QuanticaAlgebra IOpen in the courseAn inequality says one quantity is larger than another rather than equal to it. Nearly every move chapter 4 used on equations, adding the same number to both sides or scaling both sides, carries over unchanged. Exactly one does not, and finding which one, and why it fails, is the work of this lesson.
Problem
Three warehouses each store a whole number of pallets. Warehouse A stores more than warehouse B, and warehouse B stores more than warehouse C. Warehouse C stores \(47\) pallets. Enter the smallest number of pallets warehouse A could store.
Show a hint
- Comparing \(A\) with \(C\) in one step gives a true statement, but a weaker one than the two given comparisons allow. Work out what \(B\) must be first.
- \(B\) is a whole number above \(47\), so the smallest \(B\) can be is \(48\). Now take the same step one more time, from \(B\) to \(A\).
Show the full solution
Since \(B>47\) and \(B\) is a whole number, \(B\) is at least \(48\), and since \(A>B\), \(A\) is at least \(49\). The triple \(49,48,47\) satisfies every condition, so the smallest count is \(\boxed{49}\). Comparing \(A\) with \(C\) directly gives the weaker \(A>47\), which is true but allows \(A=48\), a value the step through \(B\) rules out.
Problem
At the start of a shift the first bin holds \(415\) bolts and the second holds \(392\). A machine then drops \(168\) bolts into each bin. Enter how many more bolts the first bin holds than the second after the drop.
Show a hint
- Ask what happens to the gap between the two bins, not to either count on its own.
- The two counts become \(415+168\) and \(392+168\). Subtract one from the other.
Show the full solution
After the drop the counts are \(415+168=583\) and \(392+168=560\), and \(583-560=\boxed{23}\), the same gap as \(415-392\) before it. Both counts go up by the same \(168\), so the gap is untouched. Adding \(168\) to the gap instead gives \(191\), which would be right only if the bolts all went into the first bin.
Problem
A hiking club logged two trips. The first covered more than \(34\) kilometers and the second covered more than \(57\) kilometers, and neither distance has to be a whole number. Enter the largest number of kilometers that the two trips together are guaranteed to exceed.
Show a hint
- Each trip has its own number to beat. The two comparisons point the same way, so they can be combined into one comparison about the total.
- Write each trip as a comparison, then add the two left sides together and the two right sides together. Test the result on a sample pair such as \(34.2\) and \(57.3\).
Show the full solution
The first distance is more than \(34\) and the second is more than \(57\), and two comparisons pointing the same way may be added, so the total is more than \(34+57=\boxed{91}\). Writing the two distances as \(f\) and \(s\), both \(f-34\) and \(s-57\) are positive, so \((f+s)-91\) is positive as well. Distances just above \(34\) and \(57\) give a total just above \(91\), so no larger number works.
Problem
A swim club beat a tennis club in wins two years running, \(41\) to \(12\) in the first year and \(44\) to \(39\) in the second. Subtracting the first comparison from the second, the way two equations are subtracted, gives the conclusion that the swim club's gain in wins was the larger one. Enter how many more wins the tennis club gained from the first year to the second than the swim club gained.
Show a hint
- Both stated comparisons are true, so if subtracting one from the other were a legal move, its conclusion would have to be true as well. Work out each club's gain on its own.
- The swim club went from \(41\) to \(44\) and the tennis club went from \(12\) to \(39\). Subtract within each club, then compare the two gains.
Show the full solution
The swim club gained \(44-41=3\) wins and the tennis club gained \(39-12=27\), so the tennis club's gain is larger by \(27-3=\boxed{24}\). The subtraction gives \(3>27\), which is false even though both starting comparisons are true, so one inequality may not be subtracted from another.
Adding shifts both sides by the same amount and leaves the difference alone. Multiplying does not, since \(cx-cy=c(x-y)\) scales that difference by \(c\). Scaling a positive number by a positive \(c\) keeps it positive, and scaling by a negative \(c\) makes it negative, so multiplying splits into two cases. The next two problems take one of each.
Problem
A press stamped \(t\) parts on Monday, with \(t>185\). On Tuesday it stamped \(3\) times Monday's number and then \(40\) more. Apply those same two operations to both sides of \(t>185\). The result says Tuesday's count is greater than a certain number. Enter that number.
Show a hint
- Tuesday's count is \(3t+40\). Build it up from \(t\gt 185\) one operation at a time.
- Multiplying both sides by \(3\) is multiplying by a positive number, so the direction holds. Now add \(40\) to both sides.
Show the full solution
Multiplying \(t>185\) by the positive number \(3\) keeps the direction, giving \(3t>555\), and adding \(40\) to both sides keeps it as well, so Tuesday's count satisfies \(3t+40>\boxed{595}\). The direction holds through both steps because \(3\) is positive and adding \(40\) changes both sides by the same amount.
Problem
A student divides both sides of \(-7x>91\) by \(-7\), leaves the inequality symbol pointing the same way, and concludes that the smallest integer \(x\) can be is \(-12\). Enter the largest integer \(x\) can actually be.
Show a hint
- Test the student's answer against the original statement before doing anything else. Put \(x=-12\) into \(-7x\) and compare the result with \(91\).
- Dividing both sides by \(-7\) reverses the direction, so the allowed values sit below \(-13\) rather than above it.
Show the full solution
Dividing \(-7x\gt 91\) by \(-7\) reverses the direction to \(x\lt -13\), so the largest integer available is \(\boxed{-14}\). Checking, \(-7(-14)=98\gt 91\) while \(-7(-13)=91\) is not above \(91\). The student's \(-12\) fails outright, since \(-7(-12)=84\) sits below \(91\).
Choosing between those two cases takes the sign of the multiplier. With a number written out the sign is there to read. With a letter it is not, and there are three cases rather than two, since multiplying by zero gives \(0\) on both sides. The next problem asks what that leaves.
Problem
It is given that \(a>b\), and \(k\) is a real number whose sign is not stated. Enter one word, greater, less, equal or unknown, for how \(ka\) compares with \(kb\).
Show a hint
- Write \(ka-kb=k(a-b)\). Since \(a-b\) is positive, the sign of \(ka-kb\) matches the sign of \(k\).
- Fix \(a=9\) and \(b=2\), then try \(k=4\), \(k=-4\) and \(k=0\) in turn, comparing \(ka\) with \(kb\) each time.
Show the full solution
Here \(ka-kb=k(a-b)\) and \(a-b\) is positive, so the sign of that difference is the sign of \(k\). With \(a=9\) and \(b=2\), \(k=4\) gives \(36\gt 8\), \(k=-4\) gives \(-36\lt -8\), and \(k=0\) gives \(0=0\), so the comparison is \(\boxed{\text{unknown}}\). All three cases are possible, so an inequality cannot be multiplied by a letter whose sign is not known.
Problem
It is given that \(m>n\), with nothing said about the signs of \(m\) and \(n\), and \(t\) is a real number. Statement 1 is \(m-9>n-9\). Statement 2 is \(-2m>-2n\). Statement 3 is \(\frac{m}{5}>\frac{n}{5}\). Statement 4 is \(mt>nt\). Statement 5 is \(m+n>2n\). Statement 6 is \(m-n>0\). Enter how many of the six must be true.
Show a hint
- Each statement is one move applied to both sides of \(m\gt n\). Name the move first, then ask whether that move preserves the direction.
- Statement 5 is \(m\gt n\) with \(n\) added to both sides, and statement 6 is what \(m\gt n\) means written out. Check the other four against what each operation does to the direction.
Show the full solution
Statement 1 subtracts \(9\) from both sides, statement 3 divides by the positive number \(5\), statement 5 adds \(n\) to both sides, and statement 6 restates \(m>n\), so those four hold. Statement 2 multiplies by \(-2\) without reversing the direction and statement 4 multiplies by a letter whose sign is not stated, so the count is \(\boxed{4}\). The signs of \(m\) and \(n\) never matter here. Only the sign of each multiplier does.
A square is never negative, whatever real number is put into it. Squaring means multiplying a number by itself, so the two factors are the same number and have the same sign. A positive times a positive is positive, a negative times a negative is positive, and \(0\times 0\) is \(0\). So \(x^2\) is never negative, whatever real \(x\) is.
Problem
For every real number \(x\), the quantity \(x^2+82-18x\) is equal to \((x-9)^2+1\). How does \(x^2+82\) compare with \(18x\) at every real \(x\)? Enter one word, greater, less, equal or unknown.
Show a hint
- Deciding which is larger is the same as deciding the sign of the difference \(x^2+82-18x\), and the problem already gives that difference in a second form.
- \((x-9)^2\) is the square of a real number, so it is never negative. Add \(1\) and ask how small the whole thing can get.
Show the full solution
The difference \(x^2+82-18x\) equals \((x-9)^2+1\), and the square of a real number is never negative, so that difference is at least \(1\) at every real \(x\). Since the difference is always positive, \(x^2+82\) is \(\boxed{\text{greater}}\). Rearranging the two quantities is not enough on its own. You need \(x^2\ge 0\), applied to \(x-9\).
Problem
Each of these four pairs has \(x\gt y\). Pair 1 is \(x=9\), \(y=4\). Pair 2 is \(x=-2\), \(y=-5\). Pair 3 is \(x=7\), \(y=-1\). Pair 4 is \(x=-4\), \(y=-8\). Enter how many of the four pairs satisfy \(\frac{1}{x}\lt \frac{1}{y}\).
Show a hint
- Work out both reciprocals in each pair and compare them. A negative number has a negative reciprocal, and of two negative numbers the one further from zero is the smaller.
- Watch for the pair where \(x\) is positive and \(y\) is negative. There \(\frac{1}{x}\) is positive while \(\frac{1}{y}\) is negative, and no positive number is less than a negative one.
Show the full solution
Pair 1 gives \(\frac{1}{9}\lt \frac{1}{4}\), pair 2 gives \(-\frac{1}{2}\lt -\frac{1}{5}\) and pair 4 gives \(-\frac{1}{4}\lt -\frac{1}{8}\), while pair 3 gives \(\frac{1}{7}\gt -1\), so the count is \(\boxed{3}\). The three that flip have \(x\) and \(y\) on the same side of zero, and pair 3 does not.
Every rule in this lesson is one question about a sign. Adding or subtracting the same number never changes a direction. Multiplying or dividing changes it exactly when the multiplier is negative, and leaves it undetermined when the multiplier is a letter of unstated sign. The last two problems run several of these moves in one chain.
Problem
Two numbers satisfy \(p>11\) and \(q>-3\), and nothing else is known about either one. Enter the smallest number that \(4-6(p+q)\) must stay below.
Show a hint
- Two inequalities pointing the same way may be added, so pin down what \(p+q\) must exceed before touching anything else.
- That gives \(p+q\gt 8\). Multiplying both sides by \(-6\) is multiplying by a negative, so settle the direction before adding \(4\).
Show the full solution
Adding \(p\gt 11\) and \(q\gt -3\) gives \(p+q\gt 8\), multiplying by \(-6\) reverses the direction to \(-6(p+q)\lt -48\), and adding \(4\) to both sides gives \(4-6(p+q)\lt \boxed{-44}\). Skipping the reversal would give \(4-6(p+q)\gt -44\), which is false at \(p=12\) and \(q=-2\), since those give \(-56\).
Problem
Two numbers satisfy \(15-8s>15-8t\), and their product \(st\) is positive. Neither one is known to be positive on its own. Enter one word, greater, less, equal or unknown, for how \(\frac{1}{t}\) compares with \(\frac{1}{s}\).
Show a hint
- Two moves. First turn \(15-8s\gt 15-8t\) into a comparison between \(s\) and \(t\) alone, watching the step that reverses the direction.
- That gives \(s\lt t\). The reciprocal rule needs the two numbers to share a sign, and \(st\gt 0\) says exactly that even though neither sign is known on its own.
Show the full solution
Subtracting \(15\) from both sides gives \(-8s\gt -8t\), and dividing by \(-8\) reverses the direction to \(s\lt t\). Since \(st\gt 0\) the two share a sign, so the reciprocal rule applies and \(\frac{1}{t}\) compared with \(\frac{1}{s}\) is \(\boxed{\text{less}}\). Both sign cases check out, \(s=1,t=2\) giving \(\frac{1}{2}\lt 1\) and \(s=-2,t=-1\) giving \(-1\lt -\frac{1}{2}\). Drop \(st\gt 0\) and the answer is unknown, since \(s=-1,t=2\) gives \(\frac{1}{t}\gt \frac{1}{s}\).
Every rule here says what one move does to a comparison already in hand. Lesson 8.2, Which Is Bigger, uses them in the other direction. It starts with two specific quantities too awkward to compute outright and looks for the move that settles which one is larger.
Practice these ideas
Practice
\(w\) is an integer and \(w-18>45\). Enter the smallest value \(w\) can be.
Show the solution
Adding \(18\) to both sides leaves the direction unchanged, giving \(w>63\), so the smallest integer available is \(\boxed{64}\). Checking, \(64-18=46>45\) while \(63-18=45\) is not above \(45\).
Practice
Integers \(a\), \(b\), and \(c\) satisfy \(a>b\) and \(b>c\), with \(c=61\). Enter the smallest possible value of \(a\).
Show the solution
Since \(b\gt 61\) and \(b\) is an integer, \(b\) is at least \(62\), and \(a\gt b\) makes \(a\) at least \(63\). The triple \(63,62,61\) meets every condition, so the answer is \(\boxed{63}\). Chaining to \(a\gt 61\) first is true but weaker, since \(a=62\) would need an integer \(b\) with \(61\lt b\lt 62\), and there is none.
Practice
A number \(g\) satisfies \(g>16\). Enter the largest number that \(7g\) must exceed.
Show the solution
Multiplying \(g>16\) by the positive number \(7\) leaves the direction unchanged, so \(7g>\boxed{112}\). Nothing larger works, since \(g\) can sit just above \(16\), which puts \(7g\) just above \(112\).
Practice
Real numbers \(m\) and \(n\) satisfy \(m>30\) and \(n>17\). Enter the largest number that \(m+n\) must exceed.
Show the solution
Two inequalities pointing the same way may be added, so \(m+n>30+17=\boxed{47}\). Nothing larger works, since taking \(m\) and \(n\) just above \(30\) and \(17\) makes \(m+n\) as close to \(47\) as we like.
Practice
\(q\) is an integer and \(\frac{q}{8}>13\). Enter the smallest value \(q\) can be.
Show the solution
Multiplying both sides by the positive number \(8\) keeps the direction, giving \(q>104\), so the smallest integer is \(\boxed{105}\). Checking, \(\frac{105}{8}=13.125\) is above \(13\) while \(\frac{104}{8}=13\) is not.
Practice
A number \(k\) satisfies \(k>3\), and nothing else is known about it. Enter the smallest number that \(-8k\) must stay below.
Show the solution
Multiplying \(k\gt 3\) by \(-8\) reverses the direction, so \(-8k\lt \boxed{-24}\). Keeping \(\gt \) would give the false claim \(-8k\gt -24\), since \(k=4\) gives \(-32\), and \(-32\) is not above \(-24\).
Practice
\(n\) is an integer and \(-5n>37\). Enter the largest value \(n\) can be.
Show the solution
Dividing \(-5n\gt 37\) by \(-5\) reverses the direction to \(n\lt -7.4\), so the largest integer is \(\boxed{-8}\). Checking, \(-5(-8)=40\gt 37\) while \(-5(-7)=35\) is not above \(37\).
Practice
Both \(20>3\) and \(b>1\) are true, and \(b\) is an integer. Subtracting the second comparison from the first would claim \(20-b>3-1\). Enter the smallest integer \(b\) that makes that claim false.
Show the solution
The claim reads \(20-b\gt 2\), which holds only while \(b\lt 18\). At \(b=18\) it reads \(2\gt 2\), which is false, so the answer is \(\boxed{18}\). Both starting comparisons still hold at \(b=18\), so subtracting one comparison from another is not a legal move.
Practice
It is given that \(a>b\), and \(t=0\). Enter one word, greater, less, equal or unknown, for how \(ta\) compares with \(tb\).
Show the solution
With \(t=0\), both \(ta\) and \(tb\) are \(0\), so the two are \(\boxed{\text{equal}}\). Zero is neither positive nor negative, so neither multiplication rule applies, and the given \(a>b\) makes no difference to the comparison.
Practice
It is given that \(u\gt v\). Statement 1 is \(u+3\gt v+3\). Statement 2 is \(4u\gt 4v\). Statement 3 is \(\frac{u}{-2}\gt \frac{v}{-2}\). Statement 4 is \(u-v\gt 0\). Statement 5 is \(-u\lt -v\). Statement 6 is \(u+v\gt 2v\). Enter how many of the six must be true.
Show the solution
Dividing both sides by \(-2\) reverses the direction, so statement 3 is the only one that fails and the count is \(\boxed{5}\). Statements 1, 4 and 6 come from adding or subtracting the same quantity on both sides, statement 2 from multiplying by the positive \(4\), and statement 5 from multiplying by \(-1\) with the direction reversed.
Practice
It is given that \(x>y\). For how many of \(k=-3\), \(k=0\) and \(k=5\) is \(kx>ky\) guaranteed? Enter that count.
Show the solution
The sign of \(kx-ky=k(x-y)\) is the sign of \(k\), since \(x-y\) is positive. So \(k=5\) gives \(kx\gt ky\), \(k=-3\) gives \(kx\lt ky\), and \(k=0\) gives \(kx=ky\), leaving \(\boxed{1}\). Three values of \(k\) and three different outcomes, so you may only multiply both sides by a letter when you know its sign.
Practice
A real number \(c\) satisfies \(c^2\le 0\). Enter the value of \((c+5)(c+8)\).
Show the solution
Since \(c^2\ge 0\) for every real \(c\), the condition \(c^2\le 0\) can hold only when \(c^2=0\), so \(c=0\) and \((0+5)(0+8)=\boxed{40}\). The trivial inequality pins \(c\) to a single value, so nothing else about \(c\) has to be found.
Practice
For every real number \(x\), the quantity \(x^2+121-22x\) is equal to \((x-11)^2\). Enter the value of \(x^2+121\) at the one real number \(x\) where it equals \(22x\).
Show the solution
\(x^2+121\) equals \(22x\) exactly when \(x^2+121-22x=0\), which is \((x-11)^2=0\), so \(x=11\). Then \(x^2+121=121+121=\boxed{242}\), and \(22(11)=242\) agrees. A square is never negative, so \((x-11)^2\) is \(0\) at one value of \(x\) and positive everywhere else, which is why \(x^2+121\) is never below \(22x\).
Practice
The numbers \(x=15\) and \(y=6\) are both positive, and \(x>y\). Enter the value of \(\frac{1}{y}-\frac{1}{x}\) as a fraction in lowest terms.
Show the solution
With a common denominator of \(30\), \(\frac{1}{6}-\frac{1}{15}=\frac{5}{30}-\frac{2}{30}=\frac{3}{30}=\boxed{\frac{1}{10}}\). The reciprocal rule says that when \(x\gt y\) and both are positive, \(\frac{1}{x}\lt \frac{1}{y}\), so this difference had to come out positive.
Practice
The numbers \(x=10\) and \(y=-4\) satisfy \(x\gt y\), so the reciprocal rule would claim \(\frac{1}{x}\lt \frac{1}{y}\). Enter the value of \(\frac{1}{x}-\frac{1}{y}\) as a fraction in lowest terms.
Show the solution
The reciprocals are \(\frac{1}{10}\) and \(-\frac{1}{4}\), so \(\frac{1}{10}-\left(-\frac{1}{4}\right)=\frac{2}{20}+\frac{5}{20}=\boxed{\frac{7}{20}}\). The difference is positive, which says \(\frac{1}{x}>\frac{1}{y}\), the opposite of the claim. These two numbers sit on opposite sides of zero, and the rule needs them on the same side.
Practice
A number \(z\) satisfies \(z\lt 6\), and \(y=9-4z\). Enter the largest number that \(y\) must exceed.
Show the solution
Multiplying \(z\lt 6\) by \(-4\) reverses the direction to \(-4z\gt -24\), and adding \(9\) to both sides gives \(9-4z\gt \boxed{-15}\). Skipping the reversal gives \(y\lt -15\) instead, which is false at \(z=0\), where \(y=9\).
Practice
An integer \(m\) satisfies \(\frac{1}{m}\lt \frac{1}{7}\) and \(m\gt 0\). Enter the smallest value \(m\) can be.
Show the solution
Both numbers are positive, so taking reciprocals reverses the comparison and \(\frac{1}{m}\lt \frac{1}{7}\) becomes \(m\gt 7\), making the smallest integer \(\boxed{8}\). Checking, \(\frac{1}{8}\lt \frac{1}{7}\). The condition \(m\gt 0\) matters, since \(m=-3\) also satisfies \(\frac{1}{m}\lt \frac{1}{7}\) but is not greater than \(7\).