Linear Inequalities: Free Response

Apply each step to both sides of $-9 < 4$ separately, then decide the symbol from the truth of the result.

Adding $11$: $-9 + 11 = 2$ and $4 + 11 = 15$.

$2 < 15$

Multiplying by $6$, a positive number: $-9(6) = -54$ and $4(6) = 24$.

$-54 < 24$

Multiplying by $-2$, a negative number: $-9(-2) = 18$ and $4(-2) = -8$. Keeping the symbol as $<$ would claim $18 < -8$, which is false, so the true statement reverses it.

$18 > -8$

Start from what $p < q$ means: the difference $q - p$ is a positive number. The claim is about the products $kp$ and $kq$, so compare them through their own difference, the way the lesson compared $-a$ and $-b$.

$kq - kp = k(q - p)$

Because $k$ is negative and $q - p$ is positive, their product is negative.

$k(q - p) < 0 \;\Longrightarrow\; kq - kp < 0 \;\Longrightarrow\; kq < kp$

The statement $kq < kp$ is exactly $kp > kq$, so the reversal holds for every choice of $p < q$ and every negative $k$, not merely for the specific numbers in part A.

Reversibility cannot be the deciding property, because every one of the three steps in part A can be undone by an opposite step of the same family, so all three are equally reversible, and still only the negative multiplication flipped the direction. What actually decides a flip is the sign of the number multiplying both sides, not whether the step has an inverse.

A reversal, like the one in part A, is still an accurate report about the original two numbers: $18 > -8$ genuinely reflects what happened to $-9$ and $4$ under multiplication by $-2$, and dividing by $-2$ recovers them exactly. Multiplying by $0$ is not that.

$-9 < 4 \;\xrightarrow{\;\times 0\;}\; 0 < 0$

The result $0 < 0$ is false outright and says nothing at all about $-9$ or $4$ specifically: any pair of numbers multiplied by $0$ produces that identical statement (or $0 \le 0$, for a non-strict symbol), so no trace of the starting numbers survives.

The property every step in part A shares is that each one is invertible: addition of $11$ is undone by subtracting $11$, multiplication by $6$ by dividing by $6$, and multiplication by $-2$ by dividing by $-2$. Multiplying by $0$ has no such partner, because dividing by $0$ is not defined, so there is no step that could take $0 < 0$ back to $-9 < 4$. That is the real difference: a reversal changes the symbol but preserves recoverability, while multiplying by $0$ destroys recoverability altogether.

The total adjustment is the per-unit amount $r$ times the number of units $x$, and exceeding $60$ dollars means that total is strictly greater than $60$.

$rx > 60$

When $r > 0$, dividing both sides by the positive number $r$ keeps the direction.

$x > \frac{60}{r}$

Divide both sides by the negative number $r$, which reverses the direction.

$rx > 60 \;\Rightarrow\; x < \frac{60}{r}$

Since $r$ is negative, dividing the positive number $60$ by a negative number gives a negative result, so $\frac{60}{r} < 0$.

Part B's solution set is $x < \frac{60}{r}$, and $\frac{60}{r}$ was shown there to be a negative number. A set of the form 'less than a negative number' contains only negative numbers, since anything at or above $0$ automatically fails to be below a negative boundary.

$x < \frac{60}{r} < 0 \;\Longrightarrow\; x < 0$

A genuine count of units sold is never negative, so no realistic $x$ can belong to this case's solution set: whatever the deduction rate turns out to be, it can never push the adjustment above $60$ dollars for an actual sale.

With $r = 0$, the left side $rx$ is $0$ for every $x$, so the inequality has no variable left in it.

$0 > 60$

This statement is false regardless of $x$, so the solution set is empty: dropping the renegotiation can never itself push the adjustment above $60$ dollars.

The general rule for $ax > b$ at $a = 0$ says the outcome is every real number when $b < 0$, and empty when $b \ge 0$. Here $b = 60$ is positive, squarely in the $b \ge 0$ half, so the empty outcome found directly is exactly what the general rule predicts.

Every line after the second one follows correctly from the line before it: distributing, combining, and isolating $x$ are all ordinary algebra. The break is at the very first move.

$\frac{12}{x+1} > 3 \;\xrightarrow{\;\times(x+1)\;}\; 12 > 3(x+1)$

That step is licensed only once the sign of $x + 1$ is known: positive keeps the direction, negative reverses it, and $x + 1 = 0$ is not even allowed, since the original expression is undefined there. Marcus's work fixes no sign at all and simply keeps the direction, which silently assumes $x + 1 > 0$, and it never sets aside $x = -1$ before starting. Both omissions belong to the same first step.

Substitute each value into the original inequality as it was given, before any of Marcus's steps.

At $x = -5$:

$\frac{12}{-5+1} = \frac{12}{-4} = -3, \qquad -3 > 3 \text{ is false.}$

So $x = -5$ fails the original, even though $-5 < 3$ places it squarely inside Marcus's claimed set. That is enough on its own to rule his set out.

At $x = 2$:

$\frac{12}{2+1} = \frac{12}{3} = 4, \qquad 4 > 3 \text{ is true.}$

Here the number checked out and does lie in his claimed set, so this test alone does not expose anything, but the failure at $x = -5$ already settles the matter.

Exclude $x = -1$ immediately, since the original expression is undefined there.

Case $x + 1 > 0$, that is $x > -1$. Multiply by the positive $x + 1$, keeping the direction:

$12 > 3(x+1) \;\Rightarrow\; 9 > 3x \;\Rightarrow\; x < 3.$

Combined with $x > -1$, this case contributes $-1 < x < 3$.

Case $x + 1 < 0$, that is $x < -1$. Multiply by the negative $x + 1$, reversing the direction:

$12 < 3(x+1) \;\Rightarrow\; 9 < 3x \;\Rightarrow\; x > 3.$

This asks for $x > 3$, but the case assumed $x < -1$, and no number is both, so this case contributes nothing.

The full solution set is $-1 < x < 3$, the interval where the first case's result actually held.

The case $x + 1 < 0$ assumed $x < -1$ and, after flipping the direction for the negative multiplier, reduced to $x > 3$.

$x < -1 \quad\text{and}\quad x > 3$

Those two conditions cannot both hold for any real number, since no number is simultaneously less than $-1$ and greater than $3$. The case is not empty because of some general fact about cases; it is empty because its own conclusion, $x > 3$, directly contradicts the assumption, $x < -1$, that opened it.

Solve each piece on its own first. From $5x - 4 > 41$: add $4$, then divide by $5$.

$5x > 45 \;\Rightarrow\; x > 9.$

From $3x + 19 < 4$: subtract $19$, then divide by $3$.

$3x < -15 \;\Rightarrow\; x < -5.$

An 'and' keeps only numbers both pieces accept. The first piece needs $x$ above $9$; the second needs $x$ below $-5$. Since $9$ is not below $-5$, no number satisfies both, so the intersection is empty.

Solve each piece on its own. From $3x - 4 < 26$: add $4$, then divide by $3$.

$3x < 30 \;\Rightarrow\; x < 10.$

From $6x - 5 \ge 7$: add $5$, then divide by $6$.

$6x \ge 12 \;\Rightarrow\; x \ge 2.$

An 'or' keeps any number accepted by either piece. Take any real number: either it is below $10$, in which case the first piece accepts it, or it is at least $10$, and every such number is certainly at least $2$ as well, so the second piece accepts it. No number escapes both, so the union is every real number.

Solve the first piece. Subtracting $8x$ from both sides cancels every copy of the variable.

$8x + 3 < 8x - 6 \;\Longrightarrow\; 3 < -6.$

This leftover statement has no $x$ in it, and it is false, so no value of $x$ satisfies this piece; it contributes the empty set. Uniting the empty set with the second piece leaves the second piece unchanged.

$\varnothing \cup \{x : x < 20\} = \{x : x < 20\}$

The full solution set is exactly $x < 20$.

The first piece still reduces to the false statement $3 < -6$, contributing the empty set $\varnothing$, regardless of which connective it is joined by.

Under 'and', the combined set is the intersection $\varnothing \cap \{x : x < 20\}$.

$\varnothing \cap \{x : x < 20\} = \varnothing$

An intersection can never contain more than either piece, so intersecting with the empty set always gives the empty set, whatever the other piece happens to be: the combined solution set here is empty.

Under 'or', by contrast, the combined set was the union $\varnothing \cup \{x : x < 20\}$, and a union can never contain less than either piece, so uniting with the empty set changes nothing: the other piece survives whole. The same false, variable-free piece is fatal under 'and' and silent under 'or', because intersecting with nothing keeps nothing, while uniting with nothing keeps everything the other piece already had.

Everything turns on the sign of $a$, exactly as in the lesson's own classification of $ax > b$.

If $a > 0$, divide both sides by the positive number $a$. Dividing by a positive keeps the direction.

$ax < b \;\Rightarrow\; x < \frac{b}{a}.$

If $a < 0$, divide both sides by the negative number $a$. Dividing by a negative reverses the direction.

$ax < b \;\Rightarrow\; x > \frac{b}{a}.$

If $a = 0$, there is nothing to divide by, and the left side $ax$ is $0$ for every $x$, so the inequality collapses to a statement with no $x$ in it.

$0 < b.$

When $b > 0$, the claim $0 < b$ is true, so every real number is a solution and the solution set is all of $\mathbb{R}$. When $b \le 0$, the claim $0 < b$ is false (it fails both at $b = 0$, where $0 < 0$ is false, and at any $b < 0$), so no value of $x$ works and the solution set is $\varnothing$.

These three cases, $a > 0$, $a < 0$, and $a = 0$, are exhaustive, and within $a = 0$ the two possibilities for $b$ are exhaustive too, so the classification covers every real $a$ and $b$, not merely chosen instances.

Here $a = -11$, a negative number, and $b = 33$. Part A gives $x > \frac{b}{a}$ for $a < 0$.

$\frac{b}{a} = \frac{33}{-11} = -3 \;\Longrightarrow\; x > -3.$

Verify both test values directly in $-11x < 33$. At $x = -4$: $-11(-4) = 44$, and $44 < 33$ is false, consistent with $-4$ lying below $-3$ and so outside the solution set. At $x = -2$: $-11(-2) = 22$, and $22 < 33$ is true, consistent with $-2$ lying above $-3$ and so inside the solution set.

A guess that every branch simply swaps would predict the opposite of the $ax > b$ result at this boundary, so it would predict all of $\mathbb{R}$ for $ax < b$ at $a = 0, b = 0$.

Checking directly: at $a = 0, b = 0$, the inequality $ax < b$ becomes

$0 < 0,$

which is false for every $x$, exactly as false as $0 > 0$ was in the lesson's own case. So the actual outcome is $\varnothing$, the same conclusion as $ax > b$ reached at this same boundary, not its opposite.

The reason the mirror breaks here is that $0 < 0$ and $0 > 0$ are both false statements, for the identical reason: a strict inequality never counts two equal numbers as satisfying it, whichever direction it points. So the equality boundary $b = 0$ is on the empty-set side of BOTH classifications, and only the branches with $b$ strictly positive or strictly negative actually swap between the two families.

The lesson's rule for $ax > b$ at $a = 0$ gives all of $\mathbb{R}$ (in $x$) exactly when $b < 0$, and part A's rule for $ax < b$ at $a = 0$ gives all of $\mathbb{R}$ exactly when $b > 0$. The compound '$0 > b$ or $0 < b$' is true for a given $b$ precisely when at least one of those two conditions on $b$ holds.

$\{b : b < 0\} \cup \{b : b > 0\} = \{b : b \neq 0\}.$

So the compound is true for every real number $b$ except $b = 0$. This lines up exactly with part C: at $b = 0$ neither $0 > b$ nor $0 < b$ can be true, since both reduce to a false strict inequality between two equal numbers, so $b = 0$ is the one value left out of the union, consistent with it sitting on the empty-set side of both classifications.