Chapter test · nothing is marked until you submit

Linear Equations: 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, 129 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. Two equations, and a different way to begin . 10 points. Question 1 of 10.

Both equations below come apart with the ordinary moves. The last part looks at a different way to begin the second one.
1. Part A.
  Solve $7 - 2x = 3x + 22$ , and check your value in the equation as it is written above.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Solve $4(x + 3) - 2(x - 1) = 10$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Dividing both sides of part B's equation by $2$ as the very first move, before anything is expanded, is also legal. Carry that route out to the end. Then justify why the two routes could not have reached different values, and state what the division had to reach for that to be true.
  
  Carry your own answer forward Compare this route against whatever value you reached in part B, even if it was not the expected one, and say honestly whether the two agree. The credit here is for the account of why an early division is safe and what it must cover, not for landing on a particular number.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer
Part A

$x = -3$ , and both sides come to $13$ there.
- $x = -3$ is the value; an answer of $x = 3$ has lost the sign in the final division
Part B

$x = -2$ .
- $x = -2$ only; an answer of $x = 0$ comes from writing $-2(x - 1)$ as $-2x - 2$
Part C

The division gives $2(x + 3) - (x - 1) = 5$ , which finishes at the same $x = -2$ . The routes agree because dividing both sides by $2$ is a property of equality, so the new equation has exactly the same solutions, and the move is only that if the $2$ divides every term on BOTH sides, the $10$ included.
Worked solution

Part A

Collect the variable on the side where its coefficient is larger, so add $2x$ to both sides, then subtract $22$ .

$7 - 2x = 3x + 22 \quad\Longrightarrow\quad 7 = 5x + 22 \quad\Longrightarrow\quad -15 = 5x \quad\Longrightarrow\quad x = -3$

The check uses the original: the left is $7 - 2(-3) = 13$ and the right is $3(-3) + 22 = 13$ .

Part B

Expand both brackets first, remembering that $-2 \times (-1) = +2$ , then combine like terms.

$4x + 12 - 2x + 2 = 10 \quad\Longrightarrow\quad 2x + 14 = 10 \quad\Longrightarrow\quad 2x = -4 \quad\Longrightarrow\quad x = -2$

Checking in the original: $4(1) - 2(-3) = 4 + 6 = 10$ .

Part C

Dividing both sides by $2$ must reach every term, including the constant on the right:

$2(x + 3) - (x - 1) = 5 \quad\Longrightarrow\quad 2x + 6 - x + 1 = 5 \quad\Longrightarrow\quad x + 7 = 5 \quad\Longrightarrow\quad x = -2.$

The two routes agree because the division property of equality holds for any nonzero divisor, and $2$ is nonzero once and for all, whatever $x$ turns out to be. So the divided equation is equivalent to the original: every number satisfying one satisfies the other, and the move is undone by multiplying back by $2$ .

What makes it that move, rather than a different equation altogether, is that the $2$ divides the WHOLE of each side. Dividing $4(x + 3)$ and $2(x - 1)$ but leaving the $10$ alone would produce an equation with a different solution, and no property of equality would license it.

In one line

$7 - 2x = 3x + 22$ gives $x = -3$ , where both sides come to $13$ , and $4(x + 3) - 2(x - 1) = 10$ gives $x = -2$ . Dividing that second equation by $2$ first turns it into $2(x + 3) - (x - 1) = 5$ and reaches the same $x = -2$ : the division property of equality produces an equivalent equation whenever the divisor is nonzero, provided the $2$ is applied to every term on both sides, the $10$ included.

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

Collects the variable terms on one side and the constants on the other, changing the sign of each term that crosses. . Worth 2 points.

Reports the value and confirms it by evaluating each side of the original equation separately. . Worth 1 point.

Part B 3 points

Distributes each factor across both terms inside its brackets, with the correct sign on the product of the two negatives. . Worth 2 points.

Reports the value and confirms it in the equation as it was given. . Worth 1 point.

Part C 4 points

Carries the divided equation through to a value rather than stopping at the first line. . Worth 1 point.

Grounds the agreement in something stronger than the two answers happening to match on this occasion. . Worth 2 points. needs an explanation, not just an answer

Names what the division had to cover for the two routes to agree. . Worth 1 point.
2. Two equations, one constant apart . 12 points. Question 2 of 10.

These two equations differ in a single constant:

$\frac{x}{2} + \frac{x - 3}{4} = \frac{3x - 3}{4}, \qquad \frac{x}{2} + \frac{x - 3}{4} = \frac{3x - 1}{4}.$

Both clear with the same multiplier.
1. Part A.
  Clear the fractions from the FIRST equation and say how many numbers satisfy it. Support your verdict by testing it at two values of your own choosing.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Do the same for the SECOND equation, and again test your verdict at two values.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  One constant separates these two equations, and their answers are as far apart as two answers can be. Say what the cleared line of each one records, and what a reader should take from the fact that a single constant moves an equation between those two extremes.
  
  Carry your own answer forward Interpret whichever cleared lines and verdicts you produced in parts A and B, even if they were not the expected ones. The credit here is for saying what a leftover statement about constants decides, not for having reached one particular pair.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
The answer

Part A

Clearing gives $3x - 3 = 3x - 3$ , so every number satisfies it: infinitely many solutions. Two tests: at $x = 0$ both sides are $-\tfrac{3}{4}$ , and at $x = 5$ both sides are $3$ .

Part B

Clearing gives $3x - 3 = 3x - 1$ , so there is no solution. At $x = 0$ the sides are $-\tfrac{3}{4}$ and $-\tfrac{1}{4}$ ; at $x = 5$ they are $3$ and $\tfrac{7}{2}$ .

Part C

Each cleared line compares constants only, because the variable terms match on both sides. They agree in the first, making it an identity; they differ in the second, making it a contradiction. Once the variable has cancelled, the constants are all that is left to be true or false.

Worked solution

Part A

Multiply every term by $4$ , the smallest number both denominators divide:

$2x + (x - 3) = 3x - 3 \quad\Longrightarrow\quad 3x - 3 = 3x - 3 \quad\Longrightarrow\quad -3 = -3.$

The variable cancels and leaves a statement that is true, so no value of $x$ can break it: every number is a solution. Testing at $x = 0$ gives $0 - \tfrac{3}{4} = -\tfrac{3}{4}$ on both sides, and at $x = 5$ gives $\tfrac{5}{2} + \tfrac{1}{2} = 3$ on the left and $\tfrac{12}{4} = 3$ on the right.

Part B

The same multiplier of $4$ works, because only a constant has changed:

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

The variable cancels again, but this time the statement left behind is false, and no choice of $x$ can repair a falsehood about numbers. At $x = 0$ the two sides are $-\tfrac{3}{4}$ and $-\tfrac{1}{4}$ ; at $x = 5$ they are $3$ and $\tfrac{7}{2}$ . In both tests the right side runs exactly $\tfrac{1}{2}$ ahead.

Part C

Both equations have the same variable part on each side, $3x$ against $3x$ , so in both the variable was always going to cancel. What the cleared line records is the comparison of the constants alone:

$-3 = -3 \quad \text{(true, so every number works)}, \qquad -3 = -1 \quad \text{(false, so no number works)}.$

The reader's lesson is about where the information sits. While the variable is present, changing a constant usually slides the solution along; here the variable is gone by the time the constants meet, so the constants are no longer adjusting an answer, they are deciding whether there is one at all. That is why the outcome jumps straight from everything to nothing with nothing in between: an equation whose variable cancels has no room for a single value.

In one line

Clearing with the multiplier $4$ turns the first equation into $3x - 3 = 3x - 3$ and the second into $3x - 3 = 3x - 1$ . The first leaves the true statement $-3 = -3$ , so every number satisfies it; the second leaves the false statement $-3 = -1$ , so nothing does. The variable terms are identical on both sides in each equation, so the constants are not adjusting a solution, they are deciding whether one exists, which is why one constant carries the answer from everything to nothing.

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 every term, both fractions and the whole right-hand side, by a common multiplier of the denominators. . Worth 2 points.

Reads the verdict off the statement left behind rather than off the tests, and confirms it at two different values, not one. . Worth 2 points.

Part B 4 points

Clears the second equation with the same multiplier and reaches a statement with no variable in it. . Worth 2 points.

States the number of solutions on the strength of the leftover statement, and exhibits two values that bear the verdict out. . Worth 2 points.

Part C 4 points

Says what each cleared line is comparing, and why nothing else is left to compare. . Worth 2 points. needs an explanation, not just an answer

Explains what the leftover statement decides once the variable has gone, and why that leaves no room for a single value. . Worth 2 points. needs an explanation, not just an answer
3. One formula, three letters it can be aimed at . 12 points. Question 3 of 10.

A trapezoid with parallel sides $b_1$ and $b_2$ and height $h$ has area

$A = \frac{h(b_1 + b_2)}{2}.$

As written, the formula computes the area from the other three quantities. It does not have to be used that way.
1. Part A.
  Solve the formula for $b_1$ , and state the restriction your final step requires.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  A trapezoid has area $42$ square centimetres, height $6$ centimetres, and one parallel side of $5$ centimetres. Find the other parallel side, and confirm the result in the formula as it was given.
  
  Carry your own answer forward Substitute into whichever rearrangement you produced in part A, even if it was not the expected one, and then run the numbers forward through the original formula and report honestly whether they agree. The credit here is for substituting correctly and testing the result, not for arriving at a particular length.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Aimed at the height instead, the same formula becomes $h = \dfrac{2A}{b_1 + b_2}$ . Explain what is the same about that statement and the one you wrote in part A, and what is different about them, and say which of the two differences a person choosing between them would actually care about.
  
  Carry your own answer forward Compare against whichever rearrangement you produced in part A, even if it was not the expected one, and describe its condition honestly. The credit here is for the account of what a rearrangement does and does not change, not for having matched a particular form.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer
Part A

$b_1 = \dfrac{2A}{h} - b_2$ , valid when $h \neq 0$ .
- $b_1 = \dfrac{2A - hb_2}{h}$ is the same expression over a common denominator; $b_1 = \dfrac{2A}{h} - \dfrac{b_2}{h}$ is not, because $b_2$ was never divided by $h$
Part B

The other parallel side is $9$ centimetres.

Part C

Both are the same statement: the same values satisfy each, subject to the condition each carries. What differs is which letter stands alone, and what the division needs, $h \neq 0$ against $b_1 + b_2 \neq 0$ . A user cares about the isolated letter, since that decides which quantity arrives with no algebra attached.
Worked solution

Part A

Undo the operations wrapped around $b_1$ from the outside in: the division by $2$ , then the multiplication by $h$ , then the addition of $b_2$ .

$2A = h(b_1 + b_2) \quad\Longrightarrow\quad \frac{2A}{h} = b_1 + b_2 \quad\Longrightarrow\quad b_1 = \frac{2A}{h} - b_2$

The middle step divides by $h$ , so the result holds only when $h \neq 0$ . That step is the only one that needs a condition; multiplying by $2$ and subtracting $b_2$ need none.

Part B

Substitute into the rearranged formula, keeping every length in centimetres:

$b_1 = \frac{2(42)}{6} - 5 = 14 - 5 = 9.$

Running it forward through the original confirms it: $\dfrac{6(9 + 5)}{2} = \dfrac{84}{2} = 42$ square centimetres, which is the area given.

Part C

What is the same. Every step from $A = \dfrac{h(b_1 + b_2)}{2}$ to either form is a property of equality, so the forms are equivalent: a set of values fitting one fits the others. Neither says anything new about a trapezoid.

What differs. One letter stands alone, and that letter is the one you can compute:

$b_1 = \frac{2A}{h} - b_2 \;\;(h \neq 0), \qquad h = \frac{2A}{b_1 + b_2} \;\;(b_1 + b_2 \neq 0).$

The conditions differ too, because a different expression sits under the fraction bar in each.

Which difference matters in use. The conditions are technically different but neither can be violated by a real trapezoid, whose height and sides are positive lengths. What a user notices is the first difference: the form decides which quantity arrives with no algebra attached, so you rearrange once, toward the letter you will be computing again and again.

In one line

$b_1 = \dfrac{2A}{h} - b_2$ , valid when $h \neq 0$ , which gives a second parallel side of $9$ centimetres for the trapezoid in part B, confirmed by $\dfrac{6(9 + 5)}{2} = 42$ square centimetres. That form and $h = \dfrac{2A}{b_1 + b_2}$ are the same statement reached by properties of equality, differing only in which letter stands alone and in what each division requires; the isolated letter is the difference that decides which form is worth writing down.

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

Undoes the operations around $b_1$ in reverse order, clearing the fraction before dividing by $h$ . . Worth 2 points.

Subtracts $b_2$ from the whole of the other side rather than from part of it. . Worth 1 point.

States the restriction the final step requires and attaches it to the step that requires it. . Worth 1 point.

Part B 4 points

Substitutes the three given values into the rearranged formula correctly. . Worth 2 points.

Reports the answer as a length in centimetres and confirms it by computing the area forward from the original formula. . Worth 2 points.

Part C 4 points

Explains what the two forms have in common by appealing to the steps that connect them, not to how they look. . Worth 2 points. needs an explanation, not just an answer

Names more than one difference between the two forms. . Worth 1 point.

Chooses one difference as the one that matters in use, and gives a reason that refers to what the user is computing. . Worth 1 point. needs an explanation, not just an answer
4. Two price lists that cross once . 13 points. Question 4 of 10.

A print shop charges a fixed setup fee of $45$ dollars for a job and then $2$ dollars for each poster printed. A rival shop charges no setup fee at all and $3.50$ dollars for each poster. A customer wants $n$ posters, all from one shop.
1. Part A.
  Say what $n$ counts, write an expression for what each shop charges for $n$ posters, and write one equation saying the two shops charge the same.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
2. Part B.
  Solve your equation, and report the size of order at which the two shops charge the same and what that order costs. Check both figures against the words of the problem.
  
  Carry your own answer forward Solve whichever equation you wrote in part A, even if it was not the expected one, and read the cost back through the shops' own price lists rather than through the equation. The credit here is for solving your own model and reporting both quantities with their units.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Work out what each shop would charge for $20$ posters and for $50$ posters. Compare the two shops as a customer would have to, and say what the order size you found in part B does, and does not, tell that customer.
  
  Carry your own answer forward Compare against whichever crossing order you found in part B, even if it was not the expected one, and say honestly whether your two test orders fall on the same side of it or on opposite sides. The credit here is for the comparison and for what a crossing point does and does not settle.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 5 points
The answer

Part A

$n$ is the number of posters in the order. The print shop charges $45 + 2n$ dollars and the rival charges $3.5n$ dollars, so the two are equal when $45 + 2n = 3.5n$ .

Part B

The charges agree at an order of $30$ posters, and the order costs $105$ dollars at either shop.

Part C

At $20$ posters the rival charges $70$ dollars against the print shop's $85$ ; at $50$ it charges $175$ against $145$ . So the rival is cheaper below the crossing order and dearer above it. That order is where the choice stops mattering; it names no winner, only where the verdict changes.

Worked solution

Part A

Only one quantity is unknown, the size of the order, so one letter carries the whole problem. Each shop's charge is its setup fee plus its per-poster price times the number of posters, and the rival's setup fee is zero:

$\text{print shop: } 45 + 2n, \qquad \text{rival: } 3.5n, \qquad 45 + 2n = 3.5n.$

The equation says the two charges are the same number of dollars, which is what "charge the same" means.

Part B

Collect the variable on the side where its coefficient is larger, then divide:

$45 + 2n = 3.5n \quad\Longrightarrow\quad 45 = 1.5n \quad\Longrightarrow\quad n = 30.$

The question asks for two things, and $n$ is only the first. Reading the cost back off each shop's own words: the print shop charges $45 + 2(30) = 105$ dollars and the rival charges $3.5(30) = 105$ dollars, which agree, as they must.

Part C

Evaluate both price lists at both order sizes:

$20: \; 45 + 40 = 85 \;\text{ against }\; 70, \qquad 50: \; 45 + 100 = 145 \;\text{ against }\; 175.$

The rival wins the small order by $15$ dollars and loses the large one by $30$ . That is the shape of the comparison: the rival starts ahead because it charges nothing to set up, and falls behind because it charges $1.50$ more on every poster, so the setup fee is eaten away at $1.50$ a poster and runs out after $30$ of them.

The crossing order answers a narrow question. It is the single order size at which the choice is free, and it is the boundary between the two verdicts, but on its own it names no winner: a customer still has to know whether their order is above it or below it. Testing two order sizes, one on each side, is what turns the boundary into advice.

In one line

With $n$ the number of posters, $45 + 2n = 3.5n$ gives $n = 30$ , and an order of $30$ posters costs $105$ dollars at either shop. Either side of that the verdict reverses: at $20$ posters the rival is cheaper, $70$ dollars against $85$ , and at $50$ posters it is dearer, $175$ against $145$ . The crossing order marks where the cheaper shop changes; it does not name one, so a customer still has to say which side of $30$ their order falls on.

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

States what the letter counts, so the expressions can be read. . Worth 1 point.

Builds each charge as a fixed amount plus a per-poster amount times the count, with the rival's fixed amount zero. . Worth 2 points.

Sets the two expressions equal to each other, rather than equal to a number the problem never gave. . Worth 1 point.

Part B 4 points

Solves the equation correctly, keeping the decimal coefficient intact. . Worth 2 points.

Reports the order as a number of posters and the cost in dollars, both asked for and both labelled. . Worth 1 point.

Checks the cost by running the number of posters through each shop's price list separately. . Worth 1 point.

Part C 5 points

Evaluates both charges at both order sizes and names which shop is cheaper in each case. . Worth 2 points.

Accounts for the pattern in the two comparisons using the two shops' fees and per-poster prices, not just the two totals. . Worth 2 points. needs an explanation, not just an answer

Says what the crossing order does and does not tell a customer choosing a shop. . Worth 1 point. needs an explanation, not just an answer
5. A wrong value that came from somewhere . 11 points. Question 5 of 10.

The equation

$5(x - 2) = 3(x + 4) - 2$

has exactly one solution. An attempt at it, since lost, ended with $x = 12$ , and every piece of arithmetic in it was correct.
1. Part A.
  Solve the equation, and check your value in it as it is written above.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Substitute $x = 12$ into the equation as it is written above, evaluating each side on its own. Report the two values, and say what the substitution settles and what it leaves open.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Two accounts of where $12$ came from are on the table. The first is that one constant in the equation was copied with the wrong sign and everything after that was legal. The second is that a move was made which no property of equality licenses. Show that the first account can produce exactly $12$ , by naming the constant and writing the equation it produces. Then say why the two accounts are different kinds of failure.
  
  Carry your own answer forward Argue from whichever solution you found in part A, even if it was not the expected one, so that the comparison is between your own value and $12$ . The credit here is for locating a constant that accounts for the gap and for the distinction between the two kinds of failure.
  
  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

$x = 10$ , and both sides come to $40$ there.
- $x = 10$ only; $x = 6$ comes from expanding one of the two brackets onto its first term alone
Part B

The left comes to $50$ and the right to $46$ , so $x = 12$ does not satisfy the equation. That settles the verdict on the value, but says nothing about which step produced it.

Part C

Copying the right-hand $-2$ as $+2$ gives $5(x - 2) = 3(x + 4) + 2$ , whose solution is exactly $12$ . The accounts differ in what the value solves: a mis-copied constant leaves a sound chain that solves a different equation, while an unlicensed move breaks the chain, so its value may solve nothing at all.
Worked solution

Part A

Expand both brackets, then collect:

$5x - 10 = 3x + 12 - 2 = 3x + 10 \quad\Longrightarrow\quad 2x = 20 \quad\Longrightarrow\quad x = 10.$

Checking in the original: the left is $5(10 - 2) = 40$ and the right is $3(14) - 2 = 40$ .

Part B

Evaluate the two sides independently, without simplifying the equation first:

$5(12 - 2) = 50, \qquad 3(12 + 4) - 2 = 48 - 2 = 46.$

The two sides differ by $4$ , so $12$ is not a solution. A failed substitution is a verdict on a number and nothing more: it rules the value out, and it does not say whether the fault was an unlicensed move, a mis-copied constant, or a slip of the pen.

Part C

The mis-copied constant. Flip the sign of the $-2$ on the right:

$5(x - 2) = 3(x + 4) + 2 \quad\Longrightarrow\quad 5x - 10 = 3x + 14 \quad\Longrightarrow\quad 2x = 24 \quad\Longrightarrow\quad x = 12.$

So that one sign accounts for the whole discrepancy. Notice the size of the effect: the constant moved by $4$ , and the solution moved by $2$ , because the net coefficient of $x$ is $2$ and it is that coefficient which converts a change in the constants into a change in the answer.

Why the accounts differ. Under the first, every move was a property of equality, so the chain is sound and $12$ genuinely solves the equation the work was done on. The error is confined to the first line, and the value can be certified by checking it against that altered equation. Under the second, a move that is not a property of equality breaks the link between the equations, so the final line need not share solutions with anything above it, and the value it names may solve nothing at all. The first failure copies faithfully from a wrong start; the second reasons unfaithfully from a right one.

In one line

The equation is solved by $x = 10$ , where both sides come to $40$ , and $x = 12$ gives $50$ against $46$ , so it does not satisfy the equation as written. Copying the right-hand $-2$ as $+2$ produces $5(x - 2) = 3(x + 4) + 2$ , whose solution is exactly $12$ . That failure and an unlicensed move are different: a mis-copied constant leaves a sound chain that correctly solves the wrong equation, while an unlicensed move breaks the link between the lines, so the value at the end may solve nothing at all.

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

Distributes each factor across both terms inside its brackets and combines the two constants on the right. . Worth 2 points.

Reports the value and confirms it by evaluating both sides of the original separately. . Worth 1 point.

Part B 3 points

Evaluates each side separately at $x = 12$ and reports both values. . Worth 2 points.

Says what a substitution settles about a value, and what it leaves unlocated. . Worth 1 point. needs an explanation, not just an answer

Part C 5 points

Names the constant whose sign is at fault and writes the altered equation. . Worth 2 points.

Verifies that the altered equation really is solved by $12$ , rather than asserting it. . Worth 1 point.

Distinguishes the two accounts by what the value at the end is a solution of, rather than by which is the more careless. . Worth 2 points. needs an explanation, not just an answer
6. Two proportions, and what each one forbids . 14 points. Question 6 of 10.

Each equation below sets one fraction against one fraction, and each carries the variable in a denominator:

$\frac{x + 3}{x - 1} = \frac{5}{2}, \qquad \frac{3}{x + 2} = \frac{6}{2x + 4}.$

Name what an equation forbids before solving it, not afterwards.
1. Part A.
  State the value the first equation forbids, then solve it and say what becomes of the value you find.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Do the same with the second equation: state what it forbids, clear it, and report every number that satisfies it.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Here is a claim: two fractions are equal exactly when their numerators are equal and their denominators are equal. Refute it with a specific pair of fractions from part B and a specific value of $x$ . Then say which half of the claim survives the refutation, and state what a side of an equation must look like before its numerator may be multiplied by the denominator across from it.
  
  Carry your own answer forward Build the counterexample from the pair of fractions you worked with in part B and a value you choose yourself, and check the two values honestly even if part B did not come out as expected. The credit here is for a specific pair that defeats the claim and for the condition you name.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 5 points
The answer
Part A

The equation forbids $x = 1$ . Solving gives $x = \tfrac{11}{3}$ , which is not the forbidden value, so it is kept.
- $x = \tfrac{11}{3}$ may be written $3\tfrac{2}{3}$ ; what is not the same is any answer that keeps $x = 1$ as a second solution
Part B

The equation forbids $x = -2$ , because both denominators vanish there. Clearing leaves $6x + 12 = 6x + 12$ , so every number except $-2$ satisfies it.

Part C

At $x = 0$ the fractions are $\tfrac{3}{2}$ and $\tfrac{6}{4}$ , equal although $3 \neq 6$ and $2 \neq 4$ , so equality does not require matching. The other direction survives: equal numerators over equal nonzero denominators do give equal fractions. Multiplying across needs one fraction each side and nonzero denominators.
Worked solution

Part A

The denominator $x - 1$ is zero at $x = 1$ , so that value is out before any work starts. Each side is a single fraction, so each numerator may be multiplied by the denominator across from it.

$2(x + 3) = 5(x - 1) \quad\Longrightarrow\quad 2x + 6 = 5x - 5 \quad\Longrightarrow\quad 11 = 3x \quad\Longrightarrow\quad x = \frac{11}{3}$

Since $\tfrac{11}{3} \neq 1$ , the candidate survives. Checking: $\dfrac{\tfrac{11}{3} + 3}{\tfrac{11}{3} - 1} = \dfrac{20/3}{8/3} = \dfrac{20}{8} = \dfrac{5}{2}$ .

Part B

Both denominators are zero at $x = -2$ , since $2x + 4 = 2(x + 2)$ , so that one value is excluded. Multiplying each numerator by the denominator across from it:

$3(2x + 4) = 6(x + 2) \quad\Longrightarrow\quad 6x + 12 = 6x + 12.$

The variable cancels and the statement left behind is true, so every number satisfies the cleared equation. The original refuses one of them, so the answer is every number except $-2$ . Two tests: at $x = 0$ both sides are $\tfrac{3}{2}$ , and at $x = 1$ both sides are $1$ .

Part C

The refutation. Take the fractions from part B at $x = 0$ :

$\frac{3}{0 + 2} = \frac{3}{2}, \qquad \frac{6}{2(0) + 4} = \frac{6}{4} = \frac{3}{2}.$

They are the same number, and yet $3 \neq 6$ and $2 \neq 4$ . So matching term by term is not required for equality, and the claim's "only when" half is false. Part B says the same thing for every allowed $x$ at once, which is why that equation had so many solutions.

What survives. The other half stands: if two fractions have the same numerator and the same nonzero denominator, they are literally the same number. Equality of fractions is implied by matching, and does not imply it.

What multiplying across requires. Each side must already BE one fraction, with nothing added to it and nothing left outside it, and each denominator must be nonzero. When a side is a sum, it has to be combined into a single fraction first, or the terms outside the fraction never get multiplied and the equation quietly changes.

In one line

The first equation forbids $x = 1$ and is solved by $x = \tfrac{11}{3}$ , which is allowed and checks out at $\tfrac{5}{2}$ . The second forbids $x = -2$ and clears to $6x + 12 = 6x + 12$ , so every number except $-2$ satisfies it. That equation also refutes the claim: at $x = 0$ the fractions $\tfrac{3}{2}$ and $\tfrac{6}{4}$ are equal while neither their numerators nor their denominators match. Only the other direction holds, since equal numerators over equal nonzero denominators really do give equal fractions, and multiplying numerators across requires a single fraction on each side with nonzero denominators.

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

Names the excluded value from the denominator before solving. . Worth 1 point.

Multiplies each numerator by the denominator across from it and expands both sides in full. . Worth 2 points.

Compares the candidate against the excluded value and says explicitly that it is kept. . Worth 1 point.

Part B 5 points

Compares the two denominators and identifies every value at which either of them vanishes. . Worth 1 point.

Clears correctly and reads the leftover statement to decide how many numbers satisfy the cleared equation. . Worth 2 points.

Reconciles the cleared equation's verdict with what the original equation forbids, rather than reporting the first unchanged. . Worth 2 points.

Part C 5 points

Gives one specific value of $x$ and evaluates both fractions at it, showing they agree. . Worth 2 points.

Says which direction of the claim the example defeats and which direction is still true. . Worth 2 points. needs an explanation, not just an answer

States what each side of an equation must look like before numerators may be multiplied across. . Worth 1 point. needs an explanation, not just an answer
7. Two quantities, and the one they combine into . 13 points. Question 7 of 10.

Two quantities $a$ and $b$ , both positive, are combined into a third by

$T = \frac{ab}{a + b},$

so the denominator $a + b$ is never zero.
1. Part A.
  Solve the formula for $b$ , and state the condition your final step requires.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Find $b$ when $a = 12$ and $T = 4$ , and confirm the result by running your two numbers forward through the formula as it was given.
  
  Carry your own answer forward Substitute into whichever rearrangement you produced in part A, even if it was not the expected one, then run the numbers forward through the original formula and report honestly whether they agree. The credit here is for substituting and testing, not for reaching a particular number.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Your rearrangement is unusable at $a = T$ . Show, working from the original formula and not from the rearrangement, that $a = T$ can happen only when $a$ is zero, so that the condition rules out nothing a positive $a$ could ever do. Then say what this shows about the difference between a condition the algebra needs written down and a condition the quantities can actually reach.
  
  Carry your own answer forward Argue about whichever condition your part A rearrangement carries, even if it was not the expected one, and say honestly whether the quantities in this situation could ever violate it. The credit here is for the argument from the original formula and for the distinction you draw, not for matching a particular condition.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

$b = \dfrac{Ta}{a - T}$ , valid when $a \neq T$ .
- $b = \dfrac{aT}{a - T}$ is the same expression; $b = \dfrac{Ta}{T - a}$ is not, because the two terms were collected in the wrong order
Part B

$b = 6$ .

Part C

Setting $T = a$ in $T(a + b) = ab$ gives $a^2 = 0$ , so $a = 0$ . With $a$ positive that never happens, so the division is always safe here. The condition is still owed, because the rearranged formula is read on its own: it is a statement about the expression, not a warning about the situation.
Worked solution

Part A

Clear the denominator, expand, and gather the two terms holding $b$ on one side so that $b$ can be factored out. No single division reaches $b$ while it stands in two terms.

$T(a + b) = ab \quad\Longrightarrow\quad Ta + Tb = ab \quad\Longrightarrow\quad Ta = ab - Tb = b(a - T)$

Dividing by the factored expression gives $b = \dfrac{Ta}{a - T}$ , and that division requires $a - T \neq 0$ , which is to say $a \neq T$ .

Part B

Substitute into the rearranged form:

$b = \frac{4 \times 12}{12 - 4} = \frac{48}{8} = 6.$

Running $a = 12$ and $b = 6$ forward through the original gives $\dfrac{12 \times 6}{12 + 6} = \dfrac{72}{18} = 4$ , which is the value of $T$ that was given, so the rearrangement has not lost a term or a sign.

Part C

The argument. Work from the cleared original, $T(a + b) = ab$ , and put $T = a$ :

$a(a + b) = ab \quad\Longrightarrow\quad a^2 + ab = ab \quad\Longrightarrow\quad a^2 = 0 \quad\Longrightarrow\quad a = 0.$

So $T$ and $a$ can only be equal when $a$ is zero, and they are equal then. With $a$ positive, $T$ is always strictly less than $a$ , which the numbers of part B show in miniature: $T = 4$ against $a = 12$ .

What the difference is. The condition $a \neq T$ is a statement about an expression, not a forecast about which values will turn up. Once $b = \dfrac{Ta}{a - T}$ is written down it can be handed to anyone, applied to any numbers, and read without the paragraph that produced it, and nothing in the symbols records that $T < a$ here. A condition earns its place by protecting a division that could otherwise be by zero; it does not have to be reachable to be required.

In one line

$b = \dfrac{Ta}{a - T}$ , valid when $a \neq T$ , which gives $b = 6$ for $a = 12$ and $T = 4$ , confirmed by $\dfrac{12 \times 6}{18} = 4$ . The excluded case is unreachable here: putting $T = a$ into $T(a + b) = ab$ forces $a^2 = 0$ , so $a$ would have to be zero. The condition is still owed, because the rearranged formula is read on its own, away from the situation that guarantees $T < a$ .

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

Clears the fraction and expands it before attempting to isolate the target. . Worth 1 point.

Brings the target into a single term before dividing by anything. . Worth 2 points.

States the condition the final division requires and ties it to the expression divided by. . Worth 1 point.

Part B 3 points

Substitutes both given values into the rearranged formula correctly. . Worth 1 point.

Runs the pair forward through the original formula and states whether the value of $T$ comes back. . Worth 2 points.

Part C 6 points

Works from the original relation, substituting the excluded case into it, rather than reasoning from the rearranged form. . Worth 3 points. needs an explanation, not just an answer

Concludes that a positive $a$ never meets the excluded case, so the division is safe in this situation. . Worth 1 point.

Explains that the condition is still required, because a rearranged formula is read apart from the situation that produced it. . Worth 2 points. needs an explanation, not just an answer
8. A fence, and what one more metre of width costs . 14 points. Question 8 of 10.

A rectangular garden is to be enclosed by exactly $84$ metres of fencing, with the fence running along all four sides. The length is to be $6$ metres less than twice the width.
1. Part A.
  Name one unknown, with its unit, write the other dimension in terms of it, and write one equation saying the fencing comes to $84$ metres.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
2. Part B.
  Solve your equation and report both dimensions with their units. Check them against both sentences of the problem, not just against your equation.
  
  Carry your own answer forward Solve whichever equation you wrote in part A, even if it was not the expected one, and check your two dimensions against the sentences of the problem rather than against the equation you solved. The credit here is for solving your own model and reporting both lengths with units.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  The budget is raised to $90$ metres of fencing, with the length still $6$ metres less than twice the width. Give the new width. Then interpret the coefficient that multiplies your unknown once your part A equation is collected: say, in metres of fencing, what it measures, and use it to state how much extra fence buys one extra metre of width.
  
  Carry your own answer forward Interpret the coefficient that appears in your own part A equation, even if it was not the expected one, and read the new width from your own model. The credit here is for treating the coefficient as a rate and saying what it exchanges, not for a particular number.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
The answer

Part A

Let $w$ be the width in metres. Then the length is $2w - 6$ metres, and the fence runs the perimeter, so $2w + 2(2w - 6) = 84$ .

Part B

The width is $16$ metres and the length is $26$ metres.

Part C

The new width is $17$ metres. The coefficient $6$ measures metres of fencing per metre of width: widening the garden by one metre lengthens two widths by one metre each and the length by two, so the fence grows by $6$ metres. Every extra $6$ metres of fencing therefore buys exactly one extra metre of width.

Worked solution

Part A

One letter has to carry both dimensions, so name the one the other is described in terms of. With $w$ the width in metres, "six metres less than twice the width" is $2w - 6$ , with the amount taken away written second.

$2w + 2(2w - 6) = 84$

The fence runs along all four sides, so the equation sets the perimeter, two widths and two lengths, equal to the fencing available.

Part B

Expand, collect, and divide:

$2w + 4w - 12 = 84 \quad\Longrightarrow\quad 6w = 96 \quad\Longrightarrow\quad w = 16.$

The length is $2(16) - 6 = 26$ metres. Both sentences must be checked: the perimeter is $2(16) + 2(26) = 84$ metres, which is the fencing available, and $26$ is indeed $6$ less than twice $16$ . A width of $16$ metres and a length of $26$ metres are both positive, so the answer makes sense as a garden.

Part C

Only the number on the right changes, so the same collected equation serves:

$6w - 12 = 90 \quad\Longrightarrow\quad 6w = 102 \quad\Longrightarrow\quad w = 17.$

The width has moved by exactly $1$ metre for $6$ extra metres of fence, and that is the coefficient reading itself back. The $6$ arose as $2 + 4$ : one extra metre of width adds a metre to each of the two widths, and adds two metres to the length, which appears twice, for four more.

So the coefficient is a rate, metres of fencing per metre of width, and it is fixed by the shape of the model rather than by the budget. It also says what the fixed part cannot do: the $-12$ shifts the answer but never changes the exchange rate, so the same $6$ metres buys a metre of width at any budget.

In one line

With $w$ the width in metres, $2w + 2(2w - 6) = 84$ gives $w = 16$ , so the garden is $16$ metres by $26$ metres, a perimeter of $84$ metres with the length $6$ less than twice the width. Raising the budget to $90$ metres gives $6w = 102$ and a width of $17$ metres. The coefficient $6$ is a rate, metres of fencing per metre of width, made of one metre for each of the two widths and two for each of the two lengths, so every extra $6$ metres of fence buys exactly one more metre of width.

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

Names one unknown and states its unit, so the other expressions can be read as lengths. . Worth 1 point.

Writes the second dimension in terms of the unknown, with the subtraction in the order the words give. . Worth 2 points.

Sets a perimeter of two widths and two lengths equal to $84$ , rather than a sum of one of each. . Worth 1 point.

Part B 5 points

Expands and solves correctly, distributing the $2$ across both terms of the length. . Worth 2 points.

Reports both dimensions in metres, not only the letter that was solved for. . Worth 2 points.

Checks the perimeter and the length-to-width sentence separately. . Worth 1 point.

Part C 5 points

Recomputes the width from the same model with the new total, rather than starting the translation again. . Worth 1 point.

Reads the coefficient as metres of fencing per metre of width, with units on both quantities. . Worth 2 points. needs an explanation, not just an answer

Accounts for the size of the coefficient from the shape of the perimeter, not merely from the two answers. . Worth 2 points. needs an explanation, not just an answer
9. One constant left open, and a family of equations . 15 points. Question 9 of 10.

A constant $c$ is left unnamed in

$\frac{x + c}{3} - \frac{x - 1}{2} = \frac{5 - x}{6},$

so this one line stands for a whole family of equations, one for each choice of $c$ .
1. Part A.
  Multiply both sides by the smallest number all three denominators divide, expand, and gather the terms as far as they will go. Report the statement you are left with, in terms of $c$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Read the family's outcomes off that statement. For each of the three possible outcomes, exactly one solution, no solution, and every number a solution, say which values of $c$ produce it.
  
  Carry your own answer forward Read the outcomes off whichever leftover statement you produced in part A, even if it was not the expected one, and say honestly which of the three cases it makes available. The credit here is for matching each case to what the statement does, not for a particular value of $c$ .
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
3. Part C.
  Explain what it is about where $c$ sits in the equation that puts exactly one solution beyond its reach, however it is chosen. Then describe the smallest change you could make to the equation so that exactly one solution becomes the usual outcome, and say what the new family would do.
  
  Carry your own answer forward Argue from the leftover statement and outcomes you reached in parts A and B, even if they were not the expected ones. The credit here is for locating the reason in where the constant sits rather than in the arithmetic, and for a change that genuinely moves the family.
  
  Explain why it works A sentence or two. Reasons, not steps. 6 points
The answer
Part A

$2c + 3 = 5$ , equivalently $2c = 2$ .
- $2c + 3 = 5$ and $2c = 2$ and $c = 1$ all record the same leftover statement; $-x + 2c + 3 = 5 - x$ is the line before the $x$ terms are cancelled
Part B

No value of $c$ gives exactly one solution. At $c = 1$ the leftover statement is true, so every number is a solution; at every other value of $c$ it is false, so there is no solution.

Part C

The constant $c$ sits in a numerator with no $x$ beside it, so it can only move a constant term, while the coefficients of $x$ are fixed at $-1$ on each side and always cancel. Changing a coefficient that multiplies $x$ , say making the right numerator $5 - 2x$ , unbalances them, and then every $c$ gives exactly one solution.
Worked solution

Part A

The denominators $3$ , $2$ and $6$ all divide $6$ , so multiply every term by $6$ and expand, remembering that the $-3$ reaches both terms of $x - 1$ .

$2(x + c) - 3(x - 1) = 5 - x \quad\Longrightarrow\quad 2x + 2c - 3x + 3 = 5 - x \quad\Longrightarrow\quad -x + 2c + 3 = 5 - x$

Adding $x$ to both sides removes the variable from both sides at once and leaves $2c + 3 = 5$ . Notice that this happened for every $c$ : the constant never sat beside an $x$ .

Part B

The variable has already gone, so the leftover statement is the whole verdict:

$2c + 3 = 5 \quad\Longrightarrow\quad c = 1.$

At $c = 1$ the reduced line reads $5 = 5$ , which is true whatever $x$ is, so every number satisfies the equation. At any other $c$ it reads as one number equal to a different number, which no choice of $x$ can repair, so there is no solution at all. Exactly one solution is not on the menu, because a single solution requires an $x$ to survive the gathering, and none ever does.

A check at $c = 1$ and $x = 0$ : the left is $\tfrac{1}{3} + \tfrac{1}{2} = \tfrac{5}{6}$ and the right is $\tfrac{5}{6}$ . At $c = 1$ and $x = 4$ : the left is $\tfrac{5}{3} - \tfrac{3}{2} = \tfrac{1}{6}$ and the right is $\tfrac{1}{6}$ .

Part C

Why one solution is unreachable. Clearing gathers the equation into the shape $ax = b$ , and which case you land in is decided by $a$ alone: only $a \neq 0$ can name a single value. Here $c$ appears in the numerator $x + c$ with no $x$ attached to it, so it contributes to $b$ and never to $a$ . The coefficients of $x$ are settled by the other numbers, and they come out as $-1$ on the left and $-1$ on the right:

$-x + 2c + 3 = 5 - x.$

They match for every $c$ , so $a = 0$ always, and the family can only ever be an identity or a contradiction.

The smallest change. Touch a coefficient that multiplies $x$ . Writing the right-hand numerator as $5 - 2x$ leaves the left as it was and gives

$-x + 2c + 3 = 5 - 2x \quad\Longrightarrow\quad x = 2 - 2c.$

Now $a = 1$ , not $0$ , so every choice of $c$ gives exactly one solution, and the value slides with $c$ instead of the verdict flipping. The new family has no contradictions and no identities in it at all, which is the exact mirror of the old one.

In one line

Clearing with the multiplier $6$ reduces the family to $2c + 3 = 5$ , with no $x$ anywhere. So $c = 1$ makes the statement true and every number a solution, every other $c$ makes it false and no number a solution, and exactly one solution never occurs. The reason is positional: $c$ sits with no $x$ beside it, so it can only move the constant, while the coefficients of $x$ are fixed at $-1$ on each side and always cancel. Changing the right-hand numerator to $5 - 2x$ breaks that match and gives $x = 2 - 2c$ , exactly one solution for every $c$ .

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 all three terms by a common multiple of the denominators, the right-hand side included. . Worth 2 points.

Distributes each factor across both terms of the numerator it multiplies, so every sign arrives correctly. . Worth 1 point.

Cancels the matching $x$ terms and reports a statement holding $c$ and constants only. . Worth 1 point.

Part B 5 points

Identifies the value of $c$ , if any, making the leftover statement true, and names the outcome that follows. . Worth 2 points.

Says what happens at the remaining values of $c$ , reasoning from whether the leftover statement is true or false. . Worth 1 point.

Accounts for all three listed outcomes, saying explicitly if one of them never arises. . Worth 2 points. needs an explanation, not just an answer

Part C 6 points

Locates the reason in where $c$ sits, and in which part of the reduced equation it can influence. . Worth 3 points. needs an explanation, not just an answer

Proposes a change that reaches the part of the equation the constant could not, rather than another constant. . Worth 1 point.

Says what the changed family does, naming the outcome that becomes usual and what becomes of the old cases. . Worth 2 points. needs an explanation, not just an answer
10. One formula, two targets . 15 points. Question 10 of 10.

An amount $P$ left to earn simple interest at rate $r$ for time $t$ grows to

$A = P + Prt.$

The same formula is rearranged twice below, once for each of two different letters, and the two rearrangements do not cost the same amount of work.
1. Part A.
  Solve the formula for $P$ , and state the condition your final step requires.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Now solve the same formula for $r$ , and state the condition that step requires. Then confirm that your two rearrangements agree, by taking $A = 1200$ , $P = 1000$ and $t = 4$ , finding $r$ from this form, and feeding it back through the form you wrote in part A.
  
  Carry your own answer forward Feed your value of $r$ back through whichever rearrangement you produced in part A, even if it was not the expected one, and report honestly whether the deposit comes back. The credit here is for the second rearrangement, its condition, and the test, not for matching a particular expression.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  One of your two rearrangements needed the target collected and factored out, and the other did not. Compare them, saying what it is about the formula that decided which was which. Then give a test a reader can run on any formula and any target, before doing any algebra, to predict which of the two situations they are about to be in.
  
  Carry your own answer forward Compare the two rearrangements you actually produced in parts A and B, even if either was not the expected one, and let your test be one that would have predicted the work you did. The credit here is for the account of what decides the difference and for a test that can be applied in advance.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 6 points
The answer
Part A

$P = \dfrac{A}{1 + rt}$ , valid when $1 + rt \neq 0$ .
- $P = \dfrac{A}{1 + rt}$ only; $P = \dfrac{A}{rt}$ has factored $P + Prt$ as $P(rt)$ and lost the $1$ that the lone $P$ carries
Part B

$r = \dfrac{A - P}{Pt}$ , valid when $P \neq 0$ and $t \neq 0$ . The numbers give $r = 0.05$ , and $\dfrac{1200}{1 + 0.05 \times 4} = 1000$ , which is the $P$ that was given.
- $r = \dfrac{A - P}{Pt}$ may also be written $\dfrac{A}{Pt} - \dfrac{1}{t}$ ; what is not the same is $\dfrac{A - P}{P}t$ , in which $t$ has moved out of the denominator
Part C

Aimed at $P$ the target sits in both terms of $P + Prt$ , so it must be factored out before a division can reach it; aimed at $r$ it sits in one, so divisions suffice. The test, for a target appearing only to the first power: expand, clear fractions, count the terms holding it. One means divisions alone, more means factor.
Worked solution

Part A

The target stands in both terms on the right, so no division reaches it while it is in two places. Factor it out first, remembering that the lone $P$ is $P \times 1$ , so a $1$ stays inside the bracket.

$A = P(1 + rt) \quad\Longrightarrow\quad P = \frac{A}{1 + rt}$

The division requires $1 + rt \neq 0$ . Expanding the factored form back, $P(1 + rt) = P + Prt$ , confirms that nothing was dropped.

Part B

This time the target sits in one term only, so successive divisions reach it and nothing needs factoring.

$A - P = Prt \quad\Longrightarrow\quad r = \frac{A - P}{Pt}$

The division is by $Pt$ , so it requires $P \neq 0$ and $t \neq 0$ . With the given numbers, $r = \dfrac{200}{1000 \times 4} = 0.05$ , and running that through part A's form gives $\dfrac{1200}{1 + 0.2} = \dfrac{1200}{1.2} = 1000$ , the deposit it started from.
Part C
The comparison. The two rearrangements differ in one respect that decides everything:

$A = \underbrace{P}_{\text{has } P} + \underbrace{Prt}_{\text{has } P}, \qquad A - P = \underbrace{Prt}_{\text{has } r}.$

A division acts on a whole side at once, so it can only isolate a target that the side mentions once. Aimed at $P$ , both terms mention it, and dividing by $1 + rt$ before factoring is perfectly legal and gets nowhere, since the right side is still a sum and $P$ is no more isolated than before. Factoring is what turns two mentions into one, and only then does a division finish it. Aimed at $r$ , there is one mention from the start, and the work is the ordinary undoing of a multiplication.

The test, for a formula the target enters only to the first power. Expand every bracket and clear every fraction first, because a target can be hidden inside either, and then count the terms containing the target.

One term: divisions and subtractions alone will isolate it.

Two or more: gather them onto one side, factor the target out, and divide by the bracket that is left.

The count, not the look of the formula, is what decides, and it is worth making before the first move rather than after a division has failed to help.
In one line

$P = \dfrac{A}{1 + rt}$ , valid when $1 + rt \neq 0$ , and $r = \dfrac{A - P}{Pt}$ , valid when $P \neq 0$ and $t \neq 0$ . The two agree: $A = 1200$ , $P = 1000$ and $t = 4$ give $r = 0.05$ , and $\dfrac{1200}{1.2}$ returns $1000$ . The difference in work is decided by a count. Aimed at $P$ the target sits in two terms, so it must be factored out before any division can reach it; aimed at $r$ it sits in one, so divisions suffice. Where the target appears only to the first power, expand the brackets, clear the fractions and count the terms holding it, and you know which case you are in before starting.
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

Brings the target into a single term, keeping the coefficient that the lone term contributes. . Worth 2 points.

Divides by the whole factored expression rather than by part of it. . Worth 1 point.

States the condition the division requires. . Worth 1 point.

Part B 5 points

Isolates the single term holding $r$ and divides by the whole of $Pt$ . . Worth 2 points.

States every part of the condition the division requires, not only one of them. . Worth 1 point.

Runs the computed rate back through the other rearrangement and says whether the original value is recovered. . Worth 2 points.

Part C 6 points

Says that the target appears in two terms in one case and one term in the other, and identifies which is which. . Worth 2 points.

Explains why the number of terms holding the target decides whether dividing can finish the job. . Worth 2 points. needs an explanation, not just an answer

Gives a test that can be run before any algebra, and says what must be done first for its count to be trustworthy, and says what kind of formula it applies to. . Worth 2 points. 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. Two equations, and a different way to begin 10 points

Part A 3

Collects the variable terms on one side and the constants on the other, changing the sign of each term that crosses. 2
Reports the value and confirms it by evaluating each side of the original equation separately. 1

Part B 3

Distributes each factor across both terms inside its brackets, with the correct sign on the product of the two negatives. 2
Reports the value and confirms it in the equation as it was given. 1

Part C 4

Carries the divided equation through to a value rather than stopping at the first line. 1
Grounds the agreement in something stronger than the two answers happening to match on this occasion. 2
Names what the division had to cover for the two routes to agree. 1

2. Two equations, one constant apart 12 points

Part A 4

Multiplies every term, both fractions and the whole right-hand side, by a common multiplier of the denominators. 2
Reads the verdict off the statement left behind rather than off the tests, and confirms it at two different values, not one. 2

Part B 4

Clears the second equation with the same multiplier and reaches a statement with no variable in it. 2
States the number of solutions on the strength of the leftover statement, and exhibits two values that bear the verdict out. 2

Part C 4

Says what each cleared line is comparing, and why nothing else is left to compare. 2
Explains what the leftover statement decides once the variable has gone, and why that leaves no room for a single value. 2

3. One formula, three letters it can be aimed at 12 points

Part A 4

Undoes the operations around $b_1$ in reverse order, clearing the fraction before dividing by $h$ . 2
Subtracts $b_2$ from the whole of the other side rather than from part of it. 1
States the restriction the final step requires and attaches it to the step that requires it. 1

Part B 4

Substitutes the three given values into the rearranged formula correctly. 2
Reports the answer as a length in centimetres and confirms it by computing the area forward from the original formula. 2

Part C 4

Explains what the two forms have in common by appealing to the steps that connect them, not to how they look. 2
Names more than one difference between the two forms. 1
Chooses one difference as the one that matters in use, and gives a reason that refers to what the user is computing. 1

4. Two price lists that cross once 13 points

Part A 4

States what the letter counts, so the expressions can be read. 1
Builds each charge as a fixed amount plus a per-poster amount times the count, with the rival's fixed amount zero. 2
Sets the two expressions equal to each other, rather than equal to a number the problem never gave. 1

Part B 4

Solves the equation correctly, keeping the decimal coefficient intact. 2
Reports the order as a number of posters and the cost in dollars, both asked for and both labelled. 1
Checks the cost by running the number of posters through each shop's price list separately. 1

Part C 5

Evaluates both charges at both order sizes and names which shop is cheaper in each case. 2
Accounts for the pattern in the two comparisons using the two shops' fees and per-poster prices, not just the two totals. 2
Says what the crossing order does and does not tell a customer choosing a shop. 1

5. A wrong value that came from somewhere 11 points

Part A 3

Distributes each factor across both terms inside its brackets and combines the two constants on the right. 2
Reports the value and confirms it by evaluating both sides of the original separately. 1

Part B 3

Evaluates each side separately at $x = 12$ and reports both values. 2
Says what a substitution settles about a value, and what it leaves unlocated. 1

Part C 5

Names the constant whose sign is at fault and writes the altered equation. 2
Verifies that the altered equation really is solved by $12$ , rather than asserting it. 1
Distinguishes the two accounts by what the value at the end is a solution of, rather than by which is the more careless. 2

6. Two proportions, and what each one forbids 14 points

Part A 4

Names the excluded value from the denominator before solving. 1
Multiplies each numerator by the denominator across from it and expands both sides in full. 2
Compares the candidate against the excluded value and says explicitly that it is kept. 1

Part B 5

Compares the two denominators and identifies every value at which either of them vanishes. 1
Clears correctly and reads the leftover statement to decide how many numbers satisfy the cleared equation. 2
Reconciles the cleared equation's verdict with what the original equation forbids, rather than reporting the first unchanged. 2

Part C 5

Gives one specific value of $x$ and evaluates both fractions at it, showing they agree. 2
Says which direction of the claim the example defeats and which direction is still true. 2
States what each side of an equation must look like before numerators may be multiplied across. 1

7. Two quantities, and the one they combine into 13 points

Part A 4

Clears the fraction and expands it before attempting to isolate the target. 1
Brings the target into a single term before dividing by anything. 2
States the condition the final division requires and ties it to the expression divided by. 1

Part B 3

Substitutes both given values into the rearranged formula correctly. 1
Runs the pair forward through the original formula and states whether the value of $T$ comes back. 2

Part C 6

Works from the original relation, substituting the excluded case into it, rather than reasoning from the rearranged form. 3
Concludes that a positive $a$ never meets the excluded case, so the division is safe in this situation. 1
Explains that the condition is still required, because a rearranged formula is read apart from the situation that produced it. 2

8. A fence, and what one more metre of width costs 14 points

Part A 4

Names one unknown and states its unit, so the other expressions can be read as lengths. 1
Writes the second dimension in terms of the unknown, with the subtraction in the order the words give. 2
Sets a perimeter of two widths and two lengths equal to $84$ , rather than a sum of one of each. 1

Part B 5

Expands and solves correctly, distributing the $2$ across both terms of the length. 2
Reports both dimensions in metres, not only the letter that was solved for. 2
Checks the perimeter and the length-to-width sentence separately. 1

Part C 5

Recomputes the width from the same model with the new total, rather than starting the translation again. 1
Reads the coefficient as metres of fencing per metre of width, with units on both quantities. 2
Accounts for the size of the coefficient from the shape of the perimeter, not merely from the two answers. 2

9. One constant left open, and a family of equations 15 points

Part A 4

Multiplies all three terms by a common multiple of the denominators, the right-hand side included. 2
Distributes each factor across both terms of the numerator it multiplies, so every sign arrives correctly. 1
Cancels the matching $x$ terms and reports a statement holding $c$ and constants only. 1

Part B 5

Identifies the value of $c$ , if any, making the leftover statement true, and names the outcome that follows. 2
Says what happens at the remaining values of $c$ , reasoning from whether the leftover statement is true or false. 1
Accounts for all three listed outcomes, saying explicitly if one of them never arises. 2

Part C 6

Locates the reason in where $c$ sits, and in which part of the reduced equation it can influence. 3
Proposes a change that reaches the part of the equation the constant could not, rather than another constant. 1
Says what the changed family does, naming the outcome that becomes usual and what becomes of the old cases. 2

10. One formula, two targets 15 points

Part A 4

Brings the target into a single term, keeping the coefficient that the lone term contributes. 2
Divides by the whole factored expression rather than by part of it. 1
States the condition the division requires. 1

Part B 5

Isolates the single term holding $r$ and divides by the whole of $Pt$ . 2
States every part of the condition the division requires, not only one of them. 1
Runs the computed rate back through the other rearrangement and says whether the original value is recovered. 2

Part C 6

Says that the target appears in two terms in one case and one term in the other, and identifies which is which. 2
Explains why the number of terms holding the target decides whether dividing can finish the job. 2
Gives a test that can be run before any algebra, and says what must be done first for its count to be trustworthy, and says what kind of formula it applies to. 2

Multiple choice

Free response