TACHS Quiz: One Variable Inequalities
14 questions · exam conditions
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One Variable InequalitiesQuestion 1 of 14

An absolute-value inequality is shown below: ∣2p−7∣>5.|2p - 7| > 5. Which compound inequality gives the same solution set?

2p−7<−5  or  2p−7>52p - 7 < -5 \;\text{or}\; 2p - 7 > 5
2p−7≤−5  and  2p−7≥52p - 7 \le -5 \;\text{and}\; 2p - 7 \ge 5
2p−7>−5  and  2p−7<52p - 7 > -5 \;\text{and}\; 2p - 7 < 5
2p−7≥−5  or  2p−7≤52p - 7 \ge -5 \;\text{or}\; 2p - 7 \le 5
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TACHS Quiz

TACHS Quiz: One Variable Inequalities

Practice One Variable Inequalities in TACHS with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on One Variable Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for TACHS.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

An absolute-value inequality is shown below: ∣2p−7∣>5.|2p - 7| > 5. Which compound inequality gives the same solution set?

  1. 2p−7<−5  or  2p−7>52p - 7 < -5 \;\text{or}\; 2p - 7 > 5 (correct answer)
  2. 2p−7≤−5  and  2p−7≥52p - 7 \le -5 \;\text{and}\; 2p - 7 \ge 5
  3. 2p−7>−5  and  2p−7<52p - 7 > -5 \;\text{and}\; 2p - 7 < 5
  4. 2p−7≥−5  or  2p−7≤52p - 7 \ge -5 \;\text{or}\; 2p - 7 \le 5
Explanation: When you encounter an absolute value inequality like ∣2p−7∣>5|2p - 7| > 5, you need to understand what absolute value means geometrically. The expression ∣2p−7∣|2p - 7| represents the distance between 2p−72p - 7 and zero. So this inequality asks: when is that distance greater than 5? For the distance to be greater than 5, the expression 2p−72p - 7 must be either more than 5 units to the right of zero OR more than 5 units to the left of zero. This gives us two cases:
  • Case 1: 2p−7>52p - 7 > 5 (more than 5 units to the right)
  • Case 2: 2p−7<−52p - 7 < -5 (more than 5 units to the left)
Since we need either case to be true, we connect them with "OR": 2p−7<−5 or 2p−7>52p - 7 < -5 \text{ or } 2p - 7 > 5. This matches choice A. Choice B incorrectly uses "and" instead of "or," which would require both conditions to be true simultaneously - impossible since 2p−72p - 7 cannot be both ≤−5\leq -5 and ≥5\geq 5 at the same time. Choice C gives us 2p−7>−5 and 2p−7<52p - 7 > -5 \text{ and } 2p - 7 < 5, which describes when the absolute value is less than 5, not greater than 5. Choice D combines the wrong inequalities with "or," mixing elements from the "less than" case with inequality signs that don't make sense together. Strategy tip: For ∣expression∣>number|expression| > number, always use "OR" and flip the sign on the negative case. For ∣expression∣<number|expression| < number, use "AND" with the expression between the positive and negative bounds.

Question 2

Solve for yy: 5−3y2<y+4\dfrac{5 - 3y}{2} < y + 4.

  1. y>−35y > -\dfrac{3}{5} (correct answer)
  2. y<−35y < -\dfrac{3}{5}
  3. y>−135y > -\dfrac{13}{5}
  4. y<−135y < -\dfrac{13}{5}
Explanation: When you encounter an inequality with fractions, your goal is to isolate the variable while carefully tracking the inequality direction. The key difference from equations is that multiplying or dividing by a negative number flips the inequality sign. Starting with 5−3y2<y+4\frac{5 - 3y}{2} < y + 4, first eliminate the fraction by multiplying both sides by 2: 5−3y<2(y+4)5 - 3y < 2(y + 4). This gives you 5−3y<2y+85 - 3y < 2y + 8. Next, collect all terms with yy on one side. Subtract 2y2y from both sides: 5−3y−2y<85 - 3y - 2y < 8, which simplifies to 5−5y<85 - 5y < 8. Then subtract 5 from both sides: −5y<3-5y < 3. Finally, divide both sides by −5-5. Since you're dividing by a negative number, flip the inequality sign: y>−35y > -\frac{3}{5}. This matches answer choice A. Looking at the wrong answers: B gives y<−35y < -\frac{3}{5}, which results from forgetting to flip the inequality sign when dividing by −5-5. C shows y>−135y > -\frac{13}{5}, likely from an error when distributing or combining like terms. D combines both errors—wrong coefficient and wrong inequality direction. Remember: whenever you multiply or divide an inequality by a negative number, always flip the inequality sign. This is the most common mistake on inequality problems, so double-check this step every time.

Question 3

In a certain video game, you earn 4040 points plus 66 points for every level completed. To qualify for a bonus round, you must have more than 130130 points. Which inequality shows the number of levels ℓ\ell you must complete, and what is the solution?

  1. 40+6ℓ>130;  ℓ>1540 + 6\ell > 130;\; \ell > 15 (correct answer)
  2. 40+6ℓ≥130;  ℓ≥1540 + 6\ell \ge 130;\; \ell \ge 15
  3. 40+6ℓ>130;  ℓ>9040 + 6\ell > 130;\; \ell > 90
  4. 40+6ℓ≥130;  ℓ≥9040 + 6\ell \ge 130;\; \ell \ge 90
Explanation: When you encounter word problems involving inequalities, start by identifying the fixed amounts, variable rates, and comparison conditions. Here you earn a base of 40 points plus 6 points per level, and need more than 130 points total. The correct setup is 40+6ℓ>13040 + 6\ell > 130 because you start with 40 points, gain 6 points per level ℓ\ell, and must exceed 130 points (not equal to). To solve: subtract 40 from both sides to get 6ℓ>906\ell > 90, then divide by 6 to find ℓ>15\ell > 15. This means you need at least 16 complete levels. Choice A correctly uses the "greater than" symbol and arrives at ℓ>15\ell > 15. Choice B uses ≥\ge instead of >>, which would mean exactly 130 points qualifies. The problem specifically states you need "more than" 130 points, so equality doesn't work. Choice C has the right inequality symbol but incorrectly solves to ℓ>90\ell > 90. This error likely comes from forgetting to divide 90 by 6 after subtracting 40 from both sides. Choice D combines both errors: using ≥\ge when it should be >>, and getting ℓ≥90\ell \ge 90 instead of the correct solution. Study tip: Pay close attention to comparison words in inequality problems. "More than" means >>, "at least" means ≥\ge, "less than" means <<, and "at most" means ≤\le. Always double-check your algebra by substituting your boundary value back into the original inequality.

Question 4

A theater sells adult tickets for $12 each. The manager wants to earn at least $540 from adult tickets alone. If $tt $ represents the number of adult tickets sold, which inequality represents this requirement, and what is the least whole number of adult tickets that satisfies it?

  1. 12t≥540;  t=4512t \ge 540;\; t = 45 (correct answer)
  2. 12t>540;  t=4512t > 540;\; t = 45
  3. 12t≥540;  t=4412t \ge 540;\; t = 44
  4. 12t>540;  t=4412t > 540;\; t = 44
Explanation: When you encounter problems about earning "at least" a certain amount, you're dealing with inequality relationships that require careful attention to both the inequality symbol and boundary conditions. The theater needs to earn at least $540, meaning $540 or more. Since each adult ticket costs $12 and $tt representsthenumberofticketssold,thetotalrevenueisrepresents the number of tickets sold, the total revenue is 12t12t .Thephrase"atleast. The phrase "at least 540" translates to 12t≥54012t \ge 540, where the "greater than or equal to" symbol (≥\ge) captures both the minimum requirement and any amount above it. To find the minimum number of tickets, solve 12t≥54012t \ge 540 by dividing both sides by 12: t≥45t \ge 45. Since we need a whole number of tickets and tt must be greater than or equal to 45, the least value is t=45t = 45. Looking at the wrong answers: Choice B uses 12t>54012t > 540 with t=45t = 45, but this creates a contradiction—if you need more than 540,then45tickets(whichequalsexactly540, then 45 tickets (which equals exactly 540) wouldn't satisfy the requirement. Choice C has the correct inequality but claims t=44t = 44, which would only generate 528,fallingshortofthe528, falling short of the 540 minimum. Choice D combines both errors: using the wrong inequality symbol and the insufficient ticket count. Study tip: "At least" always means "greater than or equal to" (≥\ge), while "more than" means "greater than" (>>). When finding minimum whole number solutions, always check that your answer actually satisfies the original inequality.

Question 5

Determine the solution to 7−3(2x−5)>4x+17 - 3(2x - 5) > 4x + 1.

  1. x<2.1x < 2.1 (correct answer)
  2. x>2.1x > 2.1
  3. x<−2.1x < -2.1
  4. x>−2.1x > -2.1
Explanation: When you encounter an inequality with variables on both sides, your goal is to isolate the variable while carefully tracking the inequality direction. Let's solve 7−3(2x−5)>4x+17 - 3(2x - 5) > 4x + 1 step by step. First, distribute the −3-3: 7−6x+15>4x+17 - 6x + 15 > 4x + 1. Combine like terms on the left: 22−6x>4x+122 - 6x > 4x + 1. Now collect all xx terms on one side by subtracting 4x4x from both sides: 22−10x>122 - 10x > 1. Subtract 2222 from both sides: −10x>−21-10x > -21. Here's the crucial step: when you divide both sides by a negative number (−10-10), you must flip the inequality sign. So x<2110=2.1x < \frac{21}{10} = 2.1. Looking at the answer choices: Choice A (x<2.1x < 2.1) matches our solution exactly. Choice B (x>2.1x > 2.1) represents the common error of forgetting to flip the inequality when dividing by a negative. Choice C (x<−2.1x < -2.1) likely comes from a sign error during the algebra, perhaps when distributing or combining terms. Choice D (x>−2.1x > -2.1) combines both errors: the wrong sign and the failure to flip the inequality. Remember this key rule: whenever you multiply or divide both sides of an inequality by a negative number, the inequality direction must reverse. This is the most common mistake students make with inequalities, so double-check your work whenever negative coefficients are involved.

Question 6

Which inequality describes all real numbers that are no more than 8 units away from −3-3 on the number line?

  1. ∣x+3∣≤8|x + 3| \le 8 (correct answer)
  2. ∣x−8∣≤3|x - 8| \le 3
  3. ∣x−3∣<8|x - 3| < 8
  4. ∣x+8∣<3|x + 8| < 3
Explanation: When you see a question about distance on the number line, think absolute value inequalities. The phrase "no more than 8 units away from -3" is describing all points whose distance from -3 is at most 8 units. Distance problems always translate to absolute value expressions in the form ∣x−center∣≤distance|x - \text{center}| \le \text{distance}. Here, the center point is -3 and the maximum distance is 8. So you need ∣x−(−3)∣≤8|x - (-3)| \le 8, which simplifies to ∣x+3∣≤8|x + 3| \le 8. You can verify this makes sense: if x=5x = 5, then ∣5+3∣=8≤8|5 + 3| = 8 \le 8 ✓. If x=−11x = -11, then ∣−11+3∣=8≤8|-11 + 3| = 8 \le 8 ✓. The solution includes all numbers from -11 to 5. Choice B, ∣x−8∣≤3|x - 8| \le 3, describes points within 3 units of 8, not points within 8 units of -3. This represents the interval [5, 11]. Choice C, ∣x−3∣<8|x - 3| < 8, describes points within 8 units of positive 3 (not -3), and uses strict inequality instead of "no more than." This gives the interval (-5, 11). Choice D, ∣x+8∣<3|x + 8| < 3, describes points within 3 units of -8, giving the interval (-11, -5). Strategy tip: For distance problems, always write ∣x−center point∣ comparison distance|x - \text{center point}| \text{ comparison } \text{distance}. Pay attention to whether the problem says "no more than" (≤) versus "less than" (<). The phrase "away from" signals absolute value every time.

Question 7

Which of the following is equivalent to the solution of x−4≥7or2x+5<−1x - 4 \ge 7 \quad \text{or} \quad 2x + 5 < -1?

  1. x≥11  or  x<−3x \ge 11 \;\text{or}\; x < -3 (correct answer)
  2. x≥11  and  x<−3x \ge 11 \;\text{and}\; x < -3
  3. x≤11  or  x>−3x \le 11 \;\text{or}\; x > -3
  4. x<11  and  x≥−3x < 11 \;\text{and}\; x \ge -3
Explanation: When you encounter compound inequalities connected by "or," you need to solve each inequality separately, then combine the solutions using union (either condition can be true). Let's solve each inequality: For x−4≥7x - 4 \ge 7: Add 4 to both sides: x≥11x \ge 11 For 2x+5<−12x + 5 < -1: Subtract 5 from both sides: 2x<−62x < -6 Divide by 2: x<−3x < -3 Since this is an "or" compound inequality, the solution includes all values that satisfy either condition: x≥11x \ge 11 or x<−3x < -3. Looking at the answer choices: Choice A correctly states x≥11 or x<−3x \ge 11 \text{ or } x < -3, which matches our solution. Choice B uses "and" instead of "or" - this would represent intersection, meaning values that satisfy both conditions simultaneously (which is impossible since no number can be both ≥11 and <-3). Choice C has the wrong inequality directions (≤ instead of ≥, and > instead of <), completely changing which values are included. Choice D also uses "and" instead of "or" and has incorrect inequality directions. The key insight is recognizing that "or" in compound inequalities means union - you're looking for values in either region on the number line. "And" would mean intersection - values that satisfy both conditions at once. Always solve each piece separately first, then apply the correct logical connector.

Question 8

If 9<2k+1≤199 < 2k + 1 \le 19, what is the least integer value of kk that satisfies the inequality?

  1. 44
  2. 55 (correct answer)
  3. 66
  4. 99
Explanation: When you encounter compound inequalities like this one, you need to solve for the variable by working with both parts of the inequality simultaneously while preserving the relationships. Starting with 9<2k+1≤199 < 2k + 1 \le 19, subtract 1 from all three parts: 8<2k≤188 < 2k \le 18. Then divide everything by 2: 4<k≤94 < k \le 9. This means kk must be greater than 4 and less than or equal to 9. Since we need the least integer value of kk, we're looking for the smallest whole number that satisfies k>4k > 4. Since kk must be strictly greater than 4 (not equal to it), the smallest integer that works is 5. Let's verify: if k=5k = 5, then 2k+1=112k + 1 = 11, and indeed 9<11≤199 < 11 \le 19 ✓ Looking at the wrong answers: Choice A (4) fails because when k=4k = 4, we get 2k+1=92k + 1 = 9, but we need 2k+1>92k + 1 > 9, not equal to 9. Choice C (6) satisfies the inequality, but it's not the least value—5 is smaller. Choice D (9) also works since it gives us 2k+1=192k + 1 = 19, satisfying 9<19≤199 < 19 \le 19, but again, it's not the minimum. The key strategy here is to pay careful attention to whether inequalities use strict inequality signs (< or >) versus inclusive ones (≤ or ≥). The difference between "greater than" and "greater than or equal to" often determines which boundary values are acceptable in your solution set.

Question 9

Which inequality represents all real numbers mm whose double decreased by 9 is at most 3?

  1. 2m−9≤32m - 9 \le 3 (correct answer)
  2. 2m−9<32m - 9 < 3
  3. 2m+9≤32m + 9 \le 3
  4. 2m+9<32m + 9 < 3
Explanation: When you encounter word problems involving inequalities, you need to carefully translate each phrase into mathematical symbols. The key phrases here are "double decreased by 9" and "at most." Let's break down the translation step by step. "Double" means multiply by 2, so we have 2m2m. "Decreased by 9" means subtract 9, giving us 2m−92m - 9. The phrase "at most 3" means less than or equal to 3, which translates to ≤3\le 3. Putting it together: 2m−9≤32m - 9 \le 3. Choice A (2m−9≤32m - 9 \le 3) is correct because it accurately translates all parts of the word problem. Choice B (2m−9<32m - 9 < 3) makes the common error of confusing "at most" with "less than." While these seem similar, "at most 3" includes the possibility of equaling 3, so you need ≤\le, not <<. Choice C (2m+9≤32m + 9 \le 3) incorrectly interprets "decreased by 9" as "increased by 9." This is a frequent mistake—"decreased by" always means subtraction, not addition. Choice D (2m+9<32m + 9 < 3) combines both errors: using addition instead of subtraction and the wrong inequality symbol. Remember this translation guide: "at most" = ≤\le, "at least" = ≥\ge, "less than" = <<, "greater than" = >>. Also, "decreased by" always means subtract, while "increased by" means add. Practice identifying these key phrases to avoid common translation errors on inequality word problems.

Question 10

Solve 3x−54≥2\dfrac{3x - 5}{4} \ge 2.

  1. x≥133x \ge \dfrac{13}{3} (correct answer)
  2. x≤133x \le \dfrac{13}{3}
  3. x≥3x \ge 3
  4. x≤3x \le 3
Explanation: When you encounter an inequality with fractions, your goal is to isolate the variable while preserving the inequality's direction. The key principle is that you can perform the same operations on both sides of an inequality as you would with an equation, as long as you don't multiply or divide by a negative number. Starting with 3x−54≥2\dfrac{3x - 5}{4} \ge 2, first multiply both sides by 4 to eliminate the fraction: 3x−5≥83x - 5 \ge 8. Next, add 5 to both sides: 3x≥133x \ge 13. Finally, divide both sides by 3: x≥133x \ge \dfrac{13}{3}. Since we only multiplied and divided by positive numbers, the inequality sign stays the same. Looking at the wrong answers: Choice B gives x≤133x \le \dfrac{13}{3}, which flips the inequality sign incorrectly—this would happen if you mistakenly thought you needed to reverse the sign when dividing by a positive number. Choice C gives x≥3x \ge 3, which results from incorrectly adding 5 to the right side instead of adding it (getting 3x≥33x \ge 3). Choice D combines both errors: the wrong constant and the flipped inequality sign. The key strategy for inequality problems is to treat them exactly like equations until you multiply or divide by a negative number—only then do you flip the inequality sign. Also, always double-check your arithmetic when combining fractions and whole numbers, as it's easy to make calculation errors under test pressure.

Question 11

Solve the compound inequality −4<2x−6≤10-4 < 2x - 6 \le 10.

  1. 1<x≤81 < x \le 8 (correct answer)
  2. 1<x≤41 < x \le 4
  3. −1<x≤8-1 < x \le 8
  4. −1<x≤4-1 < x \le 4
Explanation: When you encounter a compound inequality with "and" logic (like this one with < and ≤), you need to solve both parts simultaneously while keeping the inequality balanced. Start by adding 6 to all three parts: −4+6<2x−6+6≤10+6-4 + 6 < 2x - 6 + 6 \le 10 + 6, which gives you 2<2x≤162 < 2x \le 16. Next, divide everything by 2: 22<2x2≤162\frac{2}{2} < \frac{2x}{2} \le \frac{16}{2}, resulting in 1<x≤81 < x \le 8. This matches answer choice A. Let's examine why the other options are incorrect. Choice B gives 1<x≤41 < x \le 4 - this gets the left boundary right but incorrectly calculates the right side. The error likely comes from dividing 10 by 2 instead of adding 6 first, then dividing by 2. Choice C shows −1<x≤8-1 < x \le 8 - this gets the right boundary correct but miscalculates the left side. This mistake happens when you divide -4 by 2 without first adding 6 to isolate the 2x term. Choice D combines both errors from choices B and C, giving −1<x≤4-1 < x \le 4. The key strategy for compound inequalities is to perform the same operation to all three parts simultaneously, just like balancing a regular equation. Always work step-by-step: first eliminate any constants added or subtracted from the variable term, then handle multiplication or division. Double-check by substituting boundary values back into the original inequality to verify your solution works.

Question 12

For which integers nn is the statement 3n+2≤173n + 2 \le 17 true?

  1. All integers n≤5n \le 5 (correct answer)
  2. All integers n≥5n \ge 5
  3. All integers n≤6n \le 6
  4. All integers n≥6n \ge 6
Explanation: When you encounter an inequality like this, you're solving for the range of values that make the statement true. Think of this as finding the boundary and determining which side of it satisfies the condition. To solve 3n+2≤173n + 2 \le 17, you need to isolate nn. First, subtract 2 from both sides: 3n≤153n \le 15. Then divide both sides by 3: n≤5n \le 5. This means the inequality is true for all integers less than or equal to 5. You can verify this by testing values. When n=5n = 5: 3(5)+2=173(5) + 2 = 17, so 17≤1717 \le 17 is true. When n=4n = 4: 3(4)+2=143(4) + 2 = 14, so 14≤1714 \le 17 is true. When n=6n = 6: 3(6)+2=203(6) + 2 = 20, so 20≤1720 \le 17 is false. Looking at the choices: Choice A correctly states n≤5n \le 5. Choice B says n≥5n \ge 5, which would include values like 6, 7, 8 that make the inequality false. Choice C says n≤6n \le 6, but we just saw that n=6n = 6 makes 20≤1720 \le 17, which is false. Choice D says n≥6n \ge 6, which includes only values that make the inequality false. The key strategy here is to always solve the inequality algebraically first, then test a few values around the boundary to confirm your work. Remember that when you have ≤\le or ≥\ge, the boundary value itself is included in the solution set.

Question 13

Which set represents the solution to −4x+10≤−6-4x + 10 \le -6?

  1. x≥4x \ge 4 (correct answer)
  2. x≤4x \le 4
  3. x≥−4x \ge -4
  4. x≤−4x \le -4
Explanation: When you encounter an inequality like this, you're solving it just like a regular equation with one crucial rule to remember: if you multiply or divide both sides by a negative number, you must flip the inequality sign. Let's solve −4x+10≤−6-4x + 10 \le -6 step by step. First, subtract 10 from both sides: −4x≤−16-4x \le -16. Now divide both sides by -4. Since we're dividing by a negative number, we flip the inequality sign from ≤\le to ≥\ge: x≥4x \ge 4. Choice A (x≥4x \ge 4) is correct because it represents our final solution after properly flipping the inequality sign. Choice B (x≤4x \le 4) is the most common trap answer. This is what you'd get if you forgot to flip the inequality sign when dividing by -4. Many students make this error because they correctly solve the algebraic steps but miss the sign flip rule. Choice C (x≥−4x \ge -4) suggests an error in the arithmetic. You might get this if you incorrectly calculated −16÷(−4)=−4-16 ÷ (-4) = -4 instead of the correct −16÷(−4)=4-16 ÷ (-4) = 4. Choice D (x≤−4x \le -4) combines both errors: the wrong arithmetic (getting -4 instead of 4) and forgetting to flip the inequality sign. Strategy tip: Always double-check your work by substituting a test value back into the original inequality. For example, try x=5x = 5: −4(5)+10=−10-4(5) + 10 = -10, and −10≤−6-10 \le -6 is true, confirming our answer.

Question 14

Solve the inequality 5x+7<2x+315x + 7 < 2x + 31. Which of the following describes all possible values of xx?

  1. x<8x < 8 (correct answer)
  2. x<−8x < -8
  3. x>8x > 8
  4. x>−8x > -8
Explanation: When you encounter linear inequalities, you solve them just like equations, with one crucial rule to remember: if you multiply or divide by a negative number, you must flip the inequality sign. Let's solve 5x+7<2x+315x + 7 < 2x + 31 step by step. First, subtract 2x2x from both sides: 5x−2x+7<315x - 2x + 7 < 31, which gives us 3x+7<313x + 7 < 31. Next, subtract 7 from both sides: 3x<243x < 24. Finally, divide both sides by 3: x<8x < 8. Since we divided by a positive number, the inequality sign stays the same. Looking at the answer choices, option A gives us x<8x < 8, which matches our solution exactly. Option B (x<−8x < -8) represents a common error where students might have made sign mistakes during the solving process. Option C (x>8x > 8) occurs when students incorrectly flip the inequality sign even though they didn't divide by a negative number. Option D (x>−8x > -8) combines both errors - flipping the sign unnecessarily and getting the wrong numerical value. You can verify the correct answer by testing a value: if x=0x = 0, then 5(0)+7=75(0) + 7 = 7 and 2(0)+31=312(0) + 31 = 31. Since 7<317 < 31, this confirms that x=0x = 0 satisfies our inequality, and since 0<80 < 8, option A is correct. Remember: only flip the inequality sign when multiplying or dividing by negative numbers. This is the most common trap in inequality problems.