HSPT Quiz: Solve Linear Equations

What this quiz covers

This quiz focuses on Solve Linear Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for HSPT.

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. $x=\dfrac85$ (correct answer)
2. $x=\dfrac{13}{5}$
3. $x=\dfrac{5}{8}$
4. $x=-\dfrac85$

Explanation: When you encounter a linear equation with parentheses and variables on both sides, your goal is to systematically isolate the variable using the order of operations in reverse. Start by distributing through the parentheses. On the left side: $5-2(x-4) = 5-2x+8 = 13-2x$. On the right side: $3(x+1)+2 = 3x+3+2 = 3x+5$. Your equation becomes $13-2x = 3x+5$. Next, collect all variable terms on one side and constants on the other. Add $2x$ to both sides: $13 = 5x+5$. Subtract 5 from both sides: $8 = 5x$. Finally, divide by 5: $x = \frac{8}{5}$. Looking at the wrong answers: Choice B gives $x = \frac{13}{5}$, which likely results from incorrectly handling the distribution or making sign errors when moving terms. Choice C gives $x = \frac{5}{8}$, which happens when you accidentally flip the fraction at the end—dividing 5 by 8 instead of 8 by 5. Choice D gives $x = -\frac{8}{5}$, which occurs from sign errors, particularly when distributing the negative through $-2(x-4)$ or when moving terms across the equals sign. The correct answer is A: $x = \frac{8}{5}$. Study tip: When solving multi-step equations, work methodically through each step and double-check your distribution of negative signs. Always verify your answer by substituting it back into the original equation—both sides should equal the same value.

Question 2

1. $8n+4\le60,\; n=7$ (correct answer)
2. $8n+4\ge60,\; n=6$
3. $8n-4\le60,\; n=8$
4. $8n+4<60,\; n=6$

Explanation: When you encounter word problems involving costs and spending limits, you're dealing with linear inequalities. The key is translating the written constraints into mathematical expressions that capture both the cost structure and the spending limitation. Here, each ticket costs $8, so $$n$ $tickets cost$ $8n$ $dollars. There's also a one-time$4 service fee regardless of how many tickets Shawn buys. This means his total cost is $8n + 4$. Since he has "at most $60" to spend, his total cost must be less than or equal to$60, giving us $8n + 4 \leq 60$. To find the maximum number of tickets, solve: $8n + 4 \leq 60$, so $8n \leq 56$, which means $n \leq 7$. The greatest whole number satisfying this is $n = 7$. Looking at the wrong answers: Choice B uses $\geq$ instead of $\leq$, which would mean Shawn must spend at least $60 rather than at most$60. Choice C subtracts the service fee ($8n - 4$) instead of adding it, misrepresenting the cost structure. Choice D uses a strict inequality ($<$) rather than "less than or equal to" ($\leq$), which incorrectly excludes the possibility of spending exactly $60, leading to$ $n = 6$ $ instead of the correct maximum of 7. Remember that "at most" translates to $$\leq$$, "at least" means $$\geq$$, and fixed fees are always added to variable costs. Double-check your inequality direction by asking whether the constraint is an upper limit or lower limit on spending.

Question 3

1. $x \le -0.5$ (correct answer)
2. $x \ge -0.5$
3. $x \le 0.5$
4. $x \le -2$

Explanation: First, distribute the constants on both sides of the inequality: $1.5 - 2x + 3 \ge 2x + 6.5$. Combine like terms on the left side: $4.5 - 2x \ge 2x + 6.5$. To gather the $x$ terms, add $2x$ to both sides: $4.5 \ge 4x + 6.5$. To isolate the $x$ term, subtract 6.5 from both sides: $-2 \ge 4x$. Finally, divide by 4: $-0.5 \ge x$. This is equivalent to $x \le -0.5$. Distractor B results from incorrectly flipping the inequality sign. Distractor D results from a sign error when distributing -0.5 to -6.

Question 4

1. $3(s + 45) = 285$ (correct answer)
2. $3s + 45 = 285$
3. $3(s - 45) = 285$
4. $s + 45 = 285/3$

Explanation: The distance formula is distance = rate × time. The first car travels for 3 hours at 45 mph, covering a distance of $45 \times 3$ miles. The second car travels for 3 hours at $s$ mph, covering a distance of $s \times 3$ miles. Since they are traveling in opposite directions, their distances from the starting point add up to the total distance between them. Therefore, $3 \times 45 + 3s = 285$. Factoring out the common factor of 3 gives $3(45 + s) = 285$, which is equivalent to $3(s + 45) = 285$. Choice D is the same equation, but it represents the next step in solving the problem, not the initial setup based on the combined rate. Choice A is the best representation of the initial physical situation, where the combined speed of separation is $(s+45)$ mph.

Question 5

1. 6 (correct answer)
2. 9
3. 12
4. 15

Explanation: When you encounter absolute value equations, remember that the expression inside the absolute value bars can be either positive or negative, leading to multiple solutions. Start by isolating the absolute value term. From $2|x - 3| + 5 = 17$, subtract 5 from both sides: $2|x - 3| = 12$. Then divide by 2: $|x - 3| = 6$. Now solve the two cases. Since $|x - 3| = 6$, either:

• Case 1: $x - 3 = 6$, so $x = 9$
• Case 2: $x - 3 = -6$, so $x = -3$

Question 6

1. The number is -15
2. The number is -12 (correct answer)
3. The number is -9
4. The number is -6

Explanation: When you see a word problem asking you to find an unknown number, you're dealing with algebraic translation. The key is converting the English phrases into mathematical expressions, then solving the resulting equation. Let's call the unknown number $x$. "A number decreased by 15" translates to $x - 15$. "3 times the number increased by 9" means $3x + 9$. Since these expressions are equal, you get: $x - 15 = 3x + 9$. To solve, collect like terms. Subtract $x$ from both sides: $-15 = 2x + 9$. Then subtract 9 from both sides: $-24 = 2x$. Finally, divide by 2: $x = -12$. You can verify this: $-12 - 15 = -27$ and $3(-12) + 9 = -36 + 9 = -27$ ✓ Looking at the wrong answers: Choice A (-15) likely comes from focusing too much on the "decreased by 15" part without solving the full equation. Choice C (-9) might result from confusing the constants in the problem or making sign errors. Choice D (-6) could come from arithmetic mistakes when combining terms or dividing incorrectly. The correct answer is B. Remember this translation pattern: "decreased by" means subtraction, "increased by" means addition, and "times" indicates multiplication. Always check your answer by substituting back into the original word problem to ensure both sides truly equal each other.

Question 7

1. Maria originally has 36 books (correct answer)
2. Maria originally has 42 books
3. Maria originally has 48 books
4. Maria originally has 54 books

Explanation: When you encounter word problems involving relationships between quantities that change, setting up equations with variables is your most reliable approach. These problems test your ability to translate verbal descriptions into mathematical expressions. Let's define variables: Let $C$ = Carlos's original number of books and $M$ = Maria's original number of books. From the problem, we know two key relationships. First, "Maria has 3 times as many books as Carlos," which gives us $M = 3C$. Second, "If Maria gives Carlos 12 books, they will have equal numbers," meaning Maria will have $M - 12$ books and Carlos will have $C + 12$ books, so $M - 12 = C + 12$. From the second equation: $M - 12 = C + 12$, so $M = C + 24$. Now substitute the first equation: $3C = C + 24$. Solving: $2C = 24$, so $C = 12$. Therefore, $M = 3(12) = 36$. Let's verify: Maria starts with 36, Carlos with 12. After the transfer, Maria has 24 and Carlos has 24. ✓ Choice A is correct with 36 books. Choice B (42 books) would mean Carlos originally had 14 books, but after the transfer they'd have 30 and 26 respectively—not equal. Choice C (48 books) implies Carlos had 16 originally, leading to 36 and 28 after transfer. Choice D (54 books) means Carlos had 18, resulting in 42 and 30 after transfer. Always verify your solution by checking it against both original conditions. This catches algebraic errors and confirms your setup was correct.

Question 8

1. $x = -50$
2. $x = -60$ (correct answer)
3. $x = -70$
4. $x = -80$

Explanation: When you encounter an equation with fractions containing the same variable, your goal is to combine like terms and isolate the variable. Start by finding a common denominator to simplify the fractions on the left side. To solve $\frac{x}{3} + \frac{x}{4} = \frac{x}{2} - 5$, first combine the left side using the common denominator 12: $\frac{4x}{12} + \frac{3x}{12} = \frac{7x}{12}$ Now your equation becomes: $\frac{7x}{12} = \frac{x}{2} - 5$ Convert $\frac{x}{2}$ to twelfths: $\frac{x}{2} = \frac{6x}{12}$ So: $\frac{7x}{12} = \frac{6x}{12} - 5$ Subtract $\frac{6x}{12}$ from both sides: $\frac{x}{12} = -5$ Multiply both sides by 12: $x = -60$ Choice B ($x = -60$) is correct. You can verify by substituting back: $\frac{-60}{3} + \frac{-60}{4} = -20 + (-15) = -35$, and $\frac{-60}{2} - 5 = -30 - 5 = -35$ ✓ Choice A ($x = -50$) gives you $-\frac{65}{6}$ on the left and $-30$ on the right—not equal. Choice C ($x = -70$) produces $-\frac{175}{12}$ versus $-40$—also unequal. Choice D ($x = -80$) yields $-\frac{200}{12}$ compared to $-45$—incorrect as well. When solving fraction equations, always find common denominators first, then systematically eliminate fractions by multiplying through. Double-check your answer by substituting back into the original equation—this catches arithmetic errors that are easy to make with fraction calculations.

Question 9

1. $a = -2$
2. $a = 0$
3. $a = 2$ (correct answer)
4. No value of $a$ exists.

Explanation: First, simplify the equation. Distribute the 2: $2ax + 6 - 4x = 6$. Subtract 6 from both sides: $2ax - 4x = 0$. Factor out $x$: $x(2a - 4) = 0$. For this equation to have infinitely many solutions, it must be true for any value of $x$. This occurs when the coefficient of $x$ is zero. So, we set $2a - 4 = 0$. Adding 4 to both sides gives $2a = 4$. Dividing by 2 gives $a = 2$. When $a=2$, the equation becomes $x(0)=0$, or $0=0$, which is always true.

Question 10

1. All real numbers (correct answer)
2. No solution
3. $y=3$
4. $y=-3$

Explanation: When you encounter an equation like this, you're dealing with a linear equation that might have one solution, no solution, or infinitely many solutions. The key is to simplify both sides and see what happens. Let's solve $5(2y-3)=10y-15$ step by step. First, distribute the 5 on the left side: $5(2y) - 5(3) = 10y - 15$, which gives us $10y - 15 = 10y - 15$. Notice that both sides are now identical. When you subtract $10y$ from both sides, you get $-15 = -15$, and when you add 15 to both sides, you get $0 = 0$. This is always true, regardless of what value you substitute for $y$. Let's check the answer choices: Choice A is correct because any real number you substitute for $y$ will make the original equation true. Choice B (no solution) would occur if we ended up with something like $0 = 5$, which is never true. Choice C ($y = 3$) suggests there's only one solution - you can verify this isn't the case by substituting other values like $y = 0$ or $y = 1$ into the original equation and seeing they work too. Choice D ($y = -3$) has the same problem as choice C - it's just one specific value when infinitely many work. Watch for this pattern on algebra problems: when both sides of an equation simplify to exactly the same expression, you have an identity that's true for all real numbers. Don't assume every equation has just one solution!

Question 11

1. $y=\dfrac{9}{5}$ (correct answer)
2. $y=\dfrac{7}{5}$
3. $y=\dfrac{11}{5}$
4. $y=2$

Explanation: When you encounter a linear equation with parentheses and multiple terms, your goal is to systematically simplify by following the order of operations, then isolate the variable. Start by distributing the 4 to both terms inside the parentheses: $4(2y-1) = 4 \cdot 2y - 4 \cdot 1 = 8y - 4$. The equation becomes: $8y - 4 - 3y = 5$. Next, combine like terms on the left side: $8y - 3y = 5y$, so you have $5y - 4 = 5$. Add 4 to both sides: $5y = 9$. Finally, divide both sides by 5: $y = \frac{9}{5}$. This confirms answer choice A is correct. Let's examine why the other answers are wrong. Answer B ($y = \frac{7}{5}$) likely results from incorrectly combining terms, perhaps getting $3y - 4 = 5$ instead of $5y - 4 = 5$. Answer C ($y = \frac{11}{5}$) might come from an error in distributing, like getting $6y - 4 = 5$ and solving for $y = \frac{9}{6} = \frac{11}{5}$ through incorrect arithmetic. Answer D ($y = 2$) could result from multiple errors, such as incorrectly distributing and then making computational mistakes. Always verify your answer by substituting back into the original equation. With $y = \frac{9}{5}$: $4(2 \cdot \frac{9}{5} - 1) - 3 \cdot \frac{9}{5} = 4(\frac{18}{5} - 1) - \frac{27}{5} = 4 \cdot \frac{13}{5} - \frac{27}{5} = \frac{52}{5} - \frac{27}{5} = \frac{25}{5} = 5$ ✓

Question 12

1. $x< -1$ (correct answer)
2. $x> -1$
3. $x<1$
4. $x>1$

Explanation: When you encounter linear inequalities, your goal is to isolate the variable on one side while maintaining the inequality's direction. The key difference from equations is that multiplying or dividing by a negative number flips the inequality sign. Starting with $7-2x>3x+12$, collect all terms with $x$ on one side and constants on the other. Subtract $3x$ from both sides: $7-2x-3x>12$, which gives $7-5x>12$. Next, subtract 7 from both sides: $-5x>5$. Now divide both sides by $-5$. Since you're dividing by a negative number, the inequality sign flips from "greater than" to "less than": $x<-1$. This matches choice A. Let's examine why the other answers are wrong. Choice B ($x>-1$) results from forgetting to flip the inequality sign when dividing by $-5$ — this is the most common error students make. Choice C ($x<1$) comes from sign errors during the algebraic manipulation, possibly treating $-2x$ as $+2x$ or making arithmetic mistakes with the constants. Choice D ($x>1$) combines both errors: incorrect algebraic manipulation AND forgetting to flip the inequality sign. You can verify your answer by testing a value: if $x=-2$ (which satisfies $x<-1$), then $7-2(-2)=11$ and $3(-2)+12=6$. Indeed, $11>6$. Remember: always flip the inequality sign when multiplying or dividing both sides by a negative number. This rule trips up many students, so double-check your work whenever negative coefficients appear.

Question 13

1. $x=\dfrac{23}{4}$ (correct answer)
2. $x=\dfrac{21}{4}$
3. $x=\dfrac{25}{4}$
4. $x=\dfrac{23}{5}$

Explanation: This equation tests your ability to solve linear equations with parentheses and variables on both sides. When you encounter equations like this, your goal is to systematically eliminate parentheses, combine like terms, and isolate the variable. Start by distributing on both sides: $3(2x-5)+4 = 2(x+7)-2$ becomes $6x-15+4 = 2x+14-2$. Simplifying the constants gives you $6x-11 = 2x+12$. Next, collect all terms with $x$ on one side and constants on the other. Subtract $2x$ from both sides: $4x-11 = 12$. Add 11 to both sides: $4x = 23$. Finally, divide by 4: $x = \frac{23}{4}$. Choice A, $x=\frac{23}{4}$, is correct. Choice B, $x=\frac{21}{4}$, likely results from an arithmetic error when combining constants—perhaps getting $4x = 21$ instead of $4x = 23$. Choice C, $x=\frac{25}{4}$, could come from incorrectly handling the signs when distributing, leading to $4x = 25$. Choice D, $x=\frac{23}{5}$, suggests an error in collecting like terms, possibly getting $5x = 23$ instead of $4x = 23$. Remember to work methodically: distribute first, combine like terms on each side, then move all variable terms to one side and constants to the other. Double-check your arithmetic at each step, especially when dealing with negative numbers and distribution. These multi-step equations reward careful, systematic work over speed.

Question 14

1. $k=\dfrac12$ (correct answer)
2. $k=2$
3. $k=-\dfrac12$
4. $k=-2$

Explanation: This equation involves a fraction set equal to an expression, so you'll need to eliminate the fraction and solve systematically. The key is to clear the denominator first, then collect like terms. Start by multiplying both sides by 3 to eliminate the fraction: $3 \cdot \frac{5-k}{3} = 3(2-k)$. This gives you $5-k = 6-3k$. Now you have a standard linear equation without fractions. Next, collect the $k$ terms on one side. Add $3k$ to both sides: $5-k+3k = 6-3k+3k$, which simplifies to $5+2k = 6$. Subtract 5 from both sides: $2k = 1$. Finally, divide by 2: $k = \frac{1}{2}$. Let's examine why the other answers are incorrect. Choice B ($k=2$) would result if you made an error when collecting terms, perhaps forgetting to distribute the negative sign properly. Choice C ($k=-\frac{1}{2}$) comes from a sign error, likely when moving terms from one side to the other. Choice D ($k=-2$) suggests multiple computational mistakes, possibly in both the fraction elimination and variable collection steps. You can verify: substituting $k=\frac{1}{2}$ gives $\frac{5-\frac{1}{2}}{3} = \frac{\frac{9}{2}}{3} = \frac{3}{2}$, and $2-\frac{1}{2} = \frac{3}{2}$ ✓. When solving equations with fractions, always clear denominators first by multiplying both sides by the denominator. This prevents fraction arithmetic errors and makes the algebra much cleaner.

Question 15

1. $n=\dfrac13$ (correct answer)
2. $n=-\dfrac13$
3. $n=1$
4. $n=-1$

Explanation: This is a linear equation that requires careful distribution and combining of like terms. When you encounter equations with parentheses on both sides, your goal is to systematically simplify each side, then isolate the variable. Start by distributing on the left side: $3(2n-1) = 6n - 3$ and $-(n+4) = -n - 4$, so the left side becomes $6n - 3 - n - 4 = 5n - 7$. On the right side: $2(n-3) = 2n - 6$. Your equation is now $5n - 7 = 2n - 6$. Subtract $2n$ from both sides: $3n - 7 = -6$. Add 7 to both sides: $3n = 1$. Therefore, $n = \frac{1}{3}$, which is choice A. Let's check why the other answers are wrong. Choice B ($n = -\frac{1}{3}$) likely comes from a sign error when distributing or combining terms. Choice C ($n = 1$) might result from incorrectly handling the fraction $\frac{1}{3}$ or making arithmetic errors in the final steps. Choice D ($n = -1$) could come from multiple sign errors, particularly when distributing the negative sign in front of $(n+4)$. You can verify A is correct by substituting back: $3(2 \cdot \frac{1}{3} - 1) - (\frac{1}{3} + 4) = 3(\frac{2}{3} - 1) - \frac{13}{3} = -1 - \frac{13}{3} = -\frac{16}{3}$, and $2(\frac{1}{3} - 3) = 2(-\frac{8}{3}) = -\frac{16}{3}$. ✓ Always double-check your distribution of negative signs and verify your answer by substitution—these are the most common sources of error in linear equations.

Question 16

1. $x\le \dfrac{11}{2}$ (correct answer)
2. $x\ge \dfrac{11}{2}$
3. $x< -\dfrac{11}{2}$
4. $x> -\dfrac{11}{2}$

Explanation: When you encounter an inequality with a fraction, 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 $\frac{2x+1}{4} \le 3$, multiply both sides by 4 to eliminate the fraction: $2x + 1 \le 12$. Next, subtract 1 from both sides: $2x \le 11$. Finally, divide both sides by 2: $x \le \frac{11}{2}$. Since we only performed operations with positive numbers, the inequality direction stays the same. Let's examine why the other choices are incorrect. Choice B gives $x \ge \frac{11}{2}$, which flips the inequality sign incorrectly—this would happen if you mistakenly thought dividing by positive 2 required flipping the sign. Choices C and D both show $x < -\frac{11}{2}$ and $x > -\frac{11}{2}$ respectively. These likely result from sign errors when subtracting 1, perhaps getting $2x \le -11$ instead of $2x \le 11$. You can verify the correct answer A by testing a value: if $x = 5$, then $\frac{2(5)+1}{4} = \frac{11}{4} = 2.75$, and indeed $2.75 \le 3$. Study tip: Remember that inequality signs only flip when you multiply or divide by a negative number. With fractions, focus on clearing denominators first by multiplying both sides by the denominator.

Question 17

1. $x=0$ (correct answer)
2. $x=6$
3. $x=-6$
4. $x=3$

Explanation: When you encounter a linear equation with decimals and parentheses like this one, your goal is to systematically simplify both sides until you isolate the variable. The key is working carefully through the order of operations and decimal arithmetic. Start by distributing $0.25$ to both terms inside the parentheses: $0.25(8x + 12) = 0.25 \cdot 8x + 0.25 \cdot 12 = 2x + 3$. Your equation becomes: $2x + 3 - 3 = 0.5x$, which simplifies to $2x = 0.5x$. Now subtract $0.5x$ from both sides: $2x - 0.5x = 0$, giving you $1.5x = 0$. When any number times $x$ equals zero, then $x = 0$. Let's verify: substituting $x = 0$ into the original equation gives $0.25(0 + 12) - 3 = 3 - 3 = 0$, and $0.5(0) = 0$. Both sides equal zero, confirming our answer. Looking at the wrong choices: Choice B ($x = 6$) likely comes from calculation errors in the distribution step. Choice C ($x = -6$) might result from sign errors when moving terms between sides. Choice D ($x = 3$) could come from confusing the simplified form $2x + 3 - 3$ and thinking the constant term somehow becomes the answer. The correct answer is A. Strategy tip: With decimal coefficients, convert them to fractions if needed ($0.25 = \frac{1}{4}$, $0.5 = \frac{1}{2}$) to avoid arithmetic mistakes, and always verify your solution by substituting back into the original equation.

Question 18

1. $3 < x \le 5$ (correct answer)
2. $x<3 \text{ or } x>5$
3. $x\le3$
4. $3\le x<5$

Explanation: When you encounter a compound inequality like this one, you're solving for values of x that satisfy both conditions simultaneously. The key is to treat each part of the inequality separately while maintaining the relationship between all three parts. To solve $4 < 3x-5 \le 10$, you need to isolate x by performing the same operations on all three parts. First, add 5 to each section: $4 + 5 < 3x - 5 + 5 \le 10 + 5$, which gives you $9 < 3x \le 15$. Next, divide everything by 3: $\frac{9}{3} < \frac{3x}{3} \le \frac{15}{3}$, resulting in $3 < x \le 5$. This means x is greater than 3 but less than or equal to 5. Choice A correctly shows $3 < x \le 5$, which matches our solution exactly. Choice B gives $x < 3$ or $x > 5$, which represents values outside our solution interval—this would be the complement of what we found. Choice C shows $x \le 3$, which only captures values less than or equal to 3, missing the entire upper portion of our solution. Choice D presents $3 \le x < 5$, which incorrectly includes 3 (should be excluded since $3 < x$) and excludes 5 (should be included since $x \le 5$). Remember that compound inequalities require you to maintain the inequality symbols' directions throughout your algebraic manipulations. Always double-check your final answer by testing a value within your solution set in the original inequality.

Question 19

1. $x=\dfrac{33}{7}$ (correct answer)
2. $x=\dfrac{7}{33}$
3. $x=-\dfrac{33}{7}$
4. $x=\dfrac{27}{7}$

Explanation: When you encounter an equation with fractions set equal to each other, the most efficient approach is to cross-multiply. This eliminates the fractions and creates a linear equation you can solve step by step. Starting with $\frac{7-x}{2}=\frac{x+1}{5}$, cross-multiply by multiplying the numerator of each fraction by the denominator of the other: $5(7-x) = 2(x+1)$. Distribute on both sides: $35 - 5x = 2x + 2$. Now collect all terms with $x$ on one side and constants on the other. Subtract $2x$ from both sides: $35 - 7x = 2$. Subtract 35 from both sides: $-7x = -33$. Finally, divide by $-7$: $x = \frac{33}{7}$. Looking at the answer choices, option A gives $x = \frac{33}{7}$, which matches our result. Option B shows $x = \frac{7}{33}$, which would result from incorrectly flipping the fraction at the end. Option C gives $x = -\frac{33}{7}$, which comes from a sign error when dividing by $-7$. Option D shows $x = \frac{27}{7}$, which results from arithmetic errors during the distribution or combining like terms steps. You can always verify your answer by substituting back into the original equation. With $x = \frac{33}{7}$, both sides equal $-\frac{19}{14}$, confirming our solution is correct. Study tip: When solving rational equations, always cross-multiply first to clear fractions, then solve the resulting linear equation systematically. Double-check by substituting your answer back into the original equation.

Question 20

1. $x< -\dfrac{2}{9}$ (correct answer)
2. $x> -\dfrac{2}{9}$
3. $x< \dfrac{2}{9}$
4. $x> \dfrac{2}{9}$

Explanation: When solving inequalities with fractions, your approach should mirror solving equations, but you must pay special attention to what happens when you multiply or divide by negative numbers. Start by clearing the fractions. Multiply both sides by 6 (the LCD of 3 and 6): $6 \cdot \frac{1-2x}{3} > 6 \cdot \frac{5x+4}{6}$ This gives you: $2(1-2x) > 5x+4$ Distribute on the left: $2-4x > 5x+4$ Now collect like terms. Subtract 2 from both sides: $-4x > 5x+2$ Subtract 5x from both sides: $-9x > 2$ Here's the crucial step: when you divide both sides by -9, you must flip the inequality sign because you're dividing by a negative number: $x < -\frac{2}{9}$ This confirms answer A is correct. Looking at the wrong answers: B gives $x > -\frac{2}{9}$, which would result from forgetting to flip the inequality sign when dividing by -9. C and D both show $\frac{2}{9}$ instead of $-\frac{2}{9}$, suggesting errors in handling the negative signs during the algebraic manipulation, possibly dropping a negative sign when moving terms or distributing. Strategy tip: The most common error in inequality problems is forgetting to flip the inequality sign when multiplying or dividing by a negative number. Always double-check this step, and consider testing a value from your solution set in the original inequality to verify your answer.