Linear Inequalities - Accuplacer Qas

Which of the following represents the solution to $-2(3x + 1) \geq 4x - 10$ on a number line?
Answer: A closed circle at $\frac{4}{5}$ with shading to the left When you encounter an inequality with variables on both sides, your goal is to isolate the variable while carefully tracking the inequality direction. The key moment comes when you multiply or divide by a negative number—this flips the inequality sign. Let's solve $-2(3x + 1) \geq 4x - 10$ step by step. First, distribute: $-6x - 2 \geq 4x - 10$ . Next, collect all terms with $x$ on one side: $-6x - 4x \geq -10 + 2$ , which gives us $-10x \geq -8$ . Now comes the critical step: divide both sides by $-10$ . Since we're dividing by a negative number, we must flip the inequality sign: $x \leq \frac{8}{10} = \frac{4}{5}$ . Since our solution is $x \leq \frac{4}{5}$ , we need a closed circle at $\frac{4}{5}$ (because the inequality includes "equal to") with shading to the left (representing all values less than or equal to $\frac{4}{5}$ ). This matches choice C. Choice A uses an open circle, which would only be correct for strict inequalities (< or >). Choice B has the wrong shading direction—it would represent $x \geq \frac{4}{5}$ . Choice D places the boundary at the wrong location ( $-\frac{4}{5}$ instead of $\frac{4}{5}$ ), likely from sign errors during solving. Study tip: Always double-check your work by testing a value from your solution region in the original inequality. If $x = 0$ (which is ≤ $\frac{4}{5}$ ), substituting confirms our answer is correct.
If $k$ is a negative constant, which of the following inequalities is equivalent to $kx < 10$ ?
Answer: $x > \frac{10}{k}$ To solve the inequality $kx < 10$ for $x$ , you must divide both sides by $k$ . The problem states that $k$ is a negative constant. A fundamental rule of inequalities is that when you multiply or divide both sides by a negative number, you must reverse the direction of the inequality sign. Therefore, dividing by the negative constant $k$ changes the ' $<$ ' to a ' $>$ '. The result is $x > \frac{10}{k}$ .
What is the solution to the inequality $\frac{2}{3}(x - 6) \le x + 4$ ?
Answer: $x \ge -24$ To solve this inequality, first eliminate the fraction by multiplying both sides by 3: $3 \cdot \frac{2}{3}(x - 6) \le 3(x + 4)$ , which simplifies to $2(x - 6) \le 3(x + 4)$ . Next, distribute on both sides: $2x - 12 \le 3x + 12$ . Then, gather the $x$ terms. Subtract $2x$ from both sides: $-12 \le x + 12$ . Finally, subtract 12 from both sides: $-24 \le x$ . This is equivalent to $x \ge -24$ .
What is the solution set for the inequality $3(2x + 5) - 10 < 6x + 7$ ?
Answer: All real numbers First, distribute the 3 on the left side: $6x + 15 - 10 < 6x + 7$ . Combine the constant terms on the left: $6x + 5 < 6x + 7$ . Next, subtract $6x$ from both sides: $5 < 7$ . The variable $x$ has been eliminated, and the resulting statement, $5 < 7$ , is always true. This means that the original inequality is true for any real number value of $x$ . Therefore, the solution set is all real numbers.
A company's monthly profit, $P$ , is given by the formula $P = 150x - 45,000$ , where $x$ is the number of units sold. How many units must be sold for the company to have a profit of at least $15,000?
Answer: $x \ge 400$ The phrase 'at least $15,000' means the profit $P$ must be greater than or equal to 15,000. So we set up the inequality: $150x - 45,000 \ge 15,000$ . To solve for $x$ , first add 45,000 to both sides: $150x \ge 15,000 + 45,000$ , which simplifies to $150x \ge 60,000$ . Finally, divide both sides by 150: $x \ge \frac{60,000}{150}$ , which gives $x \ge 400$ . The company must sell at least 400 units.
Which of the following represents the solution to $7 - 2(x - 4) > 5(3 - x)$ ?
Answer: $x > 0$ First, distribute on both sides of the inequality: $7 - 2x + 8 > 15 - 5x$ . Combine the constant terms on the left side: $15 - 2x > 15 - 5x$ . Subtract 15 from both sides: $-2x > -5x$ . To solve for $x$ , add $5x$ to both sides: $3x > 0$ . Finally, divide by 3: $x > 0$ .
Solve the inequality $3(x - 4) > 2x + 8$ and express the solution in interval notation.
Answer: $(20, \infty)$ When you encounter linear inequalities, your goal is to isolate the variable while carefully handling the inequality sign. The process mirrors solving equations, with one crucial exception regarding multiplication or division by negative numbers. Start by distributing and simplifying: $3(x - 4) > 2x + 8$ becomes $3x - 12 > 2x + 8$ . Next, collect like terms by subtracting $2x$ from both sides: $3x - 2x - 12 > 8$ , which gives you $x - 12 > 8$ . Finally, add 12 to both sides: $x > 20$ . In interval notation, this is $(20, \infty)$ , making D correct. Let's examine why the other answers are wrong. Choice A, $(-\infty, 4)$ , represents $x < 4$ , which would come from incorrectly flipping the inequality sign without justification. Choice B, $(-\infty, 20)$ , represents $x < 20$ , suggesting you flipped the inequality sign when you shouldn't have—remember, you only flip when multiplying or dividing by a negative number. Choice C, $(4, \infty)$ , represents $x > 4$ , which appears to come from an algebraic error, possibly incorrectly combining the constants or making a sign error during distribution. The key strategy for inequality problems is to treat them exactly like equations until you need to multiply or divide by a negative number—only then do you flip the inequality sign. Always double-check your final answer by testing a value from your solution set in the original inequality.
The solution to the inequality $|2x - 6| \leq 8$ can be written as $a \leq x \leq b$ . What is the value of $a + b$ ?
Answer: 6 When you encounter absolute value inequalities, remember that the absolute value represents distance on a number line. The inequality $|2x - 6| \leq 8$ means the expression inside the absolute value bars must be within 8 units of zero. To solve this, you need to consider both cases of the absolute value. The inequality $|2x - 6| \leq 8$ is equivalent to the compound inequality: $-8 \leq 2x - 6 \leq 8$ Now solve both parts simultaneously. Add 6 to all three parts: $-8 + 6 \leq 2x - 6 + 6 \leq 8 + 6$ $-2 \leq 2x \leq 14$ Divide everything by 2: $-1 \leq x \leq 7$ Therefore, $a = -1$ and $b = 7$ , so $a + b = -1 + 7 = 6$ . Looking at the wrong answers: Choice A) 4 likely comes from forgetting the negative solution and only considering the positive boundary. Choice B) 10 might result from adding the absolute values of the boundaries (|-1| + |7| = 8, though this gives 8, not 10). Choice D) 8 could come from mistakenly using the original constraint value or adding the distances incorrectly. The key strategy here is to always convert absolute value inequalities into compound inequalities. Remember that $|expression| \leq k$ becomes $-k \leq expression \leq k$ , while $|expression| \geq k$ becomes two separate inequalities. This systematic approach prevents common sign errors and ensures you capture both boundaries of the solution.
Given that $x < y$ . If both sides of the inequality are multiplied by a constant $c$ , the resulting inequality is $cx > cy$ . Which of the following must be true about $c$ ?
Answer: $c < 0$ The original inequality is $x < y$ . After multiplying by $c$ , the inequality becomes $cx > cy$ . The inequality symbol has been reversed (from less than to greater than). The rule for inequalities states that the symbol is reversed only when multiplying or dividing both sides by a negative number. Therefore, the constant $c$ must be negative ( $c < 0$ ).
A rental car company charges a base fee of $45 plus $0.25 per mile driven. If Marcus wants to spend at most $120 on his rental, what inequality represents the maximum number of miles $m$ he can drive?
Answer: $45 + 0.25m \leq 120$ The total cost is the base fee ($45) plus the cost per mile ($0.25m). Since Marcus wants to spend 'at most' $120, the total cost must be less than or equal to $120, giving us $45 + 0.25m ≤ 120$ . Choice B uses the wrong inequality direction. Choice C incorrectly subtracts the base fee. Choice D reverses the coefficients.
If $3x + 2y \leq 12$ and $y \geq 0$ , what is the maximum value of $x$ when $y = 1$ ?
Answer: $\frac{10}{3}$ When you encounter a system of inequalities with constraints, you're looking for values that satisfy all given conditions simultaneously. Here, you need to find the maximum value of $x$ when $y = 1$ , subject to the constraints $3x + 2y \leq 12$ and $y \geq 0$ . Since $y = 1$ and $y \geq 0$ , the second constraint is automatically satisfied. Now substitute $y = 1$ into the first inequality: $3x + 2(1) \leq 12$ , which simplifies to $3x + 2 \leq 12$ . Subtracting 2 from both sides gives $3x \leq 10$ , so $x \leq \frac{10}{3}$ . The maximum value of $x$ occurs when this inequality becomes an equality, making $x = \frac{10}{3}$ . Looking at the wrong answers: Choice A ( $4$ ) exceeds the maximum allowed value since $4 > \frac{10}{3} \approx 3.33$ . If $x = 4$ , then $3(4) + 2(1) = 14 > 12$ , violating the constraint. Choice C ( $5$ ) is even larger and would give $3(5) + 2(1) = 17 > 12$ . Choice D ( $\frac{14}{3}$ ) would yield $3(\frac{14}{3}) + 2(1) = 14 + 2 = 16 > 12$ , also violating the constraint. For inequality problems with constraints, always substitute the given values first, then solve for the remaining variable. Remember that "maximum value" in a constraint problem means finding the boundary where the inequality becomes an equality.
A rental company charges a flat fee of $40 plus $0.15 per mile. A second company charges a flat fee of $25 plus $0.25 per mile. For how many miles, $m$ , is the cost of the first company less than or equal to the cost of the second company?
Answer: $m \ge 150$ Let $m$ be the number of miles. The cost for the first company is $40 + 0.15m$ . The cost for the second company is $25 + 0.25m$ . We want to find when the first company's cost is less than or equal to the second's: $40 + 0.15m \le 25 + 0.25m$ . Subtract (0.15m) from both sides: $40 \le 25 + 0.10m$ . Subtract 25 from both sides: $15 \le 0.10m$ . Divide both sides by 0.10: $150 \le m$ , which is the same as $m \ge 150$ .
For which of the following inequalities is $x = -3$ a solution?
Answer: $2(x + 5) > 4x + 15$ Substitute $x = -3$ into each inequality to see which one results in a true statement. A) $2(-3 + 5) > 4(-3) + 15 \Rightarrow 2(2) > -12 + 15 \Rightarrow 4 > 3$ . This is true. B) $1 - 3(-3) < 4(-3 - 5) \Rightarrow 1 + 9 < 4(-8) \Rightarrow 10 < -32$ . This is false. C) $\frac{1}{2}(-3) + 6 \le 4 \Rightarrow -1.5 + 6 \le 4 \Rightarrow 4.5 \le 4$ . This is false. D) $5 - (-3) > 2(-3) + 14 \Rightarrow 8 > -6 + 14 \Rightarrow 8 > 8$ . This is false. Therefore, only the inequality in choice A is satisfied by $x = -3$ .
Which of the following values is in the solution set of the compound inequality $-7 < 3 - 2x \le 5$ ?
Answer: -1 To solve the compound inequality, perform operations on all three parts. First, subtract 3 from all parts: $-7 - 3 < 3 - 2x - 3 \le 5 - 3$ , which simplifies to $-10 < -2x \le 2$ . Next, divide all parts by -2. Remember to reverse the inequality signs when dividing by a negative number: $\frac{-10}{-2} > \frac{-2x}{-2} \ge \frac{2}{-2}$ , which simplifies to $5 > x \ge -1$ . This is equivalent to $-1 \le x < 5$ . Of the choices, only -1 is in this interval. The value 5 is not included because the inequality is strictly less than 5.
What is the smallest integer $n$ for which $12 - 5n \le -13$ ?
Answer: 5 First, solve the inequality for $n$ . Subtract 12 from both sides: $-5n \le -13 - 12$ , which simplifies to $-5n \le -25$ . Next, divide both sides by -5. Remember to reverse the inequality sign when dividing by a negative number: $n \ge \frac{-25}{-5}$ , which simplifies to $n \ge 5$ . The question asks for the smallest integer that satisfies this condition. The integers that are greater than or equal to 5 are 5, 6, 7, and so on. The smallest of these is 5.
Solve the compound inequality $-4 < 2x - 6 \leq 8$ and express the solution in interval notation.
Answer: $(1, 7]$ When you encounter a compound inequality like this one, you're dealing with two inequalities joined together that must both be satisfied simultaneously. Think of it as finding values of x that fall within a specific range. To solve $-4 < 2x - 6 \leq 8$ , you need to isolate x by performing the same operations on all three parts. First, add 6 to each section: $-4 + 6 < 2x - 6 + 6 \leq 8 + 6$ , which gives you $2 < 2x \leq 14$ . Next, divide everything by 2: $1 < x \leq 7$ . This means x is greater than 1 but less than or equal to 7. In interval notation, a parenthesis indicates the endpoint is not included, while a bracket means it is included. Since x > 1 (not x ≥ 1), you use a parenthesis at 1. Since x ≤ 7, you use a bracket at 7. Therefore, the answer is $(1, 7]$ , which is choice B. Choice A uses $[1, 7)$ , incorrectly including 1 and excluding 7 — the opposite of what the inequality shows. Choice C uses $(1, 7)$ , correctly excluding 1 but wrongly excluding 7 as well. Choice D uses $[1, 7]$ , incorrectly including both endpoints when only 7 should be included. Remember: when translating inequality symbols to interval notation, < and > become parentheses, while ≤ and ≥ become brackets. Always double-check which endpoints your solution actually includes by looking at the inequality symbols carefully.
A student has a budget of $100 to buy notebooks and pens. Notebooks cost $4 each and pens cost $2.50 each. The student must buy exactly 10 notebooks. What is the maximum number of pens the student can buy?
Answer: 24 First, calculate the cost of the notebooks. The student buys 10 notebooks at $4 each, so the cost is $10 \times \$4 = \$40$ . The remaining budget for pens is $\$100 - \$40 = \$60$ . Let $p$ be the number of pens. The cost of the pens is $2.50p$ . This cost must be less than or equal to the remaining budget: $2.50p \le 60$ . To find $p$ , divide both sides by 2.50: $p \le \frac{60}{2.50}$ , which is $p \le 24$ . The maximum number of pens the student can buy is 24.
What is the solution set for the inequality $-1 \le \frac{4-x}{3} \le 2$ ?
Answer: $-2 \le x \le 7$ To solve this compound inequality, apply each operation to all three parts. First, multiply all parts by 3 to clear the fraction: $-3 \le 4 - x \le 6$ . Next, subtract 4 from all parts: $-3 - 4 \le 4 - x - 4 \le 6 - 4$ , which simplifies to $-7 \le -x \le 2$ . Finally, multiply all parts by -1 to isolate $x$ . Remember to reverse the inequality signs when multiplying by a negative number: $7 \ge x \ge -2$ . This is equivalent to $-2 \le x \le 7$ .
Which of the following best describes the solution set for the inequality $4(x - 3) + 7 \ge 2(2x + 1)$ ?
Answer: There is no solution. First, use the distributive property on both sides of the inequality: $4x - 12 + 7 \ge 4x + 2$ . Combine the constant terms on the left side: $4x - 5 \ge 4x + 2$ . Next, subtract $4x$ from both sides: $-5 \ge 2$ . This statement is false. Since the variable $x$ has been eliminated and the resulting statement is a contradiction, there is no value of $x$ that can make the original inequality true. Therefore, there is no solution.
Which of the following describes the solution to the inequality $3(1 - 2x) \ge 2(x + 13)$ ?
Answer: $x \le -\frac{23}{8}$ First, distribute on both sides of the inequality: $3 - 6x \ge 2x + 26$ . Next, gather the $x$ terms on one side and the constants on the other. Subtract $2x$ from both sides: $3 - 8x \ge 26$ . Subtract 3 from both sides: $-8x \ge 23$ . Finally, divide both sides by -8. Because you are dividing by a negative number, you must reverse the inequality sign: $x \le -\frac{23}{8}$ .

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