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Equations and Inequalities: 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, 105 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. Reading the answer off two reduced equations . 9 points. Question 1 of 10.

This question classifies two linear equations by reducing each to the form $ax=b$ , then asks what the reduced form itself guarantees about every linear equation.
1. Part A.
  Solve $9x-14=3x+22$ , reducing it fully to the form $ax=b$ , and state its solution set.
  
  Solve and show your work Write each step out, and end with the value and its units. 2 points
2. Part B.
  Solve $6(x+5)-4x=2x+30$ , reducing it fully to the form $ax=b$ , and state its solution set.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Using the general form $ax=b$ , prove that a linear equation can never have a solution set of exactly two numbers, and say what has to be true of the reduced coefficient and constant together for an equation to land on every real number the way part B did, rather than on the single number found in part A.
  
  Carry your own answer forward Illustrate the general argument using your own results from parts A and B, whatever they were.
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 4 points
The answer

Part A

The equation reduces to $6x=36$ ; $x=6$ , so the solution set is $\{6\}$ .

Part B

The equation reduces to $0x=0$ ; the solution set is $\mathbb{R}$ .

Part C

Every linear equation reduces to $ax=b$ : one solution when $a\neq0$ , all of $\mathbb{R}$ when $a=0$ and $b=0$ , none when $a=0$ and $b\neq0$ . These three exhaust every case, so two solutions never occurs. Part B reached $\mathbb{R}$ because its coefficient and constant vanished together.

Worked solution

Part A

Collect the $x$ terms on one side and the constants on the other.

$9x-14=3x+22 \quad\Longrightarrow\quad 6x=36, \qquad x=6.$

The solution set is $\{6\}$ .

Part B

Distribute and combine like terms on the left.

$6(x+5)-4x = 6x+30-4x = 2x+30.$

The equation now reads $2x+30=2x+30$ . Subtracting $2x$ leaves $30=30$ , true for every $x$ , so the reduced form is $0x=0$ and the solution set is $\mathbb{R}$ .

Part C

Every linear equation reachable by distributing and combining like terms reduces to $ax=b$ for fixed numbers $a$ and $b$ .

If $a\neq0$ , dividing gives the single value $x=\frac{b}{a}$ : exactly one solution.

If $a=0$ , the equation reads $0\cdot x=b$ , a bare numeric statement with no $x$ left in it. Either $b=0$ , true for every real number, giving $\mathbb{R}$ , or $b\neq0$ , false for every real number, giving $\varnothing$ .

$a\neq0 \Rightarrow \{b/a\}, \qquad a=0,\,b=0 \Rightarrow \mathbb{R}, \qquad a=0,\,b\neq0 \Rightarrow \varnothing$

Those three cases are exhaustive and mutually exclusive, and none of them names a set of exactly two numbers. Part B landed on $\mathbb{R}$ because its reduction sent BOTH the coefficient and the constant to zero at once, $0x=0$ ; part A's coefficient stayed nonzero, so it landed on the single-solution branch instead.

In one line

$9x-14=3x+22$ reduces to $6x=36$ , giving the single solution $x=6$ . $6(x+5)-4x=2x+30$ reduces to $0x=0$ , an identity with solution set $\mathbb{R}$ . Every linear equation reduces to $ax=b$ : one solution when $a\neq0$ , $\mathbb{R}$ when $a=b=0$ , $\varnothing$ when $a=0,b\neq0$ ; these three exhaust every case, so no linear equation has a solution set of exactly two numbers, and reaching $\mathbb{R}$ specifically requires the coefficient and the constant to vanish together.

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 2 points

Collects like terms correctly and reduces the equation to the form ax = b. . Worth 1 point.

Solves for x and reports the solution set using set notation. . Worth 1 point.

Part B 3 points

Distributes and combines like terms correctly before comparing the two sides. . Worth 1 point.

Reduces the equation to a bare numeric statement and reports the solution set that statement implies. . Worth 2 points.

Part C 4 points

Argues that a nonzero coefficient and a zero coefficient exhaust every possibility for a linear equation, rather than checking only the two given examples. . Worth 2 points. needs an explanation, not just an answer

States that landing on every real number requires both the reduced coefficient and the reduced constant to vanish together, referencing part B's own reduction. . Worth 2 points.
2. A factor that is a sum, not a single letter . 10 points. Question 2 of 10.

A savings account without compounding grows by simple interest, $A = P + Prt$ , where $P$ is the principal, $r$ the annual rate, and $t$ the time in years.
1. Part A.
  Solve $A=P+Prt$ for the principal $P$ , and state the restriction your final step requires.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Use your formula to find the principal $P$ that grows to $A=1200$ dollars after $t=4$ years at a rate of $r=0.05$ .
  
  Carry your own answer forward Substitute into whichever formula you produced in part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  The restriction from part A is $rt \neq -1$ . Since $r$ and $t$ here stand for a positive interest rate and a positive number of years, explain whether that excluded value can ever actually arise in this situation, and say what the restriction is protecting against in the formula, as opposed to protecting against $r$ or $t$ individually equaling some particular number.
  
  Carry your own answer forward Refer to the restriction you stated in part A, whatever form it took.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
The answer

Part A

$P=\dfrac{A}{1+rt}$ , valid whenever $rt \neq -1$ .

Part B

$P=1000$ dollars.

Part C

Since $r>0$ and $t>0$ here, $rt$ is always positive, so $rt=-1$ never arises in this situation. The restriction protects against the whole product $rt$ reaching $-1$ , not against any single value of $r$ or $t$ alone.

Worked solution

Part A

The target $P$ appears in both terms on the right, so factor it out before dividing.

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

The divisor is the whole factor $1+rt$ , so the rearrangement holds whenever $1+rt\neq0$ , that is $rt\neq-1$ .

Part B

Substitute $A=1200$ , $r=0.05$ , $t=4$ :

$P = \frac{1200}{1+(0.05)(4)} = \frac{1200}{1.2} = 1000.$

The principal was $1000$ dollars.

Part C

In this situation $r$ is a positive rate and $t$ a positive number of years, so their product $rt$ is always positive:

$r>0,\ t>0 \ \Longrightarrow\ rt>0 \ \Longrightarrow\ rt \neq -1 \ \text{always}.$

So the excluded case can never actually occur for a real savings account. The restriction is still a true fact about the FORMULA, not about this situation alone: it protects against the single combined quantity $rt$ landing on $-1$ , whatever $r$ and $t$ individually are, not against $r$ or $t$ separately taking any particular value.

In one line

Solving $A=P+Prt$ for $P$ gives $P=\dfrac{A}{1+rt}$ , valid whenever $rt\neq-1$ ; at $A=1200$ , $r=0.05$ , $t=4$ , the principal is $1000$ dollars. Because $r$ and $t$ are both positive in this situation, $rt$ is always positive, so $rt=-1$ never actually arises, though the restriction remains a true fact about the formula, protecting the combined quantity $rt$ , not $r$ or $t$ individually.

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

Factors P out of both terms on the right before dividing. . Worth 2 points.

States the restriction rt not equal to -1 as a single condition on the factor 1+rt. . Worth 1 point.

Part B 3 points

Substitutes the three given values into the rearranged formula and evaluates the denominator before dividing. . Worth 2 points.

Reports the principal with the correct unit, dollars, not as a bare number. . Worth 1 point.

Part C 4 points

Correctly determines that rt=-1 cannot arise when r and t are both positive, and justifies it from their signs. . Worth 2 points.

States that the restriction is on the combined quantity rt, not on r or t individually, tying it back to part A's factor. . Worth 2 points. needs an explanation, not just an answer
3. A fraction on one side, and a sign not yet known . 10 points. Question 3 of 10.

Consider the inequality $\dfrac{9}{x+2} < 3$ .
1. Part A.
  State the value that must be excluded from $\dfrac{9}{x+2}<3$ , then solve the inequality in the case $x+2>0$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Solve $\dfrac{9}{x+2}<3$ in the remaining case, $x+2<0$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Explain what would go wrong if a solver multiplied both sides of $\dfrac{9}{x+2}<3$ by $x+2$ in one single step, without splitting into the two cases above, and use your results from parts A and B to name one specific value of $x$ that single step would mishandle.
  
  Carry your own answer forward Use whichever two case-results you found in parts A and B to pick a value the naive single step would get wrong.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

Excluded value $x=-2$ ; for $x+2>0$ , the inequality gives $x>1$ .

Part B

For $x+2<0$ , the inequality holds for every such $x$ , contributing $x<-2$ in full.

Part C

A single step assumes one fixed sign for $x+2$ , so it can report only one of the two cases and silently drops the other. Assuming positive, it would report only $x>1$ and lose the whole ray $x<-2$ ; for instance $x=-3$ solves the original but not that claimed set.

Worked solution

Part A

The expression is undefined where $x+2=0$ , so exclude $x=-2$ .

For $x+2>0$ (that is $x>-2$ ), multiplying both sides by the positive $x+2$ keeps the direction.

$9 < 3(x+2) \quad\Longrightarrow\quad 9<3x+6 \quad\Longrightarrow\quad x>1.$

Combined with $x>-2$ , this case contributes $x>1$ .

Part B

For $x+2<0$ (that is $x<-2$ ), multiplying by the negative $x+2$ reverses the direction.

$9 > 3(x+2) \quad\Longrightarrow\quad 9>3x+6 \quad\Longrightarrow\quad x<1.$

Every number satisfying $x<-2$ already satisfies $x<1$ , so this entire case contributes $x<-2$ .

Part C

Multiplying by $x+2$ in a single stroke commits to one direction, which is only correct for one sign of $x+2$ . Assuming $x+2$ positive keeps the direction and reports just $x>1$ , silently discarding the entire case $x<-2$ found in part B.

$x=-3:\quad \frac{9}{-1}=-9<3 \ \checkmark \qquad -3>1 \ ? \ \text{false}$

So $x=-3$ genuinely satisfies the original inequality yet is excluded from the naive single-case answer $x>1$ , exactly because that step never considered the negative case at all.

In one line

$\dfrac{9}{x+2}<3$ excludes $x=-2$ ; splitting on the sign of $x+2$ gives $x>1$ from the positive case and all of $x<-2$ from the negative case, so the full solution set is $(-\infty,-2)\cup(1,\infty)$ . A single-step multiplication that never splits by sign can report only one of these two pieces, for instance losing $x=-3$ , which solves the original inequality but not a claimed set of $x>1$ 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 3 points

States the value excluded because it makes the expression undefined. . Worth 1 point.

Multiplies by the positive x+2 keeping the direction, and correctly combines the result with the case assumption. . Worth 2 points.

Part B 3 points

Multiplies by the negative x+2, correctly reversing the direction. . Worth 1 point.

Recognizes that every value in the case assumption already satisfies the resulting inequality, so the whole case survives. . Worth 2 points.

Part C 4 points

Explains that a single-step multiplication commits to one sign and so can report only one branch of the true solution. . Worth 2 points. needs an explanation, not just an answer

Names a specific value, using the results of parts A and B, that the naive single step mishandles. . Worth 2 points.
4. The same two pieces, joined two different ways . 9 points. Question 4 of 10.

Consider the two conditions $6x-5 \ge 19$ and $3x+2 \le -13$ .
1. Part A.
  Solve each of the two conditions, $6x-5\ge19$ and $3x+2\le-13$ , on its own.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  State the solution set of ' $6x-5\ge19$ OR $3x+2\le-13$ '.
  
  Carry your own answer forward Combine whichever two pieces you found in part A by union, since these pieces are joined by 'or'.
  
  Solve and show your work Write each step out, and end with the value and its units. 2 points
3. Part C.
  Now join the two conditions from part A with 'AND' rather than 'OR'. Report the resulting solution set, then argue from what intersection and union each do to two rays that point away from each other, and say why the connective swap can turn nearly the whole line into nothing at all.
  
  Carry your own answer forward Use your own two solved pieces from part A to decide the new combined set.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$6x-5\ge19$ gives $x\ge4$ ; $3x+2\le-13$ gives $x\le-5$ .

Part B

$x\le-5$ or $x\ge4$ .

Part C

Joined by 'and', the set is empty, since no number is both at least $4$ and at most $-5$ . An intersection can only be as large as its smallest piece, so two pieces with no overlap intersect to nothing, while a union can never be smaller than either piece alone.

Worked solution

Part A

From $6x-5\ge19$ : add $5$ , divide by $6$ .

$6x\ge24 \quad\Longrightarrow\quad x\ge4.$

From $3x+2\le-13$ : subtract $2$ , divide by $3$ .

$3x\le-15 \quad\Longrightarrow\quad x\le-5.$

Part B

An 'or' compound keeps every number accepted by at least one piece, the union of $x\ge4$ and $x\le-5$ .

$\{x\ge4\} \cup \{x\le-5\} = \{x\le-5 \ \text{or} \ x\ge4\}$

Part C

Under 'and' the combined set is the intersection $\{x\ge4\}\cap\{x\le-5\}$ . No real number is both at least $4$ and at most $-5$ , so the intersection is empty.

$\{x\ge4\} \cap \{x\le-5\} = \varnothing$

In general, an intersection can never contain more than either piece, so two pieces whose rays point away from each other, with nothing shared, intersect to nothing regardless of how reasonable each piece is on its own. A union, by contrast, can never contain less than either piece, so it never empties out this way: the same two pieces that intersect to nothing already union to almost the whole line.

In one line

$6x-5\ge19$ gives $x\ge4$ and $3x+2\le-13$ gives $x\le-5$ . Joined by 'or', the solution set is $x\le-5$ or $x\ge4$ . Joined by 'and' instead, the solution set is empty, since no number satisfies both at once; an intersection can never exceed its smallest piece, while a union can never fall short of its largest, which is why the same two pieces behave so differently under the two connectives.

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

Solves both pieces correctly on their own. . Worth 2 points.

Reports each piece's solution in consistent inequality notation, ready to be combined. . Worth 1 point.

Part B 2 points

Correctly forms the union of the two solved pieces from part A. . Worth 2 points.

Part C 4 points

States the new combined solution set under AND is the empty set. . Worth 1 point.

Explains, in general set terms, why an intersection of two non-overlapping pieces is empty while a union of the same pieces is not. . Worth 3 points. needs an explanation, not just an answer
5. Two places a nonzero condition can enter, at two different times . 12 points. Question 5 of 10.

A rational equation carries its excluded values in plain sight before any algebra begins. A literal equation can hide its own nonzero condition until after it has been rearranged.
1. Part A.
  Solve $\dfrac{4}{x-3} = \dfrac{6}{x+7}$ , listing the excluded values before clearing anything.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Solve $y = \dfrac{k(x-3)}{x+7}$ for $x$ , state the restriction your final step requires, and check whether the resulting formula could ever output the value the original excludes, $x=-7$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  The excluded values in part A and the restriction in part B both come from division by something that could be zero, yet one was found BEFORE any algebra and the other only AFTER rearranging. Explain why $x=3$ and $x=-7$ had to be listed before solving, while $k\neq y$ could only be discovered by carrying out the rearrangement.
  
  Carry your own answer forward Refer to your own excluded values from part A and your own restriction from part B, whatever they were.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

Excluded: $x=3$ and $x=-7$ . Clearing gives $x=23$ , which is not on the excluded list, so it is genuine.

Part B

$x = \dfrac{3k+7y}{k-y}$ , valid whenever $k\neq y$ and $k\neq0$ (at $k=0$ the formula lands on the excluded value $-7$ ).

Part C

Part A's excluded values sit in the ORIGINAL denominators, visible by inspection before any work. Part B's original formula has no denominator built from $k-y$ at all; that quantity is manufactured only once you collect the $x$ terms and are about to divide, so it cannot be read off before the rearrangement is carried out.

Worked solution

Part A

The denominators vanish at $x=3$ and $x=-7$ , so exclude both first. Cross-multiplying,

$4(x+7) = 6(x-3) \quad\Longrightarrow\quad 4x+28=6x-18 \quad\Longrightarrow\quad 46=2x, \quad x=23.$

$x=23$ matches neither excluded value, so it is a genuine solution.

Part B

Clear the fraction, then collect every term containing $x$ .

$y(x+7)=k(x-3) \quad\Longrightarrow\quad yx+7y=kx-3k \quad\Longrightarrow\quad x(y-k)=-(3k+7y).$

Divide by the whole factor $y-k$ (equivalently write it as $k-y$ in the numerator's favor):

$x = \frac{3k+7y}{k-y}, \qquad k\neq y.$

Check whether this formula can output the original's own excluded value $x=-7$ .

$\frac{3k+7y}{k-y}=-7 \ \Longrightarrow\ 3k+7y=-7k+7y \ \Longrightarrow\ 10k=0 \ \Longrightarrow\ k=0.$

So whenever $k=0$ , the formula's own output collides with the value the original excluded from the start; the full restriction is $k\neq y$ and $k\neq0$ .

Part C

In part A, $x=3$ and $x=-7$ are zeros of denominators already written in the ORIGINAL equation, so a solver can read them off before touching any algebra.

In part B, the original formula $y=\dfrac{k(x-3)}{x+7}$ has no $k-y$ anywhere in it; that quantity does not exist until $x$ has been collected onto one side and the equation is about to be divided by whatever is left multiplying it.

$\underbrace{x=3,\ x=-7}_{\text{read from the original}} \qquad\text{versus}\qquad \underbrace{k\neq y}_{\text{created by rearranging}}$

So the restriction $k\neq y$ is not a fact about the original formula's denominators; it is a fact about a NEW quantity the rearrangement itself creates, which is why it can only be discovered by carrying the rearrangement through, not by inspecting the starting formula.

In one line

$\dfrac{4}{x-3}=\dfrac{6}{x+7}$ excludes $x=3,-7$ and has the genuine solution $x=23$ . Solving $y=\dfrac{k(x-3)}{x+7}$ for $x$ gives $x=\dfrac{3k+7y}{k-y}$ , valid whenever $k\neq y$ . The excluded values in the first case sit visibly in the original denominators, so they can be listed before any algebra; the restriction in the second case is created only by the rearrangement itself, so it can only be discovered by carrying that rearrangement out.

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

Lists both excluded values, from each denominator separately, before clearing the equation. . Worth 1 point.

Clears the denominators correctly and solves for the one candidate, checking it against the excluded list. . Worth 2 points.

Part B 5 points

Clears the fraction and collects every term containing x on one side before factoring. . Worth 2 points.

States the restriction k not equal to y as a single condition on the whole factor divided by. . Worth 1 point.

Checks whether the formula's output can equal the original's own excluded value x = -7, and identifies k not equal to 0 as the additional condition this requires. . Worth 2 points. needs an explanation, not just an answer

Part C 4 points

States that part A's excluded values come from denominators already present in the original equation, readable before any algebra. . Worth 2 points. needs an explanation, not just an answer

States that part B's restriction comes from a quantity the rearrangement itself creates, absent from the original formula, so it can only be found after carrying out the algebra. . Worth 2 points.
6. One coefficient, two verdicts at the same boundary . 11 points. Question 6 of 10.

Consider the equation $nx+45=4x+9n$ and the inequality $nx+45>4x+9n$ , built from the identical expressions on both sides.
1. Part A.
  Reduce $nx+45=4x+9n$ to the form (coefficient) $\,x=$ (constant), and classify its solution set for every value of $n$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Reduce the inequality $nx+45>4x+9n$ the same way, and classify its solution set at the single value $n=4$ where the equation's coefficient vanished.
  
  Carry your own answer forward Reduce the inequality the same way you reduced the equation in part A, then evaluate at n = 4.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  The equation and the inequality share the identical coefficient $n-4$ , and both reduce at $n=4$ to a bare leftover statement, $0=0$ for one and $0>0$ for the other. Explain why these two leftover statements do not receive the same verdict, and say what this proves about assuming an equation's classification at a boundary carries over unchanged to the matching inequality.
  
  Carry your own answer forward Refer to your own leftover statements from parts A and B at n = 4.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$(n-4)x=9(n-4)$ . For $n\neq4$ : $x=9$ . At $n=4$ : $0x=0$ , so the solution set is $\mathbb{R}$ .

Part B

$(n-4)x>9(n-4)$ . At $n=4$ : $0x>0$ , that is $0>0$ , false for every $x$ , so the solution set is $\varnothing$ .

Part C

$0=0$ is true, since a number equals itself, but $0>0$ is false, since nothing is strictly greater than itself. The vanishing coefficient decides nothing about the comparison's truth, so an equation's boundary verdict cannot be assumed to carry over to the matching inequality.

Worked solution

Part A

Collect the $x$ terms and the constants.

$nx-4x=9n-45 \quad\Longrightarrow\quad (n-4)x=9(n-4).$

For $n\neq4$ , the factor $n-4$ cancels: $x=9$ . At $n=4$ , both sides are $0$ , so the equation reads $0x=0$ , true for every $x$ : the solution set is $\mathbb{R}$ .

Part B

The same collection gives $(n-4)x>9(n-4)$ .

$(n-4)x>9(n-4) \ \Big|_{n=4} \quad\Longrightarrow\quad 0>0$

That is false for every $x$ , so the solution set is $\varnothing$ .

Part C

Both leftover statements come from the same vanishing coefficient $n-7=0$ , but they compare $0$ using different relations. Equality is reflexive, so $0=0$ is simply true, giving every real number. Strict order is irreflexive, so $0>0$ is simply false, giving no real number at all.

$0=0 \ (\text{true}) \qquad 0>0 \ (\text{false})$

The two statements share the identical numeric content, $0$ compared with $0$ , yet the connective between them, $=$ versus $>$ , is what decides the verdict, not the fact that the coefficient vanished. So a solver cannot assume that because an equation gives $\mathbb{R}$ at some boundary, the matching strict inequality does too; each connective's own truth at the boundary must be checked separately.

In one line

$nx+45=4x+9n$ reduces to $(n-4)x=9(n-4)$ , giving $x=9$ for $n\neq4$ and $\mathbb{R}$ at $n=4$ . The matching inequality $nx+45>4x+9n$ reduces to $(n-4)x>9(n-4)$ , which at $n=4$ gives $0>0$ , false for every $x$ , so $\varnothing$ . The identical coefficient vanishing forces the identical leftover number $0$ into two different comparisons; equality is reflexive so $0=0$ is true, while strict order is irreflexive so $0>0$ is false, which is why an equation's boundary verdict cannot be assumed to carry over to its matching inequality.

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

Reduces the equation to the form (coefficient) x = (constant) in n. . Worth 1 point.

Gives the one-solution value for n not equal to 4 and correctly classifies the boundary n = 4 as every real number. . Worth 2 points.

Part B 4 points

Reduces the inequality to the same coefficient-times-x form and substitutes n = 4 correctly. . Worth 2 points.

Correctly classifies the boundary as the empty set, not every real number. . Worth 2 points.

Part C 4 points

Explains the difference using the reflexive nature of equality versus the irreflexive nature of strict order, not merely restating that one is true and the other false. . Worth 3 points. needs an explanation, not just an answer

States the general lesson: an equation's boundary verdict must not be assumed to transfer unchanged to a matching inequality. . Worth 1 point.
7. A sign you already know, and a sign you do not . 11 points. Question 7 of 10.

This question walks through the same reversal decision twice, once where the multiplier is a plain number and once where it is an expression in $x$ .
1. Part A.
  Starting from the true statement $18<25$ , multiply both sides by $-3$ , then add $9$ . Report the final true statement, and name which single step caused the direction to reverse.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Solve $\dfrac{6}{x-9}<-2$ by excluding the forbidden value and splitting into the two sign cases for $x-9$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  In part A, you decided immediately that multiplying by $-3$ would reverse the direction. In part B, you could not decide the effect of multiplying by $x-9$ until you had split into two cases. Explain what property of $-3$ let you decide its effect at once, while $x-9$ forced you to consider both possibilities first.
  
  Carry your own answer forward Refer to your own step from part A and your own two cases from part B.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 4 points
The answer

Part A

$-45>-66$ ; multiplying by $-3$ is the step that reversed the direction.

Part B

$6<x<9$ .

Part C

$-3$ is a fixed number, so its sign is known before any $x$ is chosen, letting the reversal rule apply at once. $x-9$ 's sign depends on $x$ , the very quantity being solved for, so it cannot be read in advance and both possibilities must be carried as separate cases.

Worked solution

Part A

Multiplying by $-3$ : $18(-3)=-54$ and $25(-3)=-75$ ; since $-54>-75$ , the direction reverses at this step.

$18<25 \ \xrightarrow{\times(-3)} \ -54>-75$

Adding $9$ to both sides never changes the direction: $-54+9=-45$ and $-75+9=-66$ , giving $-45>-66$ .

Part B

Exclude $x=9$ . Case $x-9>0$ ( $x>9$ ): multiplying keeps the direction, $6<-2(x-9) \Rightarrow x<6$ , which contradicts $x>9$ , contributing nothing. Case $x-9<0$ ( $x<9$ ): multiplying reverses the direction, $6>-2(x-9) \Rightarrow x>6$ , combined with $x<9$ gives $6<x<9$ .

$6 < x < 9$

Part C

The constant $-3$ in part A is a specific number, negative once and for all, independent of $x$ ; its sign is simply a fact, so the direction-reversal rule applies the moment the step is taken.

The expression $x-9$ in part B has no fixed sign: its value changes as $x$ changes, and $x$ is exactly the quantity being solved for.

$\text{sign}(-3): \text{known always} \qquad \text{sign}(x-9): \text{depends on } x$

Reading its sign in advance is not possible, since doing so would require already knowing where $x$ falls, which is the answer being sought. That is why part B had to carry both possibilities, $x-9>0$ and $x-9<0$ , as separate cases until the algebra inside each case settled which values of $x$ actually belong there.

In one line

Multiplying $18<25$ by $-3$ then adding $9$ gives $-45>-66$ , the reversal happening at the multiplication. Solving $\dfrac{6}{x-9}<-2$ by cases gives $6<x<9$ . A fixed constant like $-3$ has a sign fixed before any $x$ is chosen, so its effect is known immediately; an expression like $x-9$ has a sign that depends on $x$ itself, so it cannot be decided in advance and must be carried as separate cases instead.

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

Computes both steps correctly, reversing the direction only at the multiplication. . Worth 2 points.

Identifies the multiplication by -3, not the addition, as the step that reversed the direction. . Worth 1 point.

Part B 4 points

Excludes x = 9 and correctly handles the direction in the x - 9 > 0 case, finding it contributes nothing. . Worth 2 points.

Correctly handles the direction in the x - 9 < 0 case and combines it with that case's own assumption. . Worth 2 points.

Part C 4 points

States that a fixed constant's sign is known independent of x, letting the reversal rule apply at once. . Worth 2 points. needs an explanation, not just an answer

States that a variable expression's sign depends on the unknown being solved for, so it cannot be read in advance and both cases must be carried. . Worth 2 points. needs an explanation, not just an answer
8. One anchor, two different needs for a check . 11 points. Question 8 of 10.

Consider $\lvert x-6 \rvert = 3x-16$ and, separately, $\lvert x-6 \rvert \le 5$ , both built around the same anchor $6$ .
1. Part A.
  Solve $\lvert x-6 \rvert = 3x-16$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Solve $\lvert x-6 \rvert \le 5$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Both parts used the same anchor $6$ . Explain why part A's equation needed a further check on its candidates while part B's inequality, once translated to a compound, needed no such check at all.
  
  Carry your own answer forward Refer to your own candidates from part A and your own interval from part B.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 4 points
The answer

Part A

The solution set is $\left\{\dfrac{11}{2}\right\}$ ; the other candidate $x=5$ is extraneous.

Part B

$1 \le x \le 11$ .

Part C

Part A's right side $3x-16$ is an ordinary expression that can be negative, so the equation secretly demands $3x-16\ge0$ , a condition the bare split ignores. Part B's right side is the fixed nonnegative constant $5$ , so $-5\le x-6\le5$ is an exact equivalence with no hidden condition to drop.

Worked solution

Part A

Split with $X=x-6$ , $Y=3x-16$ . Case $X=Y$ : $x-6=3x-16 \Rightarrow x=5$ . Case $X=-Y$ : $x-6=-(3x-16) \Rightarrow 4x=22 \Rightarrow x=\frac{11}{2}$ .

$Y=3x-16:\quad x=5 \Rightarrow Y=-1<0 \ \times \qquad x=\tfrac{11}{2} \Rightarrow Y=\tfrac12 \ge0 \ \checkmark$

Only $x=\frac{11}{2}$ passes, so the solution set is $\left\{\frac{11}{2}\right\}$ .

Part B

Read the inequality as a distance: $x$ lies within $5$ of $6$ .

$-5 \le x-6 \le 5 \quad\Longrightarrow\quad 1 \le x \le 11.$

Part C

In part A, the right side $Y=3x-16$ is an expression that depends on $x$ and can be negative, so $\lvert x-6\rvert=Y$ secretly demands $Y\ge0$ , a demand the bare split into $X=Y$ or $X=-Y$ never enforces, which is why each candidate had to be checked.

In part B, the right side is the fixed constant $5$ , already known to be nonnegative before any $x$ is chosen.

$Y=3x-16: \ \text{depends on } x \qquad \text{versus} \qquad 5: \ \text{fixed, nonnegative}$

Translating $\lvert x-6\rvert\le5$ to $-5\le x-6\le5$ carries no hidden sign requirement, because there is nothing about $5$ left to determine: its sign is settled once and for all, so the translation is an exact equivalence needing no further check.

In one line

$\lvert x-6 \rvert = 3x-16$ has solution set $\left\{\frac{11}{2}\right\}$ , with $x=5$ extraneous. $\lvert x-6 \rvert \le 5$ has solution set $[1,11]$ . The equation needed a check because its right side is a variable expression that can be negative; the inequality needed none because its bound is a fixed nonnegative constant, so translating it to a compound inequality drops no hidden condition.

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

Solves both split cases correctly for their candidate values of x. . Worth 2 points.

Checks both candidates against 3x-16 >= 0 and reports the correct solution set. . Worth 2 points.

Part B 3 points

Rewrites the inequality as the compound -5 <= x-6 <= 5 before solving. . Worth 1 point.

Solves to the correct closed interval. . Worth 2 points.

Part C 4 points

Identifies that part A's right side can be negative and depends on x, creating a hidden condition the split does not enforce. . Worth 2 points. needs an explanation, not just an answer

Identifies that part B's right side is a fixed nonnegative constant, so its translation carries no hidden condition to check. . Worth 2 points. needs an explanation, not just an answer
9. The rays outside a segment, and the segment they leave behind . 10 points. Question 9 of 10.

Let $p=2$ and $q=16$ be two anchors, a gap of $g=\lvert p-q \rvert=14$ apart.
1. Part A.
  Solve $\lvert x-2 \rvert + \lvert x-16 \rvert = 20$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Solve $\lvert x-9 \rvert > 7$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Part B's boundary values, $x=2$ and $x=16$ , are exactly the anchors $p$ and $q$ from part A, even though B's strict inequality excludes both. Explain why that boundary lands on the anchors, and then say what $\lvert x-2 \rvert + \lvert x-16 \rvert = 14$ (the gap itself, instead of $20$ ) would give as its solution set, connecting that set exactly to part B's answer.
  
  Carry your own answer forward Use your own anchors from part A and your own boundary values from part B.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$x=-1$ or $x=19$ .

Part B

$x<2$ or $x>16$ .

Part C

$9$ is the midpoint of $2$ and $16$ , and $7$ is half their gap, so $9-7=2$ and $9+7=16$ are exactly the anchors, even though B excludes both. Setting the sum equal to the gap $14$ gives the closed segment $[2,16]$ , exactly the complement of part B's two open rays.

Worked solution

Part A

Since the target $d=20$ exceeds the gap $g=14$ , there are exactly two solutions, each $\frac{d-g}{2}=3$ beyond an anchor.

$x = 2-3=-1 \qquad\text{or}\qquad x=16+3=19.$

Part B

Read the inequality as two rays farther than $7$ from $9$ .

$x-9 < -7 \ \text{ or } \ x-9 > 7 \quad\Longrightarrow\quad x<2 \ \text{ or } \ x>16.$

Part C

The center $9$ used in part B is the midpoint of the anchors, $\frac{2+16}{2}=9$ , and the radius $7$ is exactly half the gap, $\frac{14}{2}=7$ .

$9-7=2, \qquad 9+7=16$

So the boundary values are not a coincidence; they are the anchors themselves, reached because the inequality's center and radius were built from the same two numbers as part A's anchors.

Setting part A's target equal to the gap, $\lvert x-2 \rvert + \lvert x-16 \rvert = 14$ , lands on the $d=g$ case, whose solution is the entire closed segment $[2,16]$ between the anchors, endpoints included. That segment is precisely the set of $x$ with $\lvert x-9 \rvert \le 7$ , and it is EXACTLY the complement of part B's two open rays $x<2$ or $x>16$ : every real number lies in one and only one of the two sets, with the two anchors themselves landing in the segment.

In one line

With anchors $p=2$ , $q=16$ (gap $14$ ), $\lvert x-2 \rvert + \lvert x-16 \rvert = 20$ gives $x=-1$ or $x=19$ , and $\lvert x-9 \rvert > 7$ gives $x<2$ or $x>16$ . The boundary values $2$ and $16$ coincide with the anchors because $9$ is their midpoint and $7$ is half their gap; setting the sum equal to the gap itself gives the closed segment $[2,16]$ , exactly the complement of the two open rays.

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

Compares the target 20 with the gap 14 and recognizes this is the two-solution case. . Worth 1 point.

Computes both solutions correctly, one step (d-g)/2 outside each anchor. . Worth 2 points.

Part B 3 points

Splits into the two ray conditions correctly. . Worth 1 point.

Solves both rays correctly and reports the union. . Worth 2 points.

Part C 4 points

Explains that 9 is the midpoint and 7 is half the gap, so the boundary values 2 and 16 coincide with the anchors for that reason, not by coincidence. . Worth 2 points. needs an explanation, not just an answer

States that the d = g case gives the closed segment [2,16], and connects it exactly as the complement of part B's two open rays. . Worth 2 points.
10. Two checks, one shared reason they exist . 12 points. Question 10 of 10.

Consider $\dfrac{3}{x+8} = \dfrac{9}{x-4}$ and $\lvert x+5 \rvert = 4x-8$ , two equations that each ask a candidate to be tested against something the derivation itself did not use.
1. Part A.
  Solve $\dfrac{3}{x+8} = \dfrac{9}{x-4}$ , listing the excluded values before clearing anything.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Solve $\lvert x+5 \rvert = 4x-8$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Both parts tested a candidate against something the derivation itself did not use: an excluded-value list built from the ORIGINAL denominators in part A, and the sign of $4x-8$ in part B. Name the one step, common to both derivations in spirit, that is not reversible, and explain what such a step can do to a solution set that a reversible step never can.
  
  Carry your own answer forward Refer to your own excluded values from part A and your own two candidates from part B.
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 5 points
The answer

Part A

Excluded: $x=-8$ and $x=4$ . Clearing gives $x=-14$ , which is not on the excluded list, so it is genuine.

Part B

The solution set is $\left\{\dfrac{13}{3}\right\}$ ; $x=\dfrac{3}{5}$ is extraneous.

Part C

Part A's non-reversible step multiplies by a denominator that could be zero; part B's splits $\lvert X\rvert=Y$ into $X=\pm Y$ , dropping $Y\ge0$ . Both can ADD a candidate the original never had, caught only by a check against the original (an excluded list or a sign test); a reversible step never adds or loses a solution.

Worked solution

Part A

Exclude $x=-8$ and $x=4$ first. Cross-multiplying,

$3(x-4) = 9(x+8) \quad\Longrightarrow\quad 3x-12=9x+72 \quad\Longrightarrow\quad -84=6x, \quad x=-14.$

$x=-14$ matches neither excluded value, so it survives.

Part B

Split with $X=x+5$ , $Y=4x-8$ . Case $X=Y$ : $x+5=4x-8 \Rightarrow x=\frac{13}{3}$ . Case $X=-Y$ : $x+5=-(4x-8) \Rightarrow 5x=3 \Rightarrow x=\frac{3}{5}$ .

$Y=4x-8:\quad x=\tfrac{13}{3} \Rightarrow Y=\tfrac{28}{3}\ge0 \ \checkmark \qquad x=\tfrac{3}{5} \Rightarrow Y=-\tfrac{28}{5}<0 \ \times$

Only $x=\frac{13}{3}$ passes, so the solution set is $\left\{\frac{13}{3}\right\}$ .

Part C

In part A, the step that is not reversible is multiplying both sides by a denominator, $x+8$ or $x-4$ , that could equal zero; that step is only undone by dividing back by the same expression, which fails exactly where it is zero. In part B, the corresponding step is splitting $\lvert X \rvert=Y$ into $X=Y$ or $X=-Y$ ; this direction is one-way, since it silently drops the requirement $Y\ge0$ that the original equation carries automatically.

$\text{multiply by a zero-able denominator} \quad\sim\quad \text{split } \lvert X\rvert=Y \text{ into } X=\pm Y$

Both steps share the same danger. A reversible step is proved, once and for all, to preserve a solution set exactly: nothing is added and nothing is lost. A step that cannot be undone carries no such guarantee, and specifically here, both steps can ADD a candidate that satisfies the DERIVED statement, the cleared equation or the bare split, without satisfying the ORIGINAL. That is precisely why each needs a check that reaches back to the original: the excluded-value list in part A comes directly from the original's own denominators, and the sign test $Y\ge0$ in part B comes directly from the fact that the original's left side can never be negative. Neither check is optional decoration; each is the one place the original equation's own information, thrown away by the one-way step, gets consulted again.

In one line

$\dfrac{3}{x+8}=\dfrac{9}{x-4}$ excludes $x=-8,4$ and has the genuine solution $x=-14$ . $\lvert x+5 \rvert = 4x-8$ has solution set $\left\{\frac{13}{3}\right\}$ , with $x=\frac{3}{5}$ extraneous. Both derivations pass through a non-reversible step, multiplying by a possibly-zero denominator or splitting an absolute-value equation into $X=\pm Y$ , and both such steps can add a candidate the original never had; the excluded-value list and the $Y\ge0$ sign test are the two places each check goes back to consult the original equation's own information.

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

Lists both excluded values before clearing the equation. . Worth 1 point.

Clears the denominators correctly, solves for the candidate, and checks it against the excluded list. . Worth 2 points.

Part B 4 points

Solves both split cases correctly for their candidate values of x. . Worth 2 points.

Checks both candidates against 4x-8 >= 0 and reports the correct solution set. . Worth 2 points.

Part C 5 points

Names the correct non-reversible step in each derivation: multiplying by a possibly-zero denominator in part A, and the bare split dropping Y >= 0 in part B. . Worth 2 points. needs an explanation, not just an answer

Explains, in general terms, that such a step can add a candidate the original never had, which a reversible step is proved never to do. . Worth 2 points. needs an explanation, not just an answer

Connects each specific check, the excluded-value list and the sign test, back to information from the original equation that the one-way step discarded. . Worth 1 point.

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. Reading the answer off two reduced equations 9 points

Part A 2

Collects like terms correctly and reduces the equation to the form ax = b. 1
Solves for x and reports the solution set using set notation. 1

Part B 3

Distributes and combines like terms correctly before comparing the two sides. 1
Reduces the equation to a bare numeric statement and reports the solution set that statement implies. 2

Part C 4

Argues that a nonzero coefficient and a zero coefficient exhaust every possibility for a linear equation, rather than checking only the two given examples. 2
States that landing on every real number requires both the reduced coefficient and the reduced constant to vanish together, referencing part B's own reduction. 2

2. A factor that is a sum, not a single letter 10 points

Part A 3

Factors P out of both terms on the right before dividing. 2
States the restriction rt not equal to -1 as a single condition on the factor 1+rt. 1

Part B 3

Substitutes the three given values into the rearranged formula and evaluates the denominator before dividing. 2
Reports the principal with the correct unit, dollars, not as a bare number. 1

Part C 4

Correctly determines that rt=-1 cannot arise when r and t are both positive, and justifies it from their signs. 2
States that the restriction is on the combined quantity rt, not on r or t individually, tying it back to part A's factor. 2

3. A fraction on one side, and a sign not yet known 10 points

Part A 3

States the value excluded because it makes the expression undefined. 1
Multiplies by the positive x+2 keeping the direction, and correctly combines the result with the case assumption. 2

Part B 3

Multiplies by the negative x+2, correctly reversing the direction. 1
Recognizes that every value in the case assumption already satisfies the resulting inequality, so the whole case survives. 2

Part C 4

Explains that a single-step multiplication commits to one sign and so can report only one branch of the true solution. 2
Names a specific value, using the results of parts A and B, that the naive single step mishandles. 2

4. The same two pieces, joined two different ways 9 points

Part A 3

Solves both pieces correctly on their own. 2
Reports each piece's solution in consistent inequality notation, ready to be combined. 1

Part B 2

Correctly forms the union of the two solved pieces from part A. 2

Part C 4

States the new combined solution set under AND is the empty set. 1
Explains, in general set terms, why an intersection of two non-overlapping pieces is empty while a union of the same pieces is not. 3

5. Two places a nonzero condition can enter, at two different times 12 points

Part A 3

Lists both excluded values, from each denominator separately, before clearing the equation. 1
Clears the denominators correctly and solves for the one candidate, checking it against the excluded list. 2

Part B 5

Clears the fraction and collects every term containing x on one side before factoring. 2
States the restriction k not equal to y as a single condition on the whole factor divided by. 1
Checks whether the formula's output can equal the original's own excluded value x = -7, and identifies k not equal to 0 as the additional condition this requires. 2

Part C 4

States that part A's excluded values come from denominators already present in the original equation, readable before any algebra. 2
States that part B's restriction comes from a quantity the rearrangement itself creates, absent from the original formula, so it can only be found after carrying out the algebra. 2

6. One coefficient, two verdicts at the same boundary 11 points

Part A 3

Reduces the equation to the form (coefficient) x = (constant) in n. 1
Gives the one-solution value for n not equal to 4 and correctly classifies the boundary n = 4 as every real number. 2

Part B 4

Reduces the inequality to the same coefficient-times-x form and substitutes n = 4 correctly. 2
Correctly classifies the boundary as the empty set, not every real number. 2

Part C 4

Explains the difference using the reflexive nature of equality versus the irreflexive nature of strict order, not merely restating that one is true and the other false. 3
States the general lesson: an equation's boundary verdict must not be assumed to transfer unchanged to a matching inequality. 1

7. A sign you already know, and a sign you do not 11 points

Part A 3

Computes both steps correctly, reversing the direction only at the multiplication. 2
Identifies the multiplication by -3, not the addition, as the step that reversed the direction. 1

Part B 4

Excludes x = 9 and correctly handles the direction in the x - 9 > 0 case, finding it contributes nothing. 2
Correctly handles the direction in the x - 9 < 0 case and combines it with that case's own assumption. 2

Part C 4

States that a fixed constant's sign is known independent of x, letting the reversal rule apply at once. 2
States that a variable expression's sign depends on the unknown being solved for, so it cannot be read in advance and both cases must be carried. 2

8. One anchor, two different needs for a check 11 points

Part A 4

Solves both split cases correctly for their candidate values of x. 2
Checks both candidates against 3x-16 >= 0 and reports the correct solution set. 2

Part B 3

Rewrites the inequality as the compound -5 <= x-6 <= 5 before solving. 1
Solves to the correct closed interval. 2

Part C 4

Identifies that part A's right side can be negative and depends on x, creating a hidden condition the split does not enforce. 2
Identifies that part B's right side is a fixed nonnegative constant, so its translation carries no hidden condition to check. 2

9. The rays outside a segment, and the segment they leave behind 10 points

Part A 3

Compares the target 20 with the gap 14 and recognizes this is the two-solution case. 1
Computes both solutions correctly, one step (d-g)/2 outside each anchor. 2

Part B 3

Splits into the two ray conditions correctly. 1
Solves both rays correctly and reports the union. 2

Part C 4

Explains that 9 is the midpoint and 7 is half the gap, so the boundary values 2 and 16 coincide with the anchors for that reason, not by coincidence. 2
States that the d = g case gives the closed segment [2,16], and connects it exactly as the complement of part B's two open rays. 2

10. Two checks, one shared reason they exist 12 points

Part A 3

Lists both excluded values before clearing the equation. 1
Clears the denominators correctly, solves for the candidate, and checks it against the excluded list. 2

Part B 4

Solves both split cases correctly for their candidate values of x. 2
Checks both candidates against 4x-8 >= 0 and reports the correct solution set. 2

Part C 5

Names the correct non-reversible step in each derivation: multiplying by a possibly-zero denominator in part A, and the bare split dropping Y >= 0 in part B. 2
Explains, in general terms, that such a step can add a candidate the original never had, which a reversible step is proved never to do. 2
Connects each specific check, the excluded-value list and the sign test, back to information from the original equation that the one-way step discarded. 1

Multiple choice

Free response