Inequalities: Free Response

A number-line graph carries exactly two pieces of information, and each is read off a different part of the picture.

The shading says which side of the boundary the solutions lie on. Graph 1 is shaded toward the smaller numbers, so its solutions are the numbers below the boundary and its symbol is a less-than one. Graph 2 is shaded toward the larger numbers, so its symbol is a greater-than one.

The circle drawn on the boundary says whether that one number is a solution. A hollow circle shuts it out, which is what the strict symbols do, and a filled circle lets it in, which is what the symbols with a bar underneath do. Reading both features together:

$\text{Graph 1: } x < -1, \qquad \text{Graph 2: } x \ge -1.$

The two readings are independent, and both are needed. The shading alone would not tell you which of $<$ and $\le$ to write, and the circle alone would not tell you which way the solutions run.

Two of the three values are settled by the shading alone, because they are nowhere near the boundary. The value $-4$ sits to the left of the boundary, inside Graph 1's shaded ray and outside Graph 2's, and $2.5$ sits to the right, inside Graph 2's ray and outside Graph 1's:

$-4 < -1 \;\checkmark \qquad 2.5 \ge -1 \;\checkmark$

The boundary value $-1$ is the interesting one. Shading cannot settle it, since the shading of each graph begins there rather than covering it, so the circle is what decides. Graph 1's circle is hollow, so $-1$ is refused there, and $-1 < -1$ is indeed false. Graph 2's circle is filled, so $-1$ is accepted there, and $-1 \ge -1$ is true because the bar under the symbol allows equality.

So each of the three values belongs to exactly one graph, but the boundary got there by a different route from the other two.

Split the line into the boundary and everything else, because those two cases are settled by different features of the pictures.

Every number other than the boundary lies either to its left or to its right, and the two graphs have taken one side each. So such a number falls inside exactly one of the two shaded rays: never inside both, and never outside both. That disposes of the whole line except for a single point.

At the boundary itself the circles decide, and the two circles disagree. The hollow one refuses the boundary and the filled one accepts it:

$-1 < -1 \;\text{is false}, \qquad -1 \ge -1 \;\text{is true}.$

So the boundary also belongs to exactly one graph. Both verdicts follow: nothing satisfies both inequalities and nothing satisfies neither. Notice how close a near miss would be. Had both circles been hollow, the boundary would have satisfied neither, and had both been filled it would have satisfied both, with the shading unchanged in either case.

Both inequalities begin with the same move. Adding $7$ to both sides clears the constant, and an addition move never disturbs the symbol, so the two arrive at the same right-hand side:

$6x \ge 18 \qquad \text{and} \qquad -6x \ge 18.$

They part company at the division. Dividing the first by the positive coefficient $6$ leaves the symbol where it stands. Dividing the second by the negative coefficient $-6$ reverses it, so the $\ge$ becomes $\le$:

$x \ge 3 \qquad \text{and} \qquad x \le -3.$

The first solution is the numbers from $3$ upward and the second is $-3$ and everything below it. Now read each graph off its own symbol. The circle comes from the strictness: both symbols carry a bar underneath, so both boundaries take a filled circle, and a reversal cannot change that, since $\ge$ and $\le$ both admit their boundary. The shading comes from the relation: the first solution runs toward the larger numbers, so it is shaded right from $3$, and the second runs toward the smaller ones, so it is shaded left from $-3$.

A solved inequality makes two claims at once: everything inside the range works, and everything outside it fails. So a check needs one value of each kind, and both must go back into the inequality as it was first written, since a later line may already be carrying the mistake you are hunting for.

Take $x = -5$, which lies inside the range $x \le -3$:

$-6(-5) - 7 = 30 - 7 = 23, \qquad 23 \ge 11 \;\checkmark$

Now take $x = 0$, which lies outside it:

$-6(0) - 7 = -7, \qquad -7 \ge 11 \;\text{is false}$

The inside value passes and the outside value fails, which is the pattern a correct range produces.

That second test is also what would have exposed a missed reversal. Leaving the symbol alone at the division would have given the numbers from $-3$ upward, a range that contains $0$; the test just run shows $0$ is not a solution of the original, so that range cannot be right. One value is enough to bring a wrong range down, which is why the outside test is worth running every time.

Follow the two solutions side by side and ask where they could possibly differ. The constant is $-7$ in both, so both are cleared by adding $7$, and both reach the same right-hand side:

$6x \ge 18 \qquad \text{and} \qquad -6x \ge 18.$

The lone change of sign is therefore doing all of its work at one step, the division. Two separate things happen there, and they are worth pulling apart.

The first is arithmetic. The same $18$ is divided by $6$ in one problem and by $-6$ in the other, and a negative divisor changes the sign of the quotient, so the two boundaries come out as opposites:

$\frac{18}{6} = 3 \qquad \text{and} \qquad \frac{18}{-6} = -3.$

The second is about order. Dividing both sides by a negative reflects them across zero, which swaps which of the two is on the left, so the symbol must reverse and the shading swings to the other side of the boundary.

What the sign does not touch is worth naming too. The constant step is identical in both, and so is the strictness: $\ge$ reversed is $\le$, and both of those admit their boundary, so a closed circle stays closed.

Build the weight of one trip before reaching for a symbol. Each sack adds $30$ pounds and there are $s$ of them, which is $30s$ pounds of feed, and the operator adds a further $155$ pounds no matter how much feed is loaded. A trip is allowed when that total stays within the rating:

$30s + 155 \le 620.$

The wording is what fixes the symbol, and it turns on one question: is a trip of exactly $620$ pounds allowed? At most $620$ pounds means no more than $620$, so a trip of exactly that weight is fine and the boundary belongs to the allowed loads. That is the symbol with a bar underneath, $\le$.

Had the lift been rated to carry less than $620$ pounds, a trip of exactly $620$ pounds would be refused, the boundary would be excluded, and the model would need the strict $<$. The two wordings differ over one weight only, but that one weight is exactly what an open circle and a closed circle disagree about.

It is worth being precise about what that difference does and does not decide here. Sacks come whole, so the total weight jumps in steps of $30$ pounds up from $155$, and no such total lands on $620$ exactly: a trip weighing the rated amount would need $15.5$ sacks. So on this lift the two wordings would in the end admit the same loads. The difference between the symbols is still real, because it is a difference between the two models, but boundary inclusion only changes an answer when the boundary is a value something in the situation can actually take.

Clear the constant first, as with any two-step problem. Subtracting $155$ from both sides is an addition move, and an addition move can never reverse a symbol, whatever the signs of the numbers involved, so it comes straight down:

$30s + 155 - 155 \le 620 - 155 \;\Rightarrow\; 30s \le 465.$

Now divide both sides by the coefficient $30$. This step is of the kind that can reverse a symbol, since dividing both sides by a negative always does, but it does not here, because the divisor $30$ is positive:

$\frac{30s}{30} \le \frac{465}{30} \;\Rightarrow\; s \le 15.5.$

So the two steps are not on the same footing even though neither reversed anything. The first could not have reversed the symbol under any circumstances; the second could have, and was checked before being carried out. That is the habit worth keeping, because it is the only step in a two-step solve where the question ever arises.

The answer is a range, not a single number: every value at or below $15.5$ sacks keeps the trip within the rating. Check it in the original inequality with one value from inside the range and one from outside. A load of $15$ sacks weighs $30(15) + 155 = 605$ pounds and $605 \le 620$ is true, while a load of $16$ sacks weighs $30(16) + 155 = 635$ pounds and $635 \le 620$ is false.

The algebra hands back a whole range of numbers, but the question is about sacks, and sacks come whole. So the last step is to ask which whole numbers lie inside the range.

The range runs up to $15.5$, so $15$ lies inside it and $16$ does not. Checking both against the rating confirms it:

$30(15) + 155 = 605 \le 620, \qquad 30(16) + 155 = 635 > 620.$

So the operator can take $15$ sacks. The edge of the range, $15.5$ sacks, is a perfectly good number and an impossible load, because half a sack is not something the lift can be asked to carry, and the boundary of a range need not be a member of anything the situation allows.

This is the ordinary shape of an applied inequality. The mathematics produces a range, the situation then decides which members of that range are available, and the answer to the question asked is the largest available one rather than the edge of the range itself. Notice too that rounding is not what settles it: the reason to go down rather than up is that $16$ sacks breaks the rating, which the check above shows directly.

Take the lines in order. The question asks for the first wrong one, and everything after a fault inherits it, so the earlier steps have to be cleared before a line can be blamed.

Line 2 comes from line 1 by subtracting $8$ from both sides. That is an addition move, and addition moves never touch the symbol, so line 2 is correct:

$-3x + 8 - 8 \ge 26 - 8 \;\Rightarrow\; -3x \ge 18.$

Line 3 comes from line 2 by dividing both sides by $-3$. The divisor is negative, so the symbol has to reverse from $\ge$ to $\le$, and the student brought it straight down instead:

$\frac{-3x}{-3} \le \frac{18}{-3} \;\Rightarrow\; x \le -6.$

So line 3 is the first wrong line. Line 4 is drawn faithfully from line 3, so it is wrong as well, but only as a consequence: the shading should run left instead of right. The circle, though, is right for the wrong reason, since a reversal turns $\ge$ into $\le$ and both of those admit the boundary, so the circle stays filled either way.

A range is a claim about every number in it, so one number inside it that fails the original inequality brings the whole claim down. The student's range accepts $-6$ and everything above it, while the corrected range accepts $-6$ and everything below, so the two agree on the boundary and disagree at every other number. Any value above $-6$ will do, and $x = 0$ is the easiest to substitute.

Put $x = 0$ into line 1, the inequality as it was first written:

$-3(0) + 8 = 8, \qquad 8 \ge 26 \;\text{is false}$

So $0$ is not a solution, yet the student's range contains it. That is decisive: their range claims something about $0$ that the original inequality denies.

For the other half of the check, take $x = -7$, which the corrected range contains and theirs does not:

$-3(-7) + 8 = 21 + 8 = 29, \qquad 29 \ge 26 \;\checkmark$

An inside value that passes together with an outside value that fails is exactly the pattern a correct range produces, and it is the pattern the student's range fails to produce.

Two steps can each have a minus sign in view and still do quite different things to an order, so ask what each one does to the two sides as positions on the number line.

Subtracting the same constant from both sides moves both of them the same distance in the same direction. Two points slid equally keep their left-to-right order, so the symbol is untouched. It makes no difference what signs are on show, whether in the terms being carried along or in the numbers the step produces: what is being done to the two sides is a slide.

Dividing both sides by a negative, here by $-3$, is not a slide. Dividing by $3$ shrinks both distances from zero and keeps the order, and the remaining factor of $-1$ sends each side to its opposite, which reflects it across zero. A reflection swaps left and right, so the side that was the smaller comes back the larger:

$2 < 5 \quad \text{becomes} \quad -2 > -5.$

That gives the rule in its exact form. Reverse the symbol when, and only when, you multiply or divide both sides by a negative number. Adding or subtracting never reverses it, whatever signs are on show, and a negative sitting somewhere else in the problem is no reason to reverse anything.

That is exactly what separates the two steps in the work above. Clearing the constant was a slide, so it could not have disturbed the symbol however the terms were signed, while removing the coefficient divided both sides by a negative and so had to turn it around.

Subtract $20$ from each side first. The left side $12$ becomes $-8$ and the right side $-6$ becomes $-26$. On the number line $-8$ sits eight ticks left of zero and $-26$ sits twenty-six ticks left, so $-8$ is still the one farther right and the statement holds as it stands:

$12 - 20 = -8, \qquad -6 - 20 = -26, \qquad -8 > -26.$

Now start again and multiply each side by $-4$. The left side becomes $-48$ and the right side becomes $24$. A negative number cannot be greater than a positive one, so keeping the symbol would make a false claim, and the true statement is the turned-around one:

$12(-4) = -48, \qquad -6(-4) = 24, \qquad -48 < 24.$

The first move is worth a second look. It produced two negative numbers and needed no reversal at all, so a minus sign showing up in the result is not by itself any reason to turn a symbol around.

One step is enough to retire a shortcut, so build one where a negative number is unmistakably present and yet nothing turns around.

Begin from $7 > 1$, which is true, and add $-9$ to both sides. The operation itself contains a negative number, so the shortcut orders the symbol to be turned around, which would produce the claim $-2 < -8$. Work the two sides out instead:

$7 + (-9) = -2, \qquad 1 + (-9) = -8, \qquad -2 > -8.$

On the number line $-2$ lies two ticks left of zero and $-8$ lies eight ticks left, so $-2$ is the one farther right and therefore the greater. The unreversed statement is true and the reversed one is false, so on this step the shortcut gives exactly the wrong instruction.

It is not merely unlucky here. Adding a number, of whatever sign, slides both sides the same distance in the same direction, and a slide cannot change which of two points is on the left. What the shortcut is missing is that the rule looks at the operation performed on both sides, not at the signs of the numbers on view.

Split the multiplication into two moves whose effects can be seen separately. Multiplying by a negative number, say by $-k$ where $k$ is positive, is the same as multiplying by $k$ and then taking the opposite of each side.

The first move scales both sides by the same positive amount. Every distance from zero is stretched or shrunk by that same factor and nothing crosses zero, so the left-to-right order of the two sides survives it untouched.

The second move sends each side to its opposite, and taking opposites reflects the whole line across zero: each point keeps its distance from zero and changes sides. A reflection turns the leftward direction into the rightward direction everywhere at once, so a pair standing in the order left then right comes back in the order right then left:

$7 < 10 \quad \text{becomes} \quad -7 > -10.$

Those two numbers are only an illustration, since the argument named no particular pair. A move that preserves order followed by a move that reverses it reverses the order overall, whatever two numbers were chosen, and that is why the symbol must turn around every time.

Division needs no separate argument. Dividing by $-k$ is multiplying by the positive $\frac{1}{k}$ and then taking opposites, so it is the same two moves in the same order, and only the last of them turns anything around.