficient $-6$ gives $33$ , with $260 - 6(33) = 62$ as the check. One equation needed a single move because only one operation is attached to its letter; the other needed two because its letter is multiplied and that product is then subtracted from a constant. Clearing the constant first is the cleaner order rather than the only one: dividing every term by $-6$ first is equally valid and also reaches $L = 33$ , but it splits the $260$ and the $62$ into thirds, while clearing first leaves every number whole.

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

Multiplies both sides by the divisor rather than dividing again, and names that operation. . Worth 2 points.

Substitutes the value back into the original equation and shows both sides coming out equal. . Worth 1 point.

States the result as a volume in litres for the whole day. . Worth 1 point.

Part B 5 points

Clears the constant before touching the coefficient, leaving the variable term alone on its side. . Worth 2 points.

Divides by the coefficient with its minus sign attached, so the result comes out positive. . Worth 2 points.

States the result as a volume lost in one hour. . Worth 1 point.

Part C 5 points

Ties the number of moves to the number of operations attached to the letter in each equation. . Worth 2 points. needs an explanation, not just an answer

Identifies the constant as the outer layer and says what clearing it first achieves for the step that follows. . Worth 2 points. needs an explanation, not just an answer

Treats dividing by the coefficient first as a valid alternative rather than an error, and says what it costs in arithmetic. . Worth 1 point.
6. One inequality, solved and drawn . 15 points. Question 6 of 10.

One inequality is to be solved and then drawn. $4 - 5x > 29$
1. Part A.
  Solve the inequality, writing each move as an operation applied to both sides. Name the move at which the symbol turned, or say that no move turned it.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Describe the number-line graph of your solution: the boundary number, the kind of circle drawn on it, and the direction the shading runs. For each of the three, say what decided it.
  
  Carry your own answer forward Draw the graph of whichever range you reached in part A, and justify its three features from that range.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
3. Part C.
  Test one value from inside your range and one from outside it in the original inequality, and report what each test gives. Then say which of your two tests would have come out differently had the opposite choice been made about the symbol at the division step.
  
  Carry your own answer forward Choose your two test values from inside and outside whichever range you reached in part A, and substitute them into the inequality as it is printed in the stem.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$x < -5$ . The symbol turned at the second move, the division of both sides by $-5$ ; subtracting the $4$ left it alone.
- $-5 > x$ is the same range read from the other end
Part B

An open circle on $-5$ with the shading running left. The boundary came from the arithmetic of the two moves, the open circle from the symbol being strict so the boundary is excluded, and the leftward shading from the relation being a "less" one, which points toward the smaller numbers.

Part C

Inside: $x = -6$ gives $4 + 30 = 34$ , and $34 > 29$ is true. Outside: $x = 0$ gives $4 > 29$ , which is false. The opposite choice would have claimed $x > -5$ , a range containing $0$ and excluding $-6$ , so both tests contradict it: the value it counts as a solution fails, and the value it rejects passes.
Worked solution

Part A

Read the left side as $4 + (-5x)$ and clear the constant first. Subtraction never disturbs the symbol.

$4 - 5x - 4 > 29 - 4 \;\Rightarrow\; -5x > 25$

Now divide both sides by the coefficient $-5$ . The divisor is negative, so the symbol reverses as you divide.

$\frac{-5x}{-5} < \frac{25}{-5} \;\Rightarrow\; x < -5$

Part B

Three separate features of the answer fix the three features of the picture.

$x < -5$

The boundary is the number the solving produced, $-5$ , and it is the one point where the shading starts. The circle is open, because a strict symbol excludes its boundary: substituting $x = -5$ into the original gives $4 + 25 = 29$ , and $29 > 29$ is false, so $-5$ is not a solution. The shading runs left, because the solutions are the numbers below the boundary, and smaller numbers lie to the left on the line. The arrowhead on that end records that they continue without stopping.

Part C

A range is checked by two tests, one from inside it and one from outside, both in the ORIGINAL inequality.

$x = -6: \quad 4 - 5(-6) = 4 + 30 = 34, \qquad 34 > 29 \;\checkmark$

$x = 0: \quad 4 - 5(0) = 4, \qquad 4 > 29 \;\text{false}$

The inside value passes and the outside value fails, which is what a correct range requires.

Now suppose the symbol had been carried straight down when both sides were divided by $-5$ . That would have given $x > -5$ , which counts $0$ as a solution and rejects $-6$ . Both of the tests above contradict it, and either one alone is enough: $0$ is claimed to be a solution but makes the original false, and $-6$ is claimed not to be one but makes the original true.

Watch out. Test in the original inequality, not in a line partway through the working. A line that already carries a mistake will happily agree with itself.

In one line

Subtracting $4$ from both sides gives $-5x > 25$ with the symbol untouched, and dividing both sides by $-5$ turns the symbol, giving $x < -5$ . The graph is an open circle on $-5$ with the shading running left: the boundary came from the arithmetic, the open circle from the strict symbol excluding it, and the direction from the relation pointing at the smaller numbers. Testing $x = -6$ gives $34 > 29$ , true, and $x = 0$ gives $4 > 29$ , false. Had the symbol been carried straight down, the claim would have been $x > -5$ , and both tests refute it, since $0$ would be counted a solution while failing and $-6$ rejected while passing.

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 5 points

Clears the constant first and leaves the symbol untouched at that step. . Worth 2 points.

Divides by the coefficient with its sign and reverses the symbol at that step. . Worth 2 points.

Names which of the two moves turned the symbol and which did not. . Worth 1 point.

Part B 5 points

Chooses the circle from whether the symbol admits its boundary, and says so. . Worth 2 points.

Chooses the shading direction from the direction of the relation rather than from the sign of the boundary. . Worth 2 points.

Names the boundary as the number the solving produced and marks the solutions as continuing without end. . Worth 1 point.

Part C 5 points

Substitutes one value from inside the range and one from outside into the original inequality and reports both outcomes. . Worth 2 points.

Explains what the opposite choice about the symbol would have claimed and which of the two tests refutes it. . Worth 3 points. needs an explanation, not just an answer
7. Six trays and what stayed in the box . 15 points. Question 7 of 10.

A gardener empties a box of seeds. Six identical trays are filled from it, every tray taking the same number of seeds, and $23$ seeds are left over in the box when no tray will take any more. The box held $275$ seeds to begin with.
1. Part A.
  Let $t$ be the number of seeds one tray takes. Write a single equation in $t$ recording what the box held. Do not solve it, and say what each piece of your equation counts.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 5 points
2. Part B.
  Solve your equation for $t$ . Set down each inverse move as you apply it, give the result with its unit, and confirm the value by putting it back into your equation.
  
  Carry your own answer forward Solve whichever equation you wrote in part A, even if it was not the expected one, and show the two moves on it honestly.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  The gardener's notebook works $275 \div 6$ out as about $45.8$ and records the tray size as $45$ seeds, rounded down. Work out what that division actually counts, and decide whether $45$ can be this tray's size. Support the decision with a calculation.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 5 points
The answer

Part A

$6t + 23 = 275$ . The term $6t$ counts the seeds that went into the six trays, the $23$ counts the seeds still in the box, and the $275$ is the two of those together, which is what the box held.
Part B

$t = 42$ seeds in one tray. Subtracting $23$ from both sides gives $6t = 252$ , and dividing both sides by $6$ gives $42$ . The check gives $6(42) + 23 = 275$ .
- $42$ seeds per tray, however the two moves are written out
Part C

The division shares the whole box among the six trays, so it counts what a tray would take had the box been emptied completely with nothing left over. It cannot be this tray's size, because $23$ seeds never entered a tray: substituting gives $6(45) + 23 = 293$ , which overshoots the $275$ the box held.
Worked solution

Part A

The box's contents ended up in exactly two places: in the trays, or still in the box. Count each.

$6t \text{ seeds in the trays}, \qquad 23 \text{ seeds left over}$

Those two amounts together are everything the box held, so the equation sets their total against $275$ .

$6t + 23 = 275$

The $23$ is added once, not once per tray, because those seeds never entered a tray at all.

Part B

The left-over seeds are the outer layer, added on after the trays were counted, so clear them first.

$6t + 23 - 23 = 275 - 23 \;\Rightarrow\; 6t = 252$

That leaves the variable term alone, ready for the division by the coefficient.

$\frac{6t}{6} = \frac{252}{6} \;\Rightarrow\; t = 42$

Check in the original equation, multiplying before adding: $6(42) + 23 = 252 + 23 = 275$ . One tray takes $42$ seeds.

Part C

Ask what $275 \div 6$ divides. It divides the whole box, all $275$ seeds, among six trays.

$275 \div 6 \text{ shares every seed in the box, including the } 23 \text{ that stayed in it}$

But the situation says $23$ seeds never went into a tray, so only $252$ seeds were ever shared out. Dividing the larger amount can only give a larger tray, which is why the quotient, about $45.8$ , came out above the true $42$ .

The check settles it in one line. Put $45$ back into the equation and see what box it describes.

$6(45) + 23 = 270 + 23 = 293$

A box of $293$ seeds, not $275$ . So $45$ is not this tray's size, and rounding was never the difficulty: the wrong quantity was divided.

Watch out. A number that is close to the answer is not evidence of a correct method. Here the near miss comes from ignoring the constant, which is exactly the step the equation exists to handle.

In one line

The box is recorded by $6t + 23 = 275$ , where $6t$ counts the seeds in the six trays and the $23$ counts what stayed behind. Clearing the constant gives $6t = 252$ , and dividing by $6$ gives $t = 42$ seeds in one tray, with $6(42) + 23 = 275$ as the check. The notebook's $275 \div 6$ shares the whole box, including the $23$ seeds that never entered a tray, so it answers a different question and comes out too large. Substituting its $45$ gives $6(45) + 23 = 293$ seeds, not $275$ , so $45$ cannot be this tray's size, and the fault is the skipped constant rather than the rounding.

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 5 points

Writes the seeds in the trays as one tray's count multiplied by the number of trays. . Worth 2 points.

Adds the left-over seeds once to the whole rather than attaching them to each tray, and sets the total against the box's contents. . Worth 2 points.

Says what each piece of the equation counts in the situation. . Worth 1 point.

Part B 5 points

Clears the constant before dividing, so the variable term stands alone at the division step. . Worth 2 points.

Divides both sides by the coefficient and substitutes the value back into the equation. . Worth 2 points.

States the result as a number of seeds in one tray. . Worth 1 point.

Part C 5 points

Says what quantity the notebook's division shares out, and names the seeds it wrongly includes. . Worth 2 points. needs an explanation, not just an answer

Rejects the recorded size and supports it with a calculation, either substituting it back or comparing against the seeds actually shared. . Worth 2 points. needs an explanation, not just an answer

Attributes the error to the step that was skipped rather than to the rounding. . Worth 1 point.
8. Two students, one expression . 15 points. Question 8 of 10.

Two students are handed the same expression and asked for its value at $k = -2$ . $6(k - 4) - 2(3k - 7) + 5k$ One of them substitutes straight away. The other shortens the expression first and substitutes into what is left.
1. Part A.
  Shorten the expression to as few terms as it will go. Write out the line with both sets of parentheses cleared before you gather anything.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Take both students' routes at $k = -2$ : work out the value from your shortened expression, and work it out again from the expression as it is printed in the stem. Report both numbers.
  
  Carry your own answer forward Use whichever shortened expression you reached in part A for the first route, and the expression printed in the stem for the second.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Your shortened expression contains a single multiple of $k$ , though the printed expression shows three. Explain what became of the other two and where the surviving term comes from, and explain why the two routes in part B were bound to agree whatever value had been chosen.
  
  Carry your own answer forward Base your explanation on whichever cleared and shortened expressions you produced in part A.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer
Part A

Cleared, the line reads $6k - 24 - 6k + 14 + 5k$ , and gathered it is $5k - 10$ .
- $-10 + 5k$ is the same expression with its terms written the other way round
Part B

Both routes give $-20$ . From the shortened form, $5(-2) - 10 = -20$ ; from the printed expression, $6(-6) - 2(-13) + 5(-2) = -36 + 26 - 10 = -20$ .
- $-20$ from both routes, however the middle stages are written
Part C

The surviving $5k$ was never inside either set of parentheses, so the cancelling pair could not touch it. The routes must agree because distributing and gathering like terms only rewrite an expression, leaving its value at every input unchanged.
Worked solution

Part A

Distribute each outside factor across every term inside its own parentheses. The second factor is $-2$ , so it reverses the sign of both terms it reaches.

$6(k - 4) = 6k - 24, \qquad -2(3k - 7) = -6k + 14$

Write the whole line with no parentheses left, keeping the $+5k$ that was never inside any.

$6k - 24 - 6k + 14 + 5k$

Now gather the multiples of $k$ and the constants separately.

$6k - 6k + 5k = 5k, \qquad -24 + 14 = -10, \qquad \text{so the expression is } 5k - 10$

Part B

From the shortened expression, one multiplication and one subtraction finish it.

$5(-2) - 10 = -10 - 10 = -20$

From the printed expression, each grouped piece is finished first, as the order of operations demands.

$(-2) - 4 = -6, \qquad 3(-2) - 7 = -13$

$6(-6) - 2(-13) + 5(-2) = -36 + 26 - 10 = -20$

The second route is the same answer through more arithmetic, and the middle term is where its signs have to be watched: $-2$ times $-13$ is $+26$ .

Part C

Track the three multiples of $k$ separately. Distributing produces $6k$ from the first group and $-6k$ from the second, and those two are opposites, so they cancel.

$6k - 6k = 0$

The $5k$ came from neither group. It was written outside both sets of parentheses and was carried down untouched, so the cancellation above had nothing to do with it.

$6k - 6k + 5k = 5k$

As for the two routes, every step of the shortening was a rewrite rather than a change. Distributing rests on $a(b + c) = ab + ac$ , and gathering like terms rests on the same property read backward, and both are true for every value of the letter, not just for the one you happen to substitute. So the two expressions produce identical numbers at every input.

$5k - 10 \text{ and the printed expression agree at } k = -2, \text{ and at every other } k$

That is exactly why simplifying first is allowed: it is a shortcut through the arithmetic, never a different question.

Watch out. Cancelling terms does not mean the letter has left the expression. It means those particular multiples of it summed to zero.

In one line

Cleared of parentheses the expression reads $6k - 24 - 6k + 14 + 5k$ , which gathers to $5k - 10$ . At $k = -2$ the shortened form gives $5(-2) - 10 = -20$ , and the printed expression gives $6(-6) - 2(-13) + 5(-2) = -36 + 26 - 10 = -20$ . The surviving multiple of $k$ is the $5k$ that sat outside both sets of parentheses, so the cancellation of $6k$ against $-6k$ never reached it. The two routes had to agree because distributing and gathering like terms only rewrite an expression, and both rest on the distributive property, which holds for every value of the letter rather than for the substituted one alone.

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 5 points

Distributes each factor across both terms inside its parentheses, carrying the minus sign of the second factor into both of its products. . Worth 3 points.

Shows the cleared line before gathering, and gathers all three multiples of the letter, including the term that was never in parentheses. . Worth 2 points.

Part B 5 points

Substitutes the negative value inside parentheses and evaluates the shortened expression correctly. . Worth 2 points.

Finishes each grouped piece before multiplying in the second route, and resolves the negative times negative in the middle term. . Worth 2 points.

Reports both values, one for each route. . Worth 1 point.

Part C 5 points

Traces the surviving multiple of the letter to the term that was outside both sets of parentheses. . Worth 2 points. needs an explanation, not just an answer

Argues that distributing and gathering like terms preserve the value at every input, so the two forms agree for any chosen value and not only for this one. . Worth 3 points. needs an explanation, not just an answer
9. Which step can turn a symbol . 17 points. Question 9 of 10.

Two inequalities are set side by side. They call for the same kind of work, and each has to be examined for whether any of its moves turns the symbol around. $\dfrac{x}{-4} + 7 \le 3 \qquad \text{and} \qquad 5x + 26 \le 6$
1. Part A.
  Solve each inequality. For each one, name the move at which the symbol turned, or state that no move turned it.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 6 points
2. Part B.
  Show that one test value settles the first inequality. Choose a number that the opposite choice about the symbol would count as a solution while yours does not, substitute it into the first inequality as printed, and report what it gives.
  
  Carry your own answer forward Compare the range you reached in part A with the range the opposite choice about the symbol would have produced, and take your test value from where the two disagree.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
3. Part C.
  Could multiplying both sides by $-4$ be a valid first move on the first inequality, before the $7$ is touched? Carry that move out carefully, decide whether what you reach is still equivalent to the printed inequality, and explain what happens to every term and to the symbol.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

$x \ge 16$ and $x \le -4$ . The first turned when both sides were multiplied by $-4$ ; the second never turned, its only multiply-or-divide step being a division by the positive $5$ .
- $16 \le x$ and $-4 \ge x$ are the same two ranges read from the other end
Part B

The opposite choice would have claimed $x \le 16$ , which counts $0$ as a solution while my range does not. Substituting $x = 0$ into the printed inequality gives $0 + 7 = 7$ , and $7 \le 3$ is false, so $0$ is not a solution and that claim is refuted by the one value.

Part C

It is valid. The multiplier reaches both terms on the left, turning $\tfrac{x}{-4}$ into $x$ and $7$ into $-28$ , and the right side becomes $-12$ . The multiplier is negative, so the symbol turns, giving $x - 28 \ge -12$ . Adding $28$ finishes at $x \ge 16$ , the same range as before.
Worked solution

Part A

In the first, clear the constant, which never disturbs the symbol, and then undo the division.

$\frac{x}{-4} + 7 - 7 \le 3 - 7 \;\Rightarrow\; \frac{x}{-4} \le -4$

The letter is divided by $-4$ , so multiply both sides by $-4$ and turn the symbol, because the multiplier is negative.

$-4 \cdot \frac{x}{-4} \ge -4 \cdot (-4) \;\Rightarrow\; x \ge 16$

In the second, clear the constant and then divide by the positive coefficient, which leaves the symbol alone.

$5x \le 6 - 26 = -20 \;\Rightarrow\; x \le -4$

Part B

The two competing answers are $x \ge 16$ and $x \le 16$ , and they overlap only at $16$ itself. So any number below $16$ tells them apart; take $0$ , which is easy to substitute.

$\frac{0}{-4} + 7 = 0 + 7 = 7, \qquad 7 \le 3 \;\text{false}$

The competing claim counts $0$ as a solution, and the inequality says it is not one, so that claim is wrong. A single value is enough for it, because one counterexample destroys a claim about a whole range.

It is worth also testing a value from inside the range you kept, since refuting one claim is not by itself a confirmation of the other. Take $x = 24$ .

$\frac{24}{-4} + 7 = -6 + 7 = 1, \qquad 1 \le 3 \;\checkmark$

Watch out. Test in the inequality as printed. A line partway through the working may already carry the error you are hunting.

Part C

Judge the move itself rather than the order it comes in. Multiplying both sides by $-4$ is a permitted balance move, and on the left it has to reach every term, not only the one carrying the letter.

$-4\left(\frac{x}{-4} + 7\right) = x - 28, \qquad -4(3) = -12$

The multiplier is negative, so the symbol reverses at that step.

$x - 28 \ge -12$

That line is still equivalent to the printed inequality, because every step taken to reach it preserves the solutions. Finishing takes one more move, adding $28$ to both sides, which leaves the symbol alone.

$x \ge 16$

The same range as the standard order gives, which is what a valid alternative route has to produce. The usual order is preferred for a different reason: clearing the constant first keeps each step's arithmetic simpler, and here it avoids having to spread the multiplier across two terms.

Watch out. A route that is not the one you were taught is not automatically an error. Judge it by whether each step is a permitted move correctly carried out.

In one line

The first inequality gives $x \ge 16$ , the symbol turning at the multiplication of both sides by $-4$ ; the second gives $x \le -4$ , with nothing to turn it, since its only multiply-or-divide step is a division by the positive $5$ . The opposite choice about the symbol in the first would have claimed $x \le 16$ , and $x = 0$ refutes that in one line: $0 + 7 = 7$ , and $7 \le 3$ is false. Multiplying both sides by $-4$ as a first move is also valid: it reaches both terms on the left, the right side becomes $-12$ , and the negative multiplier turns the symbol, giving $x - 28 \ge -12$ , which adding $28$ finishes at the same $x \ge 16$ .

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 6 points

Clears the constant first in both, leaving the symbol untouched at those steps. . Worth 2 points.

Turns the symbol in the first inequality at the step where both sides meet a negative multiplier, and reaches the correct boundary. . Worth 2 points.

Leaves the symbol alone in the second and says which move could have turned it and why it did not. . Worth 2 points.

Part B 5 points

Chooses a value that the two competing ranges disagree about, and says which range each puts it in. . Worth 2 points. needs an explanation, not just an answer

Substitutes it into the inequality as printed and reports whether the statement comes out true or false. . Worth 2 points.

States what that single outcome settles about the competing claim. . Worth 1 point. needs an explanation, not just an answer

Part C 6 points

Judges the move valid and shows the multiplier reaching both terms on the left, naming what each becomes and what the right side becomes. . Worth 3 points. needs an explanation, not just an answer

Notes that the multiplier is negative and reverses the symbol at that step. . Worth 2 points. needs an explanation, not just an answer

Carries the route to its end and compares the range it reaches with the one from the standard order. . Worth 1 point.
10. The symbol read from the other end . 18 points. Question 10 of 10.

Three finished solutions are copied from a worksheet, and two of them put the letter on the right of the symbol rather than the left. $8 \ge m, \qquad m > -4.5, \qquad -1 < m$
1. Part A.
  Rewrite each of the three so that the letter stands on the left of the symbol, and say what happens to a symbol when the two sides swap places.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 6 points
2. Part B.
  Describe the number-line graph of each of the three solutions: the boundary number, the kind of circle on it, and the direction of the shading.
  
  Carry your own answer forward Describe the graphs of the three solutions in whichever form you rewrote them in part A.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 6 points
3. Part C.
  A student graphs the third solution as an open circle on $-1$ with the shading running left, on the grounds that the symbol printed in the stem is a "less than". Decide whether that graph is right, and support the decision with one value that settles it.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

$m \le 8$ , $m > -4.5$ and $m > -1$ . On a swap the symbol is redrawn so it still opens toward the same quantity, so a "greater" one becomes a "less" one and the other way round.
- The middle one is already in this form and may be copied unchanged
Part B

First: a closed circle on $8$ , shaded left. Second: an open circle on $-4.5$ , shaded right. Third: an open circle on $-1$ , shaded right. The bar under a symbol decides the circle, and the direction of the relation once the letter is on the left decides the shading.

Part C

The graph is wrong. In $-1 < m$ the pointed end is aimed at the $-1$ , so the letter is the larger amount and the solutions lie above $-1$ , not below. The value $3$ settles it: $-1 < 3$ is true, so $3$ is a solution, and yet it sits to the right of $-1$ , outside the student's shading.
Worked solution

Part A

A symbol opens toward the larger amount and narrows to a point at the smaller one, and swapping the sides must not change which amount is claimed to be larger.

$8 \ge m \;\Longrightarrow\; m \le 8$

In both versions the $8$ is the larger of the two, so the symbol has to be redrawn pointing the other way. The second solution already has the letter on the left, so it is copied unchanged. The third swaps the same way as the first.

$-1 < m \;\Longrightarrow\; m > -1$

Here $m$ is the larger amount in both versions, which is why the strict symbol now opens toward the letter.

Part B

Read each rewritten solution and take its two decisions in turn.

$m \le 8, \qquad m > -4.5, \qquad m > -1$

The circle comes from strictness. The first symbol carries a bar, so $8$ is a solution and its circle is filled. The second and third are strict, so their boundaries are shut out and their circles are hollow.

The shading comes from the direction of the relation, once the letter is on the left. The first is a "less" relation, so it shades toward the smaller numbers, to the left of $8$ . The other two are "greater" relations, so both shade right, one from $-4.5$ and one from $-1$ .

Notice that the sign of the boundary decides nothing here. Both $-4.5$ and $-1$ are negative, and both shade right, because the direction is fixed by the symbol and not by where the boundary sits relative to zero.

Part C

The student has read the symbol as a word rather than as a picture of which side is larger. In $-1 < m$ the narrow point aims at $-1$ and the opening faces $m$ , so the claim is that $m$ is the larger of the two.

$-1 < m \;\Longrightarrow\; m > -1$

One value settles the graph. Take $m = 3$ and put it into the solution as printed.

$-1 < 3 \;\checkmark$

So $3$ is a solution, and $3$ lies to the right of $-1$ , on the side the student left unshaded. A graph that omits a solution is wrong, and one omitted solution is enough to say so.

A second value confirms the direction rather than just refuting the student's: $m = -6$ gives $-1 < -6$ , which is false, so $-6$ is not a solution, and it is exactly the sort of number the student's leftward shading claims.

Watch out. The safe habit is to put the letter on the left first, redraw the symbol as part of that move, and only then read the direction off. Reading the direction straight from a symbol whose letter is on the right is where this error starts.

In one line

With the letter on the left the three solutions read $m \le 8$ , $m > -4.5$ and $m > -1$ , the symbol being redrawn on a swap so that it still opens toward the same quantity. Their graphs are a closed circle on $8$ shaded left, an open circle on $-4.5$ shaded right, and an open circle on $-1$ shaded right, the bar deciding each circle and the direction of the relation deciding each shading. The student's leftward shading on the third is wrong: $-1 < m$ aims its point at $-1$ , so the solutions lie above it, and $m = 3$ makes $-1 < 3$ true while sitting on the unshaded side.

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 6 points

Rewrites both of the reversed solutions with the letter on the left and the symbol redrawn to face the other way. . Worth 3 points.

Leaves untouched the solution that is already written with the letter on the left. . Worth 1 point.

States the rule for a swap in terms of which quantity the symbol opens toward, rather than as a memorised flip. . Worth 2 points.

Part B 6 points

Chooses a filled circle for the non-strict solution and hollow circles for the two strict ones. . Worth 2 points.

Shades left for the "less" relation and right for both "greater" relations, with the correct boundary on each. . Worth 3 points.

Keeps the direction independent of whether the boundary number is negative. . Worth 1 point.

Part C 6 points

Rejects the graph and explains the direction from which quantity the symbol opens toward, not from the words used to read it. . Worth 3 points. needs an explanation, not just an answer

Names one value, checks it in the solution as printed, and points out which side of the boundary it lies on. . Worth 2 points.

States what a single omitted solution settles about a graph. . Worth 1 point. needs an explanation, not just an answer

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. The trolley and the shelves 15 points

Part A 5

Writes an equation in which the number removed is subtracted from the starting count, not from the amount left. 2
Applies the inverse of the attached operation to both sides and carries the arithmetic out. 2
Substitutes the value back and reports the result as a number of books. 1

Part B 5

Writes an equation in which one shelf's count is multiplied by the number of shelves to give the whole trolley. 2
Divides both sides by the coefficient rather than subtracting it, and names that operation. 2
States the result as a number of books on one shelf. 1

Part C 5

Says what the helper's subtraction computes and why it cannot be the starting count, referring to which of the two numbers is the larger. 2
Names the operation the equation calls for as the inverse of the one attached to the letter, applied to both sides. 2
Gives the balance reason that doing the same thing to both sides keeps the equation true. 1

2. One notice board, two kinds of claim 12 points

Part A 4

Writes the registration charge as the cost of one plot multiplied by the number of plots. 1
Subtracts the seven from the plot count in the order the phrase means rather than the order it is spoken. 2
Groups the reduced count so that the whole of it is charged, not just part of it. 1

Part B 4

Chooses a symbol pointing toward the smaller side for the cap and toward the larger side for the requirement. 2
Uses non-strict symbols in both, so each boundary number is allowed. 2

Part C 4

Rejects the agreement and supports it by evaluating both expressions at one stated value of the letter. 2
Says which quantity the seven is subtracted from in each expression, the count in one and the money in the other. 2

3. Two expressions from one worksheet 14 points

Part A 5

Multiplies each outside factor by every term inside its parentheses, not only by the first. 2
Carries the minus sign of the second factor into both of its products, so the second constant arrives positive. 2
Shows the cleared line before gathering, and gathers the multiples of the letter separately from the constants. 1

Part B 4

Treats the leading minus as a factor of $-1$ and reverses the sign of both terms inside the group. 2
Gathers the two multiples of the letter and reports a result of two terms. 1
States the effect of the leading minus on each term rather than only on the first. 1

Part C 5

States the test for merging as the whole variable part being shared, so that pulling it out leaves only numbers behind. 2
Names the distributive property run backward and writes the line of algebra that performs the merge. 2
Says why the remaining pair has no shared variable part, so no merge is available and the expression stops there. 1

4. Two letters and one order of operations 15 points

Part A 5

Finishes the grouped difference before applying the exponent to it. 2
Squares the difference to a positive value and multiplies by the outside factor afterwards, not before. 2
Reports a single number as the value of the rule at these inputs. 1

Part B 5

Handles the subtraction of a negative inside the group and reaches a positive difference. 2
Gets a positive product from the two negative values and then subtracts it rather than adding it. 2
Reports a single number as the value of the rule at these inputs. 1

Part C 5

Evaluates both versions at a stated pair of values and reports two numbers that differ. 2
Explains that the exponent outside the parentheses acts on the result of the subtraction, while the classmate's version squares each letter before subtracting. 3

5. Two tank records 14 points

Part A 4

Multiplies both sides by the divisor rather than dividing again, and names that operation. 2
Substitutes the value back into the original equation and shows both sides coming out equal. 1
States the result as a volume in litres for the whole day. 1

Part B 5

Clears the constant before touching the coefficient, leaving the variable term alone on its side. 2
Divides by the coefficient with its minus sign attached, so the result comes out positive. 2
States the result as a volume lost in one hour. 1

Part C 5

Ties the number of moves to the number of operations attached to the letter in each equation. 2
Identifies the constant as the outer layer and says what clearing it first achieves for the step that follows. 2
Treats dividing by the coefficient first as a valid alternative rather than an error, and says what it costs in arithmetic. 1

6. One inequality, solved and drawn 15 points

Part A 5

Clears the constant first and leaves the symbol untouched at that step. 2
Divides by the coefficient with its sign and reverses the symbol at that step. 2
Names which of the two moves turned the symbol and which did not. 1

Part B 5

Chooses the circle from whether the symbol admits its boundary, and says so. 2
Chooses the shading direction from the direction of the relation rather than from the sign of the boundary. 2
Names the boundary as the number the solving produced and marks the solutions as continuing without end. 1

Part C 5

Substitutes one value from inside the range and one from outside into the original inequality and reports both outcomes. 2
Explains what the opposite choice about the symbol would have claimed and which of the two tests refutes it. 3

7. Six trays and what stayed in the box 15 points

Part A 5

Writes the seeds in the trays as one tray's count multiplied by the number of trays. 2
Adds the left-over seeds once to the whole rather than attaching them to each tray, and sets the total against the box's contents. 2
Says what each piece of the equation counts in the situation. 1

Part B 5

Clears the constant before dividing, so the variable term stands alone at the division step. 2
Divides both sides by the coefficient and substitutes the value back into the equation. 2
States the result as a number of seeds in one tray. 1

Part C 5

Says what quantity the notebook's division shares out, and names the seeds it wrongly includes. 2
Rejects the recorded size and supports it with a calculation, either substituting it back or comparing against the seeds actually shared. 2
Attributes the error to the step that was skipped rather than to the rounding. 1

8. Two students, one expression 15 points

Part A 5

Distributes each factor across both terms inside its parentheses, carrying the minus sign of the second factor into both of its products. 3
Shows the cleared line before gathering, and gathers all three multiples of the letter, including the term that was never in parentheses. 2

Part B 5

Substitutes the negative value inside parentheses and evaluates the shortened expression correctly. 2
Finishes each grouped piece before multiplying in the second route, and resolves the negative times negative in the middle term. 2
Reports both values, one for each route. 1

Part C 5

Traces the surviving multiple of the letter to the term that was outside both sets of parentheses. 2
Argues that distributing and gathering like terms preserve the value at every input, so the two forms agree for any chosen value and not only for this one. 3

9. Which step can turn a symbol 17 points

Part A 6

Clears the constant first in both, leaving the symbol untouched at those steps. 2
Turns the symbol in the first inequality at the step where both sides meet a negative multiplier, and reaches the correct boundary. 2
Leaves the symbol alone in the second and says which move could have turned it and why it did not. 2

Part B 5

Chooses a value that the two competing ranges disagree about, and says which range each puts it in. 2
Substitutes it into the inequality as printed and reports whether the statement comes out true or false. 2
States what that single outcome settles about the competing claim. 1

Part C 6

Judges the move valid and shows the multiplier reaching both terms on the left, naming what each becomes and what the right side becomes. 3
Notes that the multiplier is negative and reverses the symbol at that step. 2
Carries the route to its end and compares the range it reaches with the one from the standard order. 1

10. The symbol read from the other end 18 points

Part A 6

Rewrites both of the reversed solutions with the letter on the left and the symbol redrawn to face the other way. 3
Leaves untouched the solution that is already written with the letter on the left. 1
States the rule for a swap in terms of which quantity the symbol opens toward, rather than as a memorised flip. 2

Part B 6

Chooses a filled circle for the non-strict solution and hollow circles for the two strict ones. 2
Shades left for the "less" relation and right for both "greater" relations, with the correct boundary on each. 3
Keeps the direction independent of whether the boundary number is negative. 1

Part C 6

Rejects the graph and explains the direction from which quantity the symbol opens toward, not from the words used to read it. 3
Names one value, checks it in the solution as printed, and points out which side of the boundary it lies on. 2
States what a single omitted solution settles about a graph. 1

Multiple choice

Free response