Solving Linear Inequalities: Free Response

Undo the addition first, then the multiplication, exactly as for an equation. Subtract $15$ from both sides, which never disturbs the direction:

$4x + 15 - 15 > 3 - 15 \quad\Longrightarrow\quad 4x > -12.$

Divide both sides by the positive number $4$, so the symbol stays put:

$\frac{4x}{4} > \frac{-12}{4} \quad\Longrightarrow\quad x > -3.$

The final answer is negative, but no flip occurred anywhere: the only division in this solve was by a positive number.

Isolate the term with $x$ before dividing anything. Subtract $20$ from both sides, which does not touch the direction:

$20 - 3x - 20 \le 5 - 20 \quad\Longrightarrow\quad -3x \le -15.$

Divide both sides by $-3$. The divisor is negative, so reverse $\le$ to $\ge$:

$\frac{-3x}{-3} \ge \frac{-15}{-3} \quad\Longrightarrow\quad x \ge 5.$

The final answer is positive this time, and a flip did happen, at the division by $-3$.

State the rule first, then test it against the two parts. Multiplying or dividing both sides of an inequality by a negative number reverses the symbol; adding, subtracting, or scaling by a positive number never does.

Part A's only scaling step divided by $4$, a positive number, so no flip was licensed there, and none occurred, even though $-3$ is a negative number.

Part B's only scaling step divided by $-3$, a negative number, so a flip was required there, and one occurred, even though $5$ is a positive number.

$\text{trigger} = \text{sign of the divisor}, \qquad \text{not the sign of the answer}.$

A rule keyed to the final answer would have gotten both parts backwards: it would flip the one that should not flip and leave alone the one that should. The two parts were built to disagree on the sign of the answer while agreeing on nothing about the rule, which is exactly what separates the real trigger from a coincidence a learner might otherwise latch onto.

Each plan's total is a base fee plus an hourly rate times the hours:

$70 + 12h < 40 + 18h.$

Collect the hourly terms on the right, subtracting $12h$ from both sides, which keeps the remaining hourly coefficient positive:

$70 < 40 + 6h.$

Subtract $40$, then divide by the positive number $6$:

$30 < 6h \quad\Longrightarrow\quad 5 < h.$

That reads $h > 5$ from the other end, reached with no flip anywhere.

Subtract $18h$ from both sides instead, which sends the hourly terms to the left with a negative coefficient:

$70 - 6h < 40.$

Subtract $70$:

$-6h < -30.$

Divide by $-6$, reversing $<$ to $>$:

$h > 5.$

The two routes land on the identical solution set, as they must: each is a chain of licensed moves applied to the same original inequality.

The solved inequality is strict, $h > 5$, so the boundary value is not automatically included; it has to be checked in the situation itself.

At $h = 5$, compare the two totals directly:

$70 + 12(5) = 130, \qquad 40 + 18(5) = 130.$

The two agree exactly. Since a customer renting for exactly $5$ hours pays the same either way, that hour count is not one where Plan A is strictly cheaper. Only rentals of more than $5$ hours make Plan A the better deal.

Everything up to $-7x < -7$ is honest algebra: distributing the $4$, collecting the variable terms, and subtracting the constant all check out. The break is in the very last move.

Going from $-7x < -7$ to a statement about $x$ alone requires dividing both sides by $-7$, a negative number. That move reverses the symbol; Priya's line kept $<$ unchanged, so the rule was not applied. Corrected, the last line should read

$\frac{-7x}{-7} > \frac{-7}{-7} \quad\Longrightarrow\quad x > 1.$

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

At $x = 0$, a member of Priya's claimed set:

$4(2-0) = 8, \qquad 3(0)+1 = 1, \qquad 8 < 1 \text{ is false.}$

So $x = 0$ does not satisfy the original, even though it lies inside Priya's set, which is enough on its own to rule that set out.

At $x = 2$, a number outside Priya's claimed set:

$4(2-2) = 0, \qquad 3(2)+1 = 7, \qquad 0 < 7 \text{ is true.}$

A number outside Priya's set satisfies the original, which is further evidence that her set is the wrong one.

The rule is stated about the move itself, not about the history of the number involved: dividing (or multiplying) both sides of an inequality by a negative number reverses the symbol, and adding or subtracting never does, whatever sign is involved.

The $-7$ dividing both sides of $-7x < -7$ is, at the moment it is used, simply a negative number, however it arose:

$-4x - 3x = -7x.$

It makes no difference to the division step that this $-7$ came from combining two terms two lines earlier rather than being handed over directly. A division by $-7$ written on its own, with no history at all, would need the identical flip.

So the source of a negative coefficient never changes what a later step must do with it; only the operation actually performed, at the moment it is performed, decides that.

Solve each piece on its own before combining them. From $3x - 1 \ge 11$: add $1$, then divide by $3$:

$3x \ge 12 \quad\Longrightarrow\quad x \ge 4.$

From $2x + 5 \le 3$: subtract $5$, then divide by $2$:

$2x \le -2 \quad\Longrightarrow\quad x \le -1.$

An 'and' keeps only numbers both pieces accept. The first piece needs $x$ at least $4$; the second needs $x$ at most $-1$. Since $4 > -1$, no number is large enough for the first requirement while also being small enough for the second, so the combined solution is empty.

Subtract $5x$ from both sides, and every copy of the variable cancels:

$5x + 9 - 5x < 5x - 2 - 5x \quad\Longrightarrow\quad 9 < -2.$

That leftover statement mentions no $x$ at all, and it is false: $9$ is not less than $-2$. A false statement with no variable in it means no value of $x$ could ever have made the original true, so the inequality has no solution. The variable cancelling is not itself the reason; the falseness of what is left behind is.

Solve each piece on its own. From $x + 6 < 10$, subtract $6$:

$x < 4.$

From $4x - 3 \ge -7$, add $3$ and divide by $4$:

$4x \ge -4 \quad\Longrightarrow\quad x \ge -1.$

An 'or' keeps any number accepted by either piece. Take any real number at all: either it is below $4$, in which case the first piece accepts it directly, or it is not below $4$, meaning it is at least $4$, and every such number is certainly at least $-1$ as well, so the second piece accepts it. There is no number left over for neither piece to catch, so the union is every real number.

Compare the four rays by where their arrows point, not by resolving anything further.

In part A, $x \ge 4$ points to the right, away from $4$, and $x \le -1$ points to the left, away from $-1$:

$x \ge 4 \; (\text{right of } 4), \qquad x \le -1 \; (\text{left of } -1).$

Between $-1$ and $4$ lies a gap that neither ray reaches into, and since an AND demands membership in both pieces at once, any number in that gap, or anywhere else for that matter, fails at least one of the two, and the combined solution is empty.

In part C, $x < 4$ points to the left, running through the gap and past $-1$, and $x \ge -1$ points to the right, running through the same gap and past $4$. The two rays now overlap across the entire gap and beyond it in both directions, and since an OR only demands membership in at least one piece, every number, including everything in what was the gap, is caught by one ray or the other.

The two situations use the identical pair of boundary numbers. What changed is only which side of each boundary the ray runs toward, and that single difference is the whole explanation for why one combination is empty and the other is everything.

A chain $p < b - kx \le q$ is shorthand for two separate inequalities sharing the same middle expression, so isolate $x$ in each on its own.

From $b - kx \le q$: subtract $b$, then divide by $-k$, reversing $\le$ to $\ge$ because $-k$ is negative:

$-kx \le q - b \quad\Longrightarrow\quad x \ge \frac{q-b}{-k} = \frac{b-q}{k}.$

From $p < b - kx$: subtract $b$, then divide by $-k$, reversing $<$ to $>$:

$p - b < -kx \quad\Longrightarrow\quad \frac{p-b}{-k} > x \quad\Longrightarrow\quad x < \frac{b-p}{k}.$

Each step is a property of equality or inequality applied to both sides and undone by its opposite move, so no solution was lost or gained in either half. Putting the two results back together, with the inclusive bound kept inclusive and the strict bound kept strict:

$\frac{b-q}{k} \le x < \frac{b-p}{k}.$

Nothing in the argument depended on a particular $k$, $b$, $p$, or $q$ beyond $k > 0$ and $p < q$, so it holds for every such choice.

Match the numbers to the letters: $k = 4$, $b = 9$, $p = -3$, $q = 13$. Substitute into the formula from part A:

$\frac{b-q}{k} = \frac{9-13}{4} = -1, \qquad \frac{b-p}{k} = \frac{9-(-3)}{4} = 3.$

So the solution is $-1 \le x < 3$.

Verify each endpoint in $-3 < 9 - 4x \le 13$ directly. At $x = -1$: $9 - 4(-1) = 13$, and $-3 < 13 \le 13$ holds, since $13 \le 13$ is true, confirming $-1$ belongs. At $x = 3$: $9 - 4(3) = -3$, and $-3 < -3$ is false, confirming $3$ is correctly excluded.

The swap is not a separate bookkeeping step; it is the flip itself, seen from the bound's point of view.

Before dividing, $q$ is an upper bound: the expression $b - kx$ is required to be no more than $q$. Dividing by $-k$ reverses $\le$ to $\ge$, so the inequality involving $q$ becomes

$b - kx \le q \quad\Longrightarrow\quad x \ge \frac{b-q}{k},$

a statement that $x$ is at least something, which is a LOWER bound. The number $q$ has not moved from where the algebra puts it; what changed is that a division by a negative number turns a 'no more than' into an 'at least', and that change of kind is precisely what turns an upper bound on one side into a lower bound on the other.

The same thing happens to $p$ in the other half, in the opposite direction: an 'at least' involving $p$ becomes a 'no more than' involving $x$.

So asking whether the swap could be avoided is asking whether the flip could be avoided while still dividing by a negative number, and the lesson's own proof already settled that it cannot: a division by a negative number reverses the symbol every time, with no exception.