QuanticaPrealgebraOpen in the courseEvery equation so far ended on one exact value, the single number that makes both sides match. Real life is rarely that tidy. A recipe wants the oven at least \(350\) degrees, so \(351\) and \(400\) work too, and a bridge bars trucks above a posted height, so countless heights pass. Each of those answers is a whole stretch of the number line, called an inequality. The balance moves from 6.1 through 6.4 all still work here, and there is exactly one new rule to add, which is that multiplying or dividing by a negative flips the sign.
Problem
A solo ride only lets you board if your ticket count is strictly more than \(5\), written \(t > 5\). The word strictly matters, because holding exactly \(5\) is not enough, you have to clear the line, not land on it. Five friends hold \(3\), \(5\), \(7\), \(12\), and \(5.5\) tickets. How many of those counts satisfy \(t > 5\)? Type the single number.
Show a hint
- The rule is \(t > 5\). Go through the friends one at a time and ask whether each ticket count is bigger than \(5\). Watch the friend holding exactly \(5\), since \(> 5\) does not include \(5\) itself.
- Mark each value yes or no for \(t > 5\). \(3\) is no, \(5\) is no because it only ties the line, \(7\) is yes, \(12\) is yes, and \(5.5\) is yes. Count the yeses.
Show the full solution
Check each count against \(t > 5\). \(3\) is no, \(5\) is no because it only ties the line, \(7\) is yes, \(12\) is yes, and \(5.5\) is yes. That is \(\boxed{3}\) friends. Strict means the boundary is out, so holding exactly \(5\) tickets fails even though it reaches the number.
Problem
A delivery cart already holds \(8\) crates, and the running total must stay under \(20\), so with \(x\) more crates the rule is \(x + 8 < 20\). Solve it the same way as \(x + 8 = 20\), by subtracting \(8\) from both sides, with no sign flip for subtraction. The boundary is strict, so a value sitting exactly on it does not count. What is the largest whole number \(x\) can be?
Show a hint
- Make the first move you would make for an equation. Subtract \(8\) from both sides to get \(x\) alone. Adding or subtracting the same amount never flips an inequality, so the \(<\) sign stays pointing the same way. Then notice the bound is strict, which matters for a whole number answer.
- Subtracting \(8\) from both sides gives \(x < 12\). Since \(x\) must be strictly less than \(12\), the value \(12\) itself is out. The largest whole number that still fits is the integer just below \(12\).
Show the full solution
Subtract \(8\) from both sides of \(x + 8 < 20\) to get \(x < 12\). Subtracting never flips the sign. The bound is strict, so \(12\) is out and the largest whole number that fits is \(\boxed{11}\). Check it, \(11 + 8 = 19\) which is under \(20\), while \(12 + 8 = 20\) is not.
Problem
A diver's gauge reads negatively below the surface. Her reading is \(x\), she rises \(5\) meters to \(x + 5\), and the safety check needs the new reading at least \(2\), so \(x + 5 \ge 2\). Subtract \(5\) from both sides, and subtracting never flips the sign. Because the sign is \(\ge\), the boundary itself is allowed. What is the smallest value \(x\) can be?
Show a hint
- Treat the inequality just like an equation for the first move. To get \(x\) by itself, undo the \(+5\) by subtracting 5 from both sides. You only flip the inequality sign when you multiply or divide both sides by a negative, and subtracting does not do that, so the \(\ge\) stays pointing the same way.
- Subtract 5 from both sides of \(x + 5 \ge 2\) and you get \(x \ge -3\). The sign is \(\ge\), which means \(-3\) itself is allowed, so the gate is closed. The smallest reading that still works is the boundary number \(-3\).
Show the full solution
Subtract \(5\) from both sides of \(x + 5 \ge 2\) to get \(x \ge -3\). Subtracting does not flip the sign. Because the sign is \(\ge\), the boundary is allowed, so the smallest reading she can start from is \(\boxed{-3}\). The sign only reverses when you multiply or divide both sides by a negative, which never happens here.
Problem
Four identical sandbags sit on a shelf rated for at most \(26\) kilograms total, so with one bag weighing \(x\), the rule is \(4x \le 26\). Solve it like an equation, keeping the inequality sign. Divide both sides by the positive number \(4\), which does not flip the sign. The word most means the boundary counts. What is the largest value \(x\) can be? Give a fraction in lowest terms.
Show a hint
- The bags share the load, so undo the \(4\) by dividing both sides by \(4\). You are dividing by a positive number here, so watch what happens to the inequality sign. When you divide by a positive, the sign keeps pointing the same way, it does not flip.
- Divide both sides of \(4x \le 26\) by \(4\). The sign stays as \(\le\), giving \(x \le \tfrac{26}{4}\). Simplify \(\tfrac{26}{4}\) to lowest terms, and since the bound is "at most" the boundary itself is allowed, so the largest value is that boundary.
Show the full solution
Divide both sides of \(4x \le 26\) by \(4\), a positive number, so the sign stays put. That gives \(x \le \tfrac{26}{4}\), which reduces to \(\tfrac{13}{2}\). At most means the boundary counts, so the heaviest one bag can be is \(\boxed{13/2}\) kilograms. If the shelf were rated for under \(26\) kilograms instead, \(\tfrac{13}{2}\) would be shut out and there would be no largest allowed weight.
Problem
Priya pays a flat \(\$45\) booth fee and clears \(\$5\) on each eraser she sells, so selling \(n\) erasers gives a profit of \(5n - 45\) dollars. She wants that profit to be at least \(\$200\), so the goal is \(5n - 45 \ge 200\). Solve it, dividing by the positive \(5\) without flipping the sign. What is the fewest whole erasers she must sell to reach her goal?
Show a hint
- Each eraser adds \(5\) to her profit, but the \(45\) booth fee gets subtracted once. Start by undoing that subtraction. Add \(45\) to both sides so the \(n\) term sits alone, and notice you are only adding and dividing by positive numbers here, so the \(\ge\) sign stays pointing the same way the whole time.
- After adding \(45\) you get \(5n \ge 245\). Divide both sides by \(5\), a positive number, so no flip, and you land on \(n \ge 49\). Since she cannot sell part of an eraser and \(49\) is already a whole number that is allowed, the fewest is \(49\) itself.
Show the full solution
Add \(45\) to both sides of \(5n - 45 \ge 200\) to get \(5n \ge 245\), then divide by \(5\), a positive, so no flip. That gives \(n \ge 49\), and \(49\) is a whole number that is allowed, so she needs \(\boxed{49}\) erasers. At \(49\) the profit is \(245 - 45 = 200\) dollars, which counts because at least includes hitting the goal exactly.
Problem
Start from a statement nobody would argue with, \(3 < 8\). Now multiply both sides by \(-1\), so the \(3\) becomes \(-3\) and the \(8\) becomes \(-8\). On a number line, the one farther left is the smaller, so \(-8\) is now the smaller value. Of the two results \(-3\) and \(-8\), which one is the larger? Type that value.
Show a hint
- Put \(-3\) and \(-8\) on a number line and remember the rule from earlier chapters. Farther to the left means smaller, farther to the right means larger. Watch what happened to the order. The statement started as \(3 < 8\), with the left number smaller, but after the move the smaller one is no longer on the left.
- Mark both numbers. \(-8\) is eight steps left of zero, and \(-3\) is only three steps left of zero, so \(-3\) sits to the RIGHT of \(-8\). The number on the right is the larger one. That is \(-3\).
Show the full solution
Multiplying \(3 < 8\) by \(-1\) turns the sides into \(-3\) and \(-8\). On the number line \(-8\) sits farther left, so it is the smaller one, and the larger of the two is \(\boxed{-3}\). Copying the old direction would give \(-3 < -8\), which is false. That is why multiplying or dividing both sides by a negative flips the sign.
Problem
A freezer alarm trips whenever \(-5x\) climbs above \(17\), so the danger zone is \(-5x > 17\). To free \(x\), divide both sides by \(-5\), and dividing by a negative flips the \(>\) to \(<\). What is the largest integer \(x\) can be?
Show a hint
- Get \(x\) alone by dividing both sides by \(-5\). The moment you divide by a negative number, something has to happen to the inequality sign. Decide what it becomes before you read off the answer.
- Dividing by \(-5\) flips \(>\) into \(<\), so \(x < -\frac{17}{5}\), which is \(x < -3.4\). The gate at \(-3.4\) is open, so \(-3.4\) itself is not allowed. Now name the largest integer that sits strictly to the left of \(-3.4\) on the number line.
Show the full solution
Divide both sides of \(-5x > 17\) by \(-5\). The divisor is negative, so the sign flips and \(x < -\frac{17}{5} = -3.4\). The bound is strict, and the largest integer sitting left of \(-3.4\) is \(\boxed{-4}\). Check it, \(-5 \cdot (-4) = 20 > 17\) is true, while \(-5 \cdot (-3) = 15 > 17\) is false.
Problem
This one pulls everything together. The variable appears on both sides and there is a group to clear. Solve \(3(2 - x) > 2x + 1\) by distributing, then gathering the variable, and stay alert because one step divides by a negative and forces the sign to flip. Once you have the range, what is the largest integer \(x\) can be?
Show a hint
- Start exactly the way you would for an equation. Distribute the \(3\) across the parentheses to get \(6 - 3x\) on the left, then move the variable terms to one side and the numbers to the other. Keep one eye out for a moment where you divide by a negative, since that is the only thing here that behaves differently from an equation.
- After distributing you have \(6 - 3x > 2x + 1\). Subtract \(2x\) from both sides to reach \(6 - 5x > 1\), then subtract \(6\) to get \(-5x > -5\). Now divide both sides by \(-5\), and because that divisor is negative, FLIP the sign. That gives \(x < 1\). The bound is strict, so \(1\) itself is not allowed. Name the largest whole number that still sits below \(1\).
Show the full solution
Distribute the \(3\), so \(3(2 - x) > 2x + 1\) becomes \(6 - 3x > 2x + 1\). Subtract \(2x\) to get \(6 - 5x > 1\), then subtract \(6\) to get \(-5x > -5\). Dividing both sides by \(-5\) flips the sign, leaving \(x < 1\). The bound is strict, so the largest integer that fits is \(\boxed{0}\). At \(x = 1\) both sides equal \(3\), and \(3 > 3\) is false, which is exactly why \(1\) is shut out.
Problem
Three runners are scored so a bigger number means faster. The sheet gives two facts. Priya beat Quinn, \(P > Q\), and Quinn was at least as fast as Rafe, \(Q \ge R\). A teammate scribbles five claims, \(P > R\), \(P \ge R\), \(P > Q\), \(Q > R\), and \(R > P\). Using only the two facts and the way inequalities chain when they point the same way, how many of the five claims are forced to be true? Type the single count.
Show a hint
- Line the two facts up so the arrows point the same way, \(P > Q \ge R\). Anything you can read straight off this chain by following greater-than signs from left to right is forced. Watch the spot where \(Q\) meets \(R\), because that link is \(\ge\) and not a strict \(>\).
- Forced because they read straight off the chain, \(P > Q\) is given outright, \(P > R\) jumps across the whole chain, and \(P \ge R\) is a weaker true statement that \(P > R\) already guarantees. Not forced, \(Q > R\) fails whenever Quinn and Rafe tie since that link is only \(\ge\), and \(R > P\) is actually backwards. Count the forced ones.
Show the full solution
Stack the two facts into one chain, \(P > Q \ge R\). Reading across it, \(P > Q\) is given outright, \(P > R\) follows because the trip across the chain crosses a strict \(>\), and \(P \ge R\) follows since it is a weaker claim than \(P > R\). The other two are not forced. \(Q > R\) fails if Quinn and Rafe tie, since that link is only \(\ge\), and \(R > P\) points backwards. That is \(\boxed{3}\) forced claims. One strict link anywhere makes the whole chain strict, but a chain of all \(\ge\) leaves ties possible.
Problem
A compound inequality traps the variable in the middle. Solve \(1 < 5 - 2x \le 11\). Whatever you do, do it to all three parts at once so the middle stays fenced in on both sides, and remember that dividing by the negative coefficient flips both signs together. Once you have \(x\) cornered, how many integers make the whole thing true? Type the single count.
Show a hint
- Peel away the constant from all three parts first. Subtract \(5\) from the left, the middle, and the right, and you should land on \(-4 < -2x \le 6\). Now you have to divide every part by \(-2\), and that negative is about to do something to the signs, so stay alert.
- Dividing all three parts by \(-2\) flips BOTH inequality signs, which gives \(2 > x \ge -3\). Read that the tidy way around and it says \(-3 \le x < 2\), closed at \(-3\) and open at \(2\). Now just list the integers from \(-3\) up to but not including \(2\) and count them.
Show the full solution
Subtract \(5\) from all three parts of \(1 < 5 - 2x \le 11\) to get \(-4 < -2x \le 6\). Divide every part by \(-2\), and since \(-2\) is negative both signs flip, giving \(2 > x \ge -3\). Written small to large that is \(-3 \le x < 2\), so the integers are \(-3, -2, -1, 0, 1\), which is \(\boxed{5}\) of them. Dividing by a negative also swaps which end is the floor and which is the ceiling, so rewrite the range in increasing order before you count.
Problem
The trickiest one, folding everything together. Solve \(5 - 2x - 8 > 4 - x\). The left side has two number pieces sitting apart, so tidy it first, then gather the variables. One move near the end forces the sign to turn around. Once you have the range, what is the largest integer \(x\) can be?
Show a hint
- Start by combining the two number terms on the left, since \(5 - 8\) is a single value. That turns the left side into \(-3 - 2x\). Then gather the variables on one side the same way you did when solving equations. Keep an eye out for the step that might flip the inequality.
- After tidying you have \(-3 - 2x > 4 - x\). Add \(x\) to both sides to get \(-3 - x > 4\), then add \(3\) to both sides to get \(-x > 7\). Now multiply both sides by \(-1\), and because the multiplier is negative you must reverse the sign, giving \(x < -7\). The bound is strict, so \(-7\) itself is not allowed. The largest integer that fits is the whole number just below \(-7\).
Show the full solution
Combine the numbers on the left, \(5 - 8 = -3\), so the inequality reads \(-3 - 2x > 4 - x\). Add \(x\) to both sides to get \(-3 - x > 4\), then add \(3\) to get \(-x > 7\). Multiplying both sides by \(-1\) flips the sign, leaving \(x < -7\). The bound is strict, so the largest integer is \(\boxed{-8}\). A bare \(-x\) carries a negative coefficient, so clearing it flips the sign just like dividing by \(-5\) would.
Problem
A metal sample starts at \(22\) degrees Celsius and the cooler pulls it down \(1.5\) degrees per minute, so after \(m\) minutes its temperature is \(22 - 1.5m\). It freezes only once that drops strictly below \(-8\), so the condition is \(22 - 1.5m < -8\). Solve it, remembering that dividing by the negative coefficient flips the sign. What is the fewest whole minutes the lab must wait until it freezes?
Show a hint
- Solve \(22 - 1.5m < -8\) just like a normal equation. Subtract \(22\) from both sides to get \(-1.5m < -30\). Now watch closely, because your next move divides both sides by a negative number, and that does something to the inequality sign.
- Divide both sides by \(-1.5\). Dividing by a negative flips the sign, so \(-1.5m < -30\) turns into \(m > 20\). The bound is strict, so \(m = 20\) does not count. Check it. At \(20\) minutes the temperature is exactly \(-8\), which is not below the line yet, and at \(21\) minutes it is \(-9.5\), which finally is. So the answer is the first whole number bigger than \(20\).
Show the full solution
Subtract \(22\) from both sides of \(22 - 1.5m < -8\) to get \(-1.5m < -30\). Divide by \(-1.5\), and the negative divisor flips the sign, giving \(m > 20\). The bound is strict, so the lab waits \(\boxed{21}\) minutes. At \(20\) minutes the sample is at exactly \(-8\), which is not yet below the line, and at \(21\) minutes it is \(-9.5\).
Practice these ideas
Practice
A rewards account stays open as long as its balance \(w\) satisfies \(w \ge -4\), since a small overdraft is forgiven. Five accounts have balances \(-6\), \(-4\), \(-3\), \(0\), and \(5\). How many satisfy \(w \ge -4\)?
Show the solution
Check each balance against \(w \ge -4\). \(-6\) fails, \(-4\) passes, and \(-3\), \(0\), and \(5\) all pass. That is \(\boxed{4}\) accounts. The equals half of \(\ge\) is what saves the account sitting exactly at \(-4\).
Practice
A scanner keeps a reading \(m\) only when \(m < 7\) and tosses anything that hits or passes the threshold, so the strict sign leaves the boundary \(7\) out. One shift gives \(2\), \(7\), \(6.9\), \(8\), and \(-1\). How many readings does the scanner keep?
Show the solution
Test each reading against \(m < 7\). \(2\) is kept, \(7\) is not since \(7 < 7\) is false, \(6.9\) is kept, \(8\) is not, and \(-1\) is kept. That is \(\boxed{3}\) readings. A strict less-than never lets a value equal the boundary, so the reading of exactly \(7\) goes out.
Practice
A ferry deck already holds \(13\) crates, and loading \(x\) more must keep the total under \(25\), so \(x + 13 < 25\). Solve it. What is the largest integer number of extra crates \(x\) can be?
Show the solution
Subtract \(13\) from both sides of \(x + 13 < 25\) to get \(x < 12\). Subtracting does not flip the sign. The bound is strict, so the largest whole number of extra crates is \(\boxed{11}\). Eleven brings the deck to \(24\), while \(12\) would hit \(25\) exactly, which the rule does not allow.
Practice
A pilot's dial setting \(x\) must satisfy \(x - 9 \ge -2\). Add \(9\) to both sides, with no flip for addition. Because the sign is \(\ge\), the boundary is allowed. What is the smallest value \(x\) can be?
Show the solution
Add \(9\) to both sides of \(x - 9 \ge -2\) to get \(x \ge 7\). Adding never flips the sign. The \(\ge\) lets \(x\) equal the boundary, so the smallest setting is \(\boxed{7}\).
Practice
A garden permit allows at most \(28\) liters, and each visit takes \(8\) liters, so over \(x\) visits the rule is \(8x \le 28\). Divide both sides by the positive \(8\), no flip. The word most means the boundary counts. What is the largest value \(x\) can be? Give a fraction in lowest terms.
Show the solution
Divide both sides of \(8x \le 28\) by \(8\), a positive, so no flip. That gives \(x \le \frac{28}{8} = \frac{7}{2}\), and at most means the boundary counts, so the largest is \(\boxed{7/2}\). At \(x = \frac{7}{2}\) the water used is exactly \(28\) liters, right on the permit limit.
Practice
Cookie packs cost \(\$7\) each, and Priya wants to spend strictly more than \(\$50\) to clear a free-shipping bar, so \(7x > 50\). Solve it. The strict sign means \(50\) itself does not count. What is the smallest integer number of packs she can buy?
Show the solution
Divide both sides of \(7x > 50\) by \(7\), a positive, so the sign stays put. That gives \(x > \frac{50}{7}\), about \(7.14\), and the first whole number above it is \(\boxed{8}\). Seven packs cost \(\$49\), which does not clear \(\$50\), so rounding down would miss the bar.
Practice
A deep-sea sensor's reading \(x\) must satisfy \(-6x < 30\). Divide both sides by \(-6\) and flip the sign because the divisor is negative. What is the smallest integer \(x\) can be?
Show the solution
Divide both sides of \(-6x < 30\) by \(-6\), and the negative divisor flips the sign, giving \(x > -5\). The bound is strict, so \(-5\) is out and the smallest integer allowed is \(\boxed{-4}\). Check it, \(-6 \cdot (-4) = 24 < 30\) is true, while \(-6 \cdot (-5) = 30 < 30\) is false.
Practice
A frost alarm trips whenever \(-5x > 12\), where \(x\) is a temperature that runs colder as it drops. Solve it, dividing by \(-5\) and flipping the sign. What is the largest integer value \(x\) can be?
Show the solution
Divide both sides of \(-5x > 12\) by \(-5\), and the negative divisor flips the sign, giving \(x < -\frac{12}{5}\), which is \(x < -2.4\). Walking left from \(-2.4\), the first integer is \(\boxed{-3}\). It is tempting to grab \(-2\), but \(-2\) is greater than \(-2.4\), so it misses the bound.
Practice
A vending machine credits Mara once it counts more than \(19\) tokens past a deposit, and \(x\) cups give \(4x - 5\) such tokens, so she needs \(4x - 5 > 19\). Add \(5\), then divide by the positive \(4\), no flip. What is the smallest integer number of cups \(x\) she can buy?
Show the solution
Add \(5\) to both sides of \(4x - 5 > 19\) to get \(4x > 24\), then divide by \(4\), a positive, so no flip. That gives \(x > 6\), and since the bound is strict the smallest whole number is \(\boxed{7}\). Six cups land on exactly \(19\) tokens, which is not more than \(19\).
Practice
A drone starts at a battery reading of \(10\) and the lamp drains \(4\) per minute, so after \(x\) minutes the reading is \(10 - 4x\), and the pilot keeps it on while \(10 - 4x \ge -6\). Subtract \(10\), then divide by \(-4\) and flip the sign. What is the largest value \(x\) can be?
Show the solution
Subtract \(10\) from both sides of \(10 - 4x \ge -6\) to get \(-4x \ge -16\). Divide by \(-4\), and the negative divisor flips \(\ge\) into \(\le\), giving \(x \le 4\). The boundary is included, so the largest value is \(\boxed{4}\). At \(x = 4\) the reading is exactly \(-6\), which the rule still allows, and at \(x = 5\) it drops to \(-10\).
Practice
One battery mode reads \(7 - x\) and another reads \(2x + 1\) after \(x\) minutes, and the first beats the second when \(7 - x > 2x + 1\). Gather the variable, then divide by a negative and flip the sign. What is the largest integer \(x\) can be?
Show the solution
Subtract \(2x\) from both sides of \(7 - x > 2x + 1\) to get \(7 - 3x > 1\), then subtract \(7\) to get \(-3x > -6\). Divide by \(-3\), which flips the sign, leaving \(x < 2\). The bound is strict, so the largest integer is \(\boxed{1}\). At \(x = 2\) the two readings tie at \(5\), and a tie is not a win.
Practice
A batch of \(x\) panels uses \(3(x - 2)\) ounces of grout because the first \(2\) panels are scrapped, and the bucket holds \(18\) ounces, so \(3(x - 2) \le 18\). Distribute, then solve, dividing by the positive \(3\) with no flip. What is the largest value \(x\) can be?
Show the solution
Distribute to get \(3x - 6 \le 18\), add \(6\) for \(3x \le 24\), then divide by \(3\), a positive, so no flip. That gives \(x \le 8\), and the bound is closed, so the largest is \(\boxed{8}\). At \(x = 8\) the grout used is \(3(6) = 18\) ounces, exactly the bucket, and \(x = 9\) needs \(21\).
Practice
A ferry logs every stop \(x\) that is at least \(-5\) and strictly west of \(4\), so \(-5 \le x < 4\). The left end is closed, so \(-5\) counts, and the right end is open, so \(4\) does not. How many integer positions get logged?
Show the solution
Since \(-5\) is included and \(4\) is not, the logged integers run \(-5, -4, -3, -2, -1, 0, 1, 2, 3\). That is \(\boxed{9}\) positions. Counting is faster as \(3 - (-5) = 8\) steps plus one for the endpoint you started on.
Practice
A candle stands \(30\) cm tall and burns down \(2.5\) cm each hour, so after \(h\) hours its height is \(30 - 2.5h\), and it must stay at least \(5\) cm tall to sit in its holder, giving \(30 - 2.5h \ge 5\). Dividing by \(-2.5\) flips the sign. What is the most whole hours the candle can burn?
Show the solution
Subtract \(30\) from both sides of \(30 - 2.5h \ge 5\) to get \(-2.5h \ge -25\). Divide by \(-2.5\), and the negative divisor flips \(\ge\) into \(\le\), giving \(h \le 10\). The boundary counts, so the candle can burn \(\boxed{10}\) whole hours. At \(10\) hours it stands exactly \(5\) cm tall, which still sits in the holder, and at \(11\) it is down to \(2.5\) cm.