Solving Inequalities - SHSAT Math

  1. The table shows values of f(x)=3x−7f(x) = 3x - 7 and g(x)=−x+9g(x) = -x + 9. Based on the table, what is the smallest integer xx for which f(x)>g(x)f(x) > g(x)?
    Answer: 55 Solve 3x−7>−x+9⇒4x>16⇒x>43x - 7 > -x + 9 \Rightarrow 4x > 16 \Rightarrow x > 4. Smallest integer is 5. From the table, at x=4x=4, f(4)=5f(4)=5 and g(4)=5g(4)=5 (equal, not greater). At x=5x=5, f(5)=8>g(5)=4f(5)=8 > g(5)=4 ✓. (A) and (B) misread 'strictly greater'. (D) is one too large.
  2. If −4≤2x+6<10-4 \leq 2x + 6 < 10, what is the range of values for xx?
    Answer: −5≤x<2-5 \leq x < 2 When you encounter compound inequalities like this one, you're solving for a variable that must satisfy two conditions simultaneously. The key is to isolate the variable by performing the same operations on all three parts of the inequality. Starting with −4≤2x+6<10-4 \leq 2x + 6 < 10, you need to isolate xx by first subtracting 6 from all three parts: −4−6≤2x+6−6<10−6-4 - 6 \leq 2x + 6 - 6 < 10 - 6, which gives you −10≤2x<4-10 \leq 2x < 4. Then divide all parts by 2: −5≤x<2-5 \leq x < 2. This means xx can equal -5 but must be strictly less than 2. Choice A (−5<x≤2-5 < x \leq 2) reverses both inequality symbols. This would mean xx cannot equal -5 but can equal 2, which contradicts our solution. Choice C (−10≤x<4-10 \leq x < 4) represents the inequality before dividing by 2 - a common error when students forget this final step. Choice D (1≤x<81 \leq x < 8) appears to come from incorrectly adding instead of subtracting 6, then making additional errors in the division step. Choice B (−5≤x<2-5 \leq x < 2) correctly shows that xx includes -5 (closed circle) and excludes 2 (open circle). Strategy tip: When solving compound inequalities, always perform identical operations on all three parts simultaneously, and pay careful attention to whether endpoints are included (≤, ≥) or excluded (<, >). Double-check by substituting boundary values back into the original inequality.
  3. The inequality 5−3x<2x−105 - 3x < 2x - 10 has the same solution set as which of the following?
    Answer: x>3x > 3 When you encounter an inequality that needs to be solved, you're working with the same algebraic principles as equations, but with one crucial difference: flipping the inequality sign when multiplying or dividing by a negative number. To solve 5−3x<2x−105 - 3x < 2x - 10, start by collecting like terms. Add 3x3x to both sides: 5<5x−105 < 5x - 10. Then add 1010 to both sides: 15<5x15 < 5x. Finally, divide both sides by 55: 3<x3 < x, which is equivalent to x>3x > 3. Let's check why the other options are incorrect. Choice A (x<−3x < -3) represents the complete opposite direction - it suggests xx values that are much smaller than what our solution requires. Choice B (x<3x < 3) has the wrong direction of the inequality; this would include values like x=0x = 0 or x=1x = 1, which don't satisfy our original inequality. Choice C (x>−3x > -3) has the right direction but the wrong boundary value - this would incorrectly include values between −3-3 and 33, such as x=0x = 0, which doesn't work in the original inequality. You can verify choice D is correct by testing a value like x=4x = 4: 5−3(4)=−75 - 3(4) = -7 and 2(4)−10=−22(4) - 10 = -2, so −7<−2-7 < -2 ✓. Remember: when solving inequalities, be extra careful with your algebraic steps and always consider testing a value from your solution set to verify your answer. The SHSAT often includes trap answers that reverse the inequality direction.
  4. Based on the diagram below, a phone plan charges a flat $12 monthly fee plus $0.08 per minute. A competing plan charges $0.20 per minute with no monthly fee. For how many whole minutes mm per month is the first plan strictly less expensive than the second plan?
    Answer: m≥101m \geq 101 Write the cost of each plan for mm minutes. The first plan costs 12+0.08m12 + 0.08m dollars, and the second costs 0.20m0.20m dollars. The first plan is strictly cheaper when 12+0.08m<0.20m12 + 0.08m < 0.20m. Subtract 0.08m0.08m: 12<0.12m12 < 0.12m, so m>100m > 100. At exactly 100 minutes both plans cost $20, so 100 does not count. The smallest whole number of minutes that works is 101, so the answer is m≥101m \geq 101. The inequality m≥100m \geq 100 includes 100 minutes, where the plans tie. The inequality m≥60m \geq 60 comes from dividing the $12 fee by the second plan's full rate, 12 ÷ 0.20 = 60, instead of by the difference in rates. The inequality m>150m > 150 comes from dividing by the first plan's rate, 12 ÷ 0.08 = 150.
  5. A rectangle has a length that is 4 units more than twice its width. If the perimeter must be less than 32 units, what is the range of possible values for the width ww?
    Answer: 0<w<40 < w < 4 When you encounter word problems involving geometric constraints, you need to translate the verbal relationships into mathematical expressions and then apply the given conditions. Start by defining the width as ww. Since the length is "4 units more than twice the width," you can express the length as 2w+42w + 4. The perimeter of a rectangle is P=2(length+width)P = 2(\text{length} + \text{width}), so: P=2(2w+4+w)=2(3w+4)=6w+8P = 2(2w + 4 + w) = 2(3w + 4) = 6w + 8 Since the perimeter must be less than 32 units: 6w+8<326w + 8 < 32. Solving this inequality: 6w<246w < 24, so w<4w < 4. However, since width represents a physical measurement, it must be positive, giving us 0<w<40 < w < 4. Choice A (w<8w < 8) incorrectly solves the inequality. This likely comes from making an error in the perimeter formula or algebraic manipulation. Choice B (0<w<80 < w < 8) makes the same algebraic error as A but correctly recognizes that width must be positive. Choice C (w<4w < 4) correctly solves the inequality but fails to account for the physical constraint that width must be positive—a rectangle cannot have zero or negative width. Choice D correctly incorporates both the mathematical constraint from the inequality and the real-world constraint that geometric measurements must be positive. Remember that word problems involving measurements require two checks: solving the mathematical relationships correctly and ensuring your answer makes sense in the physical context. Always verify that geometric measurements are positive values.
  6. A student needs to score at least 85% on a test with 40 questions to pass. If each question is worth the same number of points, what is the minimum number of questions the student must answer correctly?
    Answer: 34 questions This problem tests percentage calculations and the critical skill of rounding up when dealing with "at least" requirements in real-world contexts. To find the minimum correct answers needed, you must calculate 85% of 40 questions: 0.85×40=340.85 \times 40 = 34 questions exactly. Since the student needs to score "at least 85%," and 34 questions gives exactly 85%, the minimum number of correct answers is 34. Let's examine why each answer choice works or fails. Choice A (33 questions) gives you 3340=0.825=82.5%\frac{33}{40} = 0.825 = 82.5\%, which falls short of the required 85%. Choice C (35 questions) would work since 3540=87.5%\frac{35}{40} = 87.5\%, but the question asks for the minimum number, making this more than necessary. Choice D (32 questions) gives only 3240=80%\frac{32}{40} = 80\%, clearly insufficient. Choice B (34 questions) is correct because 3440=85%\frac{34}{40} = 85\% exactly, meeting the minimum requirement. The key insight here is recognizing that when a percentage calculation yields a whole number, that's your answer for "at least" problems. If the calculation had produced 34.2, you'd round up to 35 since you can't answer a fraction of a question correctly. Strategy tip: When you see "at least" or "minimum" with percentages, calculate the exact percentage first, then remember that partial questions always round up in real-world scenarios. However, when the calculation is exact (like here), that exact number is your minimum.
  7. The graph below shows the line y=−2x+6y = -2x + 6. For what values of xx is −2x+6<4-2x + 6 < 4? Use the graph shown to determine the answer.
    Answer: x>1x > 1 We need where the line is below y=4y=4. Setting −2x+6=4-2x + 6 = 4 gives x=1x = 1. Since the line has negative slope, it is below 4 when x>1x > 1. Algebraically: −2x<−2⇒x>1-2x < -2 \Rightarrow x > 1. (A) forgets to flip the inequality. (C) and (D) use wrong endpoint.
  8. The solution to the inequality 3x−7>2x+53x - 7 > 2x + 5 is equivalent to which of the following?
    Answer: x>12x > 12 Solving 3x−7>2x+53x - 7 > 2x + 5: subtract 2x2x from both sides to get x−7>5x - 7 > 5, then add 7 to get x>12x > 12. Choice B reverses the inequality sign. Choice C results from the error −7+5=−2-7 + 5 = -2. Choice D combines both the sign error and the arithmetic error.
  9. Refer to the figure. A rectangle has length (2x+3)(2x + 3) cm and width (x−1)(x - 1) cm, where x>1x > 1. The perimeter of the rectangle must be less than 40 cm but at least 16 cm. What is the complete set of integer values of xx that satisfies these conditions?
    Answer: {2,3,4,5}\{2, 3, 4, 5\} Perimeter = 2[(2x+3)+(x−1)]=6x+42[(2x+3)+(x-1)] = 6x + 4. Need 16≤6x+4<4016 \leq 6x + 4 < 40, so 12≤6x<3612 \leq 6x < 36, giving 2≤x<62 \leq x < 6. Combined with x>1x > 1, we get 2≤x<62 \leq x < 6. Integer values are {2,3,4,5}\{2, 3, 4, 5\}.
  10. Which inequality represents the possible numbers of miles, mm, the customer can ride?
    Answer: 0.5m+2≤100.5m+2\le10 When you encounter word problems involving costs and spending limits, you're working with inequalities that represent real-world constraints. The key is translating the language carefully into mathematical symbols. The taxi charges $2 plus $0.50 per mile, so the total cost for mm miles is 0.5m+20.5m + 2. The customer has "at most $10 to spend," which means they can spend $10 or less. This gives us the inequality 0.5m+2≤100.5m + 2 \leq 10, which is answer choice C. Let's examine why the other choices are wrong. Choice A (0.5m+2>100.5m + 2 > 10) represents situations where the cost exceeds 10,meaningthecustomercan′taffordtheride.ChoiceB(10, meaning the customer can't afford the ride. Choice B (0.5m+2<100.5m + 2 < 10$) only includes rides that cost strictly less than $10, excluding the possibility of spending exactly 10,whichthecustomercanafford.ChoiceD(10, which the customer can afford. Choice D (0.5m+2≥100.5m + 2 \geq 10$) represents costs of $10 or more, which goes beyond what the customer can afford. The critical phrase here is "at most," which always translates to "less than or equal to" (≤\leq). You can verify this makes sense: if the customer has exactly $10, they should be able to afford a ride that costs exactly $10. Remember this pattern: "at most" means ≤\leq, "at least" means ≥\geq, "more than" means >>, and "less than" means <<. Pay close attention to whether the limit value should be included in your solution set.
  11. Solve for nn: −5n≥40-5n \ge 40.
    Answer: n≤−8n \le -8 When solving inequalities involving negative coefficients, the key concept to remember is that multiplying or dividing both sides by a negative number flips the inequality sign. Starting with −5n≥40-5n \ge 40, you need to isolate nn by dividing both sides by −5-5. Since you're dividing by a negative number, the inequality sign must flip from ≥\ge to ≤\le: −5n−5≤40−5\frac{-5n}{-5} \le \frac{40}{-5} This simplifies to n≤−8n \le -8. Looking at the wrong answers: Choice A (n≥−8n \ge -8) represents the most common error—forgetting to flip the inequality sign when dividing by −5-5. This would give you the opposite relationship. Choice C (n≥8n \ge 8) makes two mistakes: not flipping the sign AND getting the wrong sign on the 8, likely from incorrectly handling the negative division. Choice D (n≤8n \le 8) correctly flips the inequality sign but incorrectly calculates 40−5\frac{40}{-5} as positive 8 instead of negative 8. You can verify the answer by testing a value: if n=−8n = -8, then −5(−8)=40-5(-8) = 40, which satisfies −5n≥40-5n \ge 40. If n=−9n = -9 (which is less than −8-8), then −5(−9)=45>40-5(-9) = 45 > 40, confirming our solution. Remember: whenever you multiply or divide an inequality by a negative number, always flip the inequality sign. This is one of the most tested concepts with inequalities on standardized exams.
  12. Solve for xx: x+7<4x + 7 < 4.
    Answer: x<−3x < -3 When you encounter an inequality like this, you're solving it just like a regular equation, but you need to pay attention to the inequality sign throughout your work. To solve x+7<4x + 7 < 4, you need to isolate xx on one side. Subtract 7 from both sides: x+7−7<4−7x + 7 - 7 < 4 - 7, which simplifies to x<−3x < -3. This matches choice A. Let's examine why the other options are incorrect. Choice B (x<11x < 11) represents the mistake of adding 7 to both sides instead of subtracting it. If you mistakenly thought x+7<4x + 7 < 4 becomes x<4+7x < 4 + 7, you'd get this wrong answer. Choice C (x>−3x > -3) shows the correct number but the wrong inequality direction - this would happen if you forgot that the original inequality was "less than" rather than "greater than." Choice D (x>11x > 11) combines both errors: adding instead of subtracting AND flipping the inequality sign incorrectly. You can verify your answer by testing a value. Since x<−3x < -3, let's try x=−4x = -4. Substituting: −4+7=3-4 + 7 = 3, and indeed 3<43 < 4 ✓. If you tried x=0x = 0 (which doesn't satisfy x<−3x < -3), you'd get 0+7=70 + 7 = 7, and 7<47 < 4 is false. Key strategy: Unlike equations, inequalities keep their direction when you add or subtract the same number from both sides. Only when multiplying or dividing by negative numbers do you flip the inequality sign.
  13. Use the table showing four students' work on the inequality −23x+5>9-\frac{2}{3}x + 5 > 9. Which student's reasoning is entirely correct?
    Answer: Carmen Correct solution: −23x>4⇒x<−6-\frac{2}{3}x > 4 \Rightarrow x < -6 (flip inequality when multiplying by −32-\frac{3}{2}). Aisha forgot to flip: x>−6x > -6. Brandon subtracted wrong: got −23x>14-\frac{2}{3}x > 14. Carmen correctly got x<−6x < -6. Dion flipped twice, ending with x>6x > 6.
  14. The graph below shows two lines y1=12x+1y_1 = \frac{1}{2}x + 1 and y2=−x+7y_2 = -x + 7. For what values of xx is y1≥y2y_1 \geq y_2?
    Answer: x≥4x \geq 4 Set 12x+1≥−x+7⇒32x≥6⇒x≥4\frac{1}{2}x + 1 \geq -x + 7 \Rightarrow \frac{3}{2}x \geq 6 \Rightarrow x \geq 4. On the graph, y1y_1 is above (or equal to) y2y_2 to the right of the intersection point at (4,3)(4, 3). (B) reverses direction. (C) and (D) use yy-coordinate of intersection.
  15. Refer to the graph shown. A student wants to buy notebooks at $3 each and pens at $1.50 each, spending no more than $30 total. The shaded region in the graph represents all valid combinations (n,p)(n, p) where nn = number of notebooks and pp = number of pens. Which inequality, combined with n≥0n \geq 0 and p≥0p \geq 0, produces this shaded region?
    Answer: 3n+1.5p≤303n + 1.5p \leq 30 Total cost is 3n+1.5p3n + 1.5p and cannot exceed 30. Spending 'no more than' means ≤\leq, so 3n+1.5p≤303n + 1.5p \leq 30. The shaded region is bounded by a solid line (including the boundary). (A) uses strict inequality with dashed boundary. (C) swaps the prices. (D) reverses direction.
  16. If −2(x+3)≥4x−6-2(x + 3) \geq 4x - 6, which of the following values of xx satisfies this inequality?
    Answer: x=0x = 0 Solve the inequality first. Distribute: −2x−6≥4x−6-2x - 6 \geq 4x - 6. Add 66 to both sides: −2x≥4x-2x \geq 4x. Subtract 4x4x: −6x≥0-6x \geq 0. Divide by −6-6 and flip the sign, because you are dividing by a negative number: x≤0x \leq 0. Of the listed values, only x=0x = 0 is less than or equal to zero. Check it: −2(0+3)=−6-2(0 + 3) = -6 and 4(0)−6=−64(0) - 6 = -6, and −6≥−6-6 \geq -6 is true. The values x=1x = 1, x=2x = 2 and x=3x = 3 are all positive; they would look correct to a student who forgets to flip the inequality and gets x≥0x \geq 0. For example, x=1x = 1 gives −8≥−2-8 \geq -2, which is false.
  17. A phone plan charges a monthly fee of $25 plus $0.15 per text message. If Maria wants to spend no more than $40 per month, what is the maximum number of text messages she can send?
    Answer: 100100 text messages Set up the inequality: 25+0.15x≤4025 + 0.15x \leq 40, where xx is the number of text messages. Solving: 0.15x≤150.15x \leq 15, so x≤100x \leq 100. Therefore, the maximum is 100 text messages. Choice B (99) assumes strict inequality. Choice C (101) would exceed the budget. Choice D (266) incorrectly uses 40÷0.1540 \div 0.15.
  18. The figure shows a triangle with sides of length xx, x+4x + 4, and 1111. Using the triangle inequality theorem, what is the complete set of possible values of xx?
    Answer: x>3.5x > 3.5 The triangle inequality says each pair of sides must add up to more than the third side. Check all three pairs. First, x+(x+4)>11x + (x + 4) > 11 gives 2x>72x > 7, so x>3.5x > 3.5. Second, x+11>x+4x + 11 > x + 4 simplifies to 11>411 > 4, which is always true. Third, (x+4)+11>x(x + 4) + 11 > x simplifies to 15>015 > 0, also always true. Only the first condition limits xx, so the complete set is x>3.5x > 3.5. The upper bound in 3.5<x<73.5 < x < 7 is invented; for example, x=10x = 10 gives sides 10, 14 and 11, a valid triangle. The bound x<15x < 15 comes from misreading the always-true third inequality, and 0<x<150 < x < 15 also drops the condition x>3.5x > 3.5.
  19. The number line shown represents the solution set to which of the following inequalities?
    Answer: −3(2x−4)≥18-3(2x - 4) \geq 18 The number line shows x≤−1x \leq -1 with a closed circle at −1-1. Solving (A): −3(2x−4)≥18⇒−6x+12≥18⇒−6x≥6⇒x≤−1-3(2x-4) \geq 18 \Rightarrow -6x + 12 \geq 18 \Rightarrow -6x \geq 6 \Rightarrow x \leq -1 ✓. (B) gives x≥−1x \geq -1 (wrong direction). (C) gives −2x≥6⇒x≤−3-2x \geq 6 \Rightarrow x \leq -3 (wrong endpoint). (D) gives x≥−1x \geq -1 (wrong direction).
  20. A taxi charges $3.50 for the first mile and $1.25 for each additional mile. If someone has $20 to spend, what is the maximum distance they can travel?
    Answer: 14.214.2 miles The cost is 3.50+1.25(x−1)≤203.50 + 1.25(x-1) \leq 20 for xx miles. Simplifying: 3.50+1.25x−1.25≤203.50 + 1.25x - 1.25 \leq 20, so 2.25+1.25x≤202.25 + 1.25x \leq 20. Thus 1.25x≤17.751.25x \leq 17.75, giving x≤14.2x \leq 14.2 miles. Choice A uses 20÷1.2520 \div 1.25. Choice B uses (20−3.50)÷1.25(20-3.50) \div 1.25. Choice D incorrectly adds instead of subtracts the first mile adjustment.

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