Inequality Basics: Free Response

'No more than $12$' allows exactly $12$ and nothing above it, so the boundary is included:

$p \le 12.$

'More than $6$' excludes $6$ itself, so the boundary is left out:

$p > 6.$

Combine the two conditions from part A into one two-sided statement, ordering the smaller number on the left:

$6 < p \le 12.$

Test each candidate against both original conditions. For $p = 6$: the capacity condition $6 \le 12$ holds, but the driver's rule $6 > 6$ is false, so $p = 6$ fails. For $p = 12$: $12 \le 12$ holds and $12 > 6$ holds, so $p = 12$ passes.

Match each end of $6 < p \le 12$ to its fence: the strict left bound $6$ takes a parenthesis, and the inclusive right bound $12$ takes a bracket:

$6 < p \le 12 \quad\Longleftrightarrow\quad (6, 12].$

Both ends are finite numbers, not $\pm\infty$, so the interval is bounded.

Add $7$ to each side of $-4 < 6$:

$-4 + 7 < 6 + 7 \;\Rightarrow\; 3 < 13.$

Multiply each side by $3$, a positive number, so the direction stays the same:

$-4(3) < 6(3) \;\Rightarrow\; -12 < 18.$

Multiply each side by $-3$, a negative number: naively keeping $<$ would claim $12 < -18$, which is false, so the true statement reverses the symbol:

$12 > -18.$

The reversing operation is (iii), multiplying by $-3$. Track where each number lands:

$-4 \xrightarrow{\;\times(-3)\;} 12, \qquad 6 \xrightarrow{\;\times(-3)\;} -18.$

Multiplying by a negative number REFLECTS every number across zero, and a reflection reverses left-right order: whichever number was further left comes back further right. That is what reverses the symbol. Adding $7$ slides both numbers the same distance in the same direction, so their order is untouched, and this is worth stating carefully, because $-4 + 7 = 3$ does cross zero and the order still holds: crossing zero is not the mechanism. Multiplying by the positive number $3$ stretches both distances from zero without swapping sides, so it does not disturb the order either.

Dividing by $-2$ is a negative divisor, the same family as operation (iii), so the prediction is that it reverses too. Carrying it out confirms it:

$\frac{-4}{-2} = 2, \qquad \frac{6}{-2} = -3, \qquad\text{so}\qquad 2 > -3.$

The prediction and the computation agree.

'At least $5.2$' includes the boundary:

$L \ge 5.2.$

'Less than $5.2$' excludes it:

$L < 5.2.$

Line the two conditions up against the same $L$:

$L \ge 5.2 \quad\text{and}\quad L < 5.2.$

For any candidate length, one of the two must fail: a length that is $5.2$ or above cannot also be strictly under $5.2$, and a length strictly under $5.2$ cannot also be $5.2$ or above. No length threads both needles, so a spring inspected against both postings can never pass, regardless of its actual length.

With Spec 2 corrected to $L \ge 5.2$, both conditions are identical:

$L \ge 5.2 \quad\text{and}\quad L \ge 5.2 \quad\Longrightarrow\quad L \ge 5.2 \;\Longleftrightarrow\; [5.2, \infty).$

Intersecting a condition with an exact copy of itself keeps every number the original condition already kept and removes none, so the combined set is exactly $[5.2, \infty)$, the same as either one alone.

Choose a true starting statement, $2 < 9$, and a nonzero multiplier, $-4$. Multiply each side separately:

$2(-4) = -8, \qquad 9(-4) = -36.$

These are the two values the shortcut must be tested against.

Compare the two computed values directly, using the ORIGINAL symbol $<$:

$-8 \;\overset{?}{<}\; -36.$

On the number line $-8$ sits to the right of $-36$, so the true relation is

$-8 > -36,$

not $-8 < -36$. The original direction fails to hold, so the shortcut's claim that the direction 'never changes' is false for this multiplier.

The three-way rule from this lesson splits multipliers into positive and negative: positive keeps the direction, negative reverses it. The shortcut is missing the word 'positive':

$\text{corrected: multiply by any POSITIVE number} \;\Rightarrow\; \text{direction unchanged}.$

The chosen multiplier $-4$ is negative, so it falls outside that corrected rule, which is exactly the case the counterexample was built from.

Divide both sides of (i) by $-3$, a negative number, so the direction reverses:

$\frac{-3x}{-3} < \frac{9}{-3} \;\Rightarrow\; x < -3.$

Divide both sides of (ii) by $2$, a positive number, so the direction stays the same:

$\frac{2x}{2} \ge \frac{14}{2} \;\Rightarrow\; x \ge 7.$

Fence each solved piece by its own symbol: $x < -3$ is strict, so it takes a parenthesis, and $x \ge 7$ is inclusive, so it takes a bracket. Join the two disjoint pieces with a union:

$(-\infty, -3) \cup [7, \infty).$

A legal chain $a < x < b$ reads as one continuous journey: values greater than $a$ AND less than $b$, which requires $a < b$ for any such value to exist. Compare the two bounds here:

$7 \;\overset{?}{<}\; -3,$

which is false, since $7$ is the larger number. Because the bounds are out of order, the chain names an empty set, not the union of two rays that the OR condition actually produces.