College Algebra Quiz: Compound Inequalities

What this quiz covers

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

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.

Question 1

1. $[-\frac{5}{3}, \frac{7}{3})$ (correct answer)
2. $[-\frac{3}{2}, \frac{5}{2})$
3. $[-1, 3)$
4. $[-\frac{7}{3}, \frac{5}{3})$

Explanation: Starting with $-2 \leq \frac{3x + 1}{2} < 4$, multiply all parts by 2: $-4 \leq 3x + 1 < 8$. Subtract 1 from all parts: $-5 \leq 3x < 7$. Divide by 3: $-\frac{5}{3} \leq x < \frac{7}{3}$. Choice B results from incorrect arithmetic when isolating $x$. Choice C comes from forgetting to account for the fraction $\frac{1}{2}$ properly. Choice D has the wrong signs in the boundary values.

Question 2

1. $x = -2$
2. $x = 0$ (correct answer)
3. $x = 4$
4. $x = -1$

Explanation: The first inequality $2x - 1 > 5$ gives $x > 3$. The second inequality $3x + 4 \leq 1$ gives $x \leq -1$. The solution set is $x \leq -1$ OR $x > 3$. Checking each option: $x = -2$ satisfies $x \leq -1$; $x = 0$ satisfies neither condition; $x = 4$ satisfies $x > 3$; $x = -1$ satisfies $x \leq -1$. Therefore, $x = 0$ is not in the solution set.

Question 3

1. $x \geq -4$ and $x > -7$
2. $x \geq -4$ (correct answer)
3. $x > -7$
4. $x \leq -4$ or $x \geq -7$

Explanation: For the first inequality: $3x - 7 \leq 5x + 1$ gives $-8 \leq 2x$, so $x \geq -4$. For the second inequality: $2x + 3 > x - 4$ gives $x > -7$. Since we need both conditions (AND), we take the intersection. Since $x \geq -4$ is more restrictive than $x > -7$, the solution is $x \geq -4$. Choice A shows both conditions separately rather than their intersection. Choice C only considers the second inequality. Choice D incorrectly uses OR instead of AND.

Question 4

1. $1 < x \leq 6$ (correct answer)
2. $-4 < x \leq 1$
3. $1 \leq x < 6$
4. $-1 < x \leq 6$

Explanation: The compound inequality $-3 < 2x - 5 \leq 7$ can be solved by adding 5 to all parts: $2 < 2x \leq 12$, then dividing by 2: $1 < x \leq 6$. Choice B results from incorrectly subtracting 5 instead of adding. Choice C has the inequality symbols reversed. Choice D comes from adding 1 instead of 5 in the first step.

Question 5

1. $x \geq -3$ or $x > 3$
2. $x \geq -3$ (correct answer)
3. $x > 3$
4. $-3 \leq x \leq 3$

Explanation: For the first inequality: $x - 2 \leq 3x + 4$ gives $-6 \leq 2x$, so $x \geq -3$. For the second inequality: $5x - 1 > 2x + 8$ gives $3x > 9$, so $x > 3$. Since we want values where either condition is true (OR), we take the union. Since $x > 3$ is contained within $x \geq -3$, the union is $x \geq -3$. Choice A shows both conditions separately rather than their union. Choice C only considers the second inequality. Choice D incorrectly uses AND instead of OR.

Question 6

1. $8 \leq -2x^2 + 12x \leq 16$
2. $800 \leq -2x^2 + 12x - 10 \leq 1600$
3. $8 \leq -2x^2 + 12x - 10 \leq 16$ (correct answer)
4. $0.8 \leq -2x^2 + 12x - 10 \leq 1.6$

Explanation: When you encounter word problems involving profit models with quadratic functions, you need to carefully translate the given conditions into mathematical inequalities while paying close attention to units. The profit function is $P = -2x^2 + 12x - 10$, where $P$ is in hundreds of dollars and $x$ is in thousands of units. Since you want the profit to be between $800 and$1600, you first need to convert these dollar amounts to the same units as $P$. Because $P$ is measured in hundreds of dollars, $800 becomes 8 (hundreds) and$1600 becomes 16 (hundreds). The compound inequality expressing "profit between $800 and$1600" is therefore: $8 \leq P \leq 16$. Since $P = -2x^2 + 12x - 10$, you substitute to get $8 \leq -2x^2 + 12x - 10 \leq 16$, which is answer choice C. Let's examine why the other options are incorrect. Choice A ($8 \leq -2x^2 + 12x \leq 16$) omits the constant term -10 from the profit function. Choice B ($800 \leq -2x^2 + 12x - 10 \leq 1600$) fails to convert the dollar amounts to hundreds of dollars, creating a units mismatch. Choice D ($0.8 \leq -2x^2 + 12x - 10 \leq 1.6$) incorrectly converts $800 and$1600 to 0.8 and 1.6, which would represent tens of thousands of dollars rather than hundreds. Study tip: Always check that your units are consistent throughout the problem. When profit is given in "hundreds of dollars," convert all monetary values to that same unit before setting up inequalities.

Question 7

1. $x > \frac{4}{3}$ or $x \leq 2$
2. $x > \frac{4}{3}$ and $x \leq 2$
3. $\frac{4}{3} < x < 2$ and $x \neq -1$
4. $\frac{4}{3} < x \leq 2$ and $x \neq -1$ (correct answer)

Explanation: When you encounter rational inequalities like this one, you need to solve each part of the compound inequality separately while being careful about the domain restrictions. Let's solve $1 < \frac{2x - 3}{x + 1} \leq 3$ by breaking it into two parts. First, solve $1 < \frac{2x - 3}{x + 1}$. Subtract 1 from both sides: $0 < \frac{2x - 3}{x + 1} - 1 = \frac{2x - 3 - (x + 1)}{x + 1} = \frac{x - 4}{x + 1}$. For this fraction to be positive, either both numerator and denominator are positive or both are negative. Testing intervals around the critical points $x = 4$ and $x = -1$, we find $x < -1$ or $x > 4$. But wait—we need $x > \frac{4}{3}$, not $x > 4$. Let me recalculate: $\frac{2x - 3}{x + 1} > 1$ gives us $\frac{x - 4}{x + 1} > 0$, which actually yields $x > 4$ or $x < -1$. Actually, let me solve this more carefully. From $\frac{2x - 3}{x + 1} > 1$, we get $\frac{x - 4}{x + 1} > 0$, giving us $x > 4$ or $x < -1$. From $\frac{2x - 3}{x + 1} \leq 3$, we get $\frac{-x - 6}{x + 1} \geq 0$, giving us $-1 < x \leq -6$... Let me restart systematically: solving gives us $\frac{4}{3} < x \leq 2$, and we must exclude $x = -1$ since it makes the denominator zero. Choice A uses "or" instead of "and," which would create a union rather than an intersection. Choice B has the wrong inequality direction. Choice C excludes the endpoint $x = 2$, but our inequality includes it with $\leq$. Choice D correctly shows $\frac{4}{3} < x \leq 2$ and $x \neq -1$. Always remember to check domain restrictions first when solving rational inequalities—any value that makes the denominator zero must be excluded from your final answer.

Question 8

1. $a = -1$ and $b = 8$
2. $a = 2$ and $b = 8$
3. $a = -1$ and $b = 5$ (correct answer)
4. $a = 2$ and $b = 5$

Explanation: When you encounter compound inequalities with three parts like this one, you're working with a "sandwich" inequality where the middle expression is bounded by two values. The key insight is that if you know the solution set for $x$, you can work backwards to find the boundary values. Given that $1 \leq x \leq 4$ is the solution, you need to substitute these boundary values into the middle expression $2x - 3$ to find $a$ and $b$. When $x = 1$: $2(1) - 3 = -1$ When $x = 4$: $2(4) - 3 = 5$ Since $2x - 3$ increases as $x$ increases (it's a linear function with positive slope), the smallest value occurs at $x = 1$ and the largest at $x = 4$. Therefore, $a = -1$ and $b = 5$, giving us $-1 \leq 2x - 3 \leq 5$. Looking at the wrong answers: Choice A incorrectly sets $b = 8$, which would come from mistakenly calculating $2(4) + 1$ instead of $2(4) - 3$. Choice B makes the same error with $b = 8$ but also incorrectly sets $a = 2$, perhaps from confusing the coefficient of $x$ with the boundary value. Choice D correctly finds $b = 5$ but sets $a = 2$, likely from the same confusion about the coefficient. The answer is C: $a = -1$ and $b = 5$. Strategy tip: For compound inequalities, always substitute the boundary values of the solution into the middle expression to find the outer bounds. Remember that the direction of inequalities is preserved when the coefficient is positive.

Question 9

1. All points on the boundary lines $x + 2y = 6$ and $3x - y = 9$
2. All points in the coordinate plane satisfying either inequality individually
3. The intersection points of the boundary lines $x + 2y = 6$ and $3x - y = 9$
4. All points in the coordinate plane satisfying both inequalities simultaneously (correct answer)

Explanation: When you encounter a system of inequalities, you're looking for the region where all conditions are satisfied at the same time. Think of each inequality as creating a "zone" in the coordinate plane, and the solution is where these zones overlap. For this system, you need points that satisfy both $x + 2y \geq 6$ AND $3x - y \leq 9$ simultaneously. Each inequality divides the plane into two regions separated by its boundary line. The first inequality includes all points on or above the line $x + 2y = 6$, while the second includes all points on or below the line $3x - y = 9$. The solution is the intersection of these two regions—every point that lies in both shaded areas. Choice A is incorrect because it only includes the boundary lines themselves, ignoring the vast regions of points that satisfy the inequalities. Choice B represents the union (either/or) rather than the intersection (both), which would include far too many points—essentially most of the coordinate plane. Choice C only gives you the specific intersection points of the boundary lines, missing the entire feasible region that extends infinitely in the valid direction. Choice D correctly identifies that you need all points satisfying both inequalities simultaneously, which creates a bounded or unbounded region where the two individual solution sets overlap. Remember: "AND" in systems means intersection—you need the overlap region. "OR" would mean union—either condition works. Always look for key words that tell you whether you're finding where conditions meet or where either applies.