Solving Linear Equations (Including Fractions) - College Algebra

1. A linear equation in the form $\frac{px+q}{r} = \frac{sx+t}{u} + v$ is solved by first finding a common denominator. If the LCD is 30, and after clearing denominators the equation becomes $6px + 6q = 5sx + 5t + 30v$, what are the possible values for the pair $(r,u)$?
   Answer: $(r,u)$ could be $(5,6)$ or $(10,15)$, among other possibilities When clearing denominators with LCD = 30, we multiply each term by $\frac{30}{\text{denominator}}$. The first fraction gets multiplied by $\frac{30}{r}$, giving coefficient 6 for the $px$ term, so $\frac{30}{r} = 6$, thus $r = 5$. The second fraction gets multiplied by $\frac{30}{u}$, giving coefficient 5 for the $sx$ term, so $\frac{30}{u} = 5$, thus $u = 6$. However, any common multiple would work: if LCD were 60, we'd have $(r,u) = (10,12)$, etc. So $(5,6)$ and $(10,15)$ are both possible.
2. Dividing quantity: solve $\frac{x}{7}-\frac{2}{7}=\frac{3}{7}$ for $x$.
   Answer: $x=5$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{7} - \frac{2}{7} = \frac{3}{7}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=5$. A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
3. Cost sharing: solve $\frac{x}{5}-\frac{1}{10}=\frac{3}{10}$ dollars for $x$.
   Answer: $x=2$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{5} - \frac{1}{10} = \frac{3}{10}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=2$. A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
4. Dividing supplies: solve $\frac{x}{8}+\frac{1}{4}=\frac{5}{8}$ for $x$.
   Answer: $x=3$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{8} + \frac{1}{4} = \frac{5}{8}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=3$. A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
5. A student is solving the equation $\frac{2x-1}{3} - \frac{x+4}{6} = \frac{x-2}{2}$. After finding a common denominator and combining fractions, what is the resulting equation before solving for $x$?
   Answer: $2(2x-1) - (x+4) = 3(x-2)$ To solve this equation, we need a common denominator of 6. Multiplying each fraction: $\frac{2(2x-1)}{6} - \frac{x+4}{6} = \frac{3(x-2)}{6}$. Multiplying both sides by 6 gives us $2(2x-1) - (x+4) = 3(x-2)$. Choice A incorrectly distributes without maintaining the structure. Choice C shows the fractions before clearing denominators. Choice D uses an incorrect common denominator approach.
6. Cost sharing: solve $\frac{1}{4}x+18=30$ dollars for $x$.
   Answer: $x=48$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation \frac{1}{4}x + 18 = 30 involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice A, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution x=48. A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as forgetting to multiply by the reciprocal. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
7. Which equation is equivalent to $0.25x + \frac{1}{3} = 0.75x - \frac{2}{3}$ when all terms are expressed with integer coefficients?
   Answer: $3x + 4 = 9x - 8$ Convert decimals to fractions: $\frac{1}{4}x + \frac{1}{3} = \frac{3}{4}x - \frac{2}{3}$. To clear denominators, multiply by LCD = 12: $12 \cdot \frac{1}{4}x + 12 \cdot \frac{1}{3} = 12 \cdot \frac{3}{4}x - 12 \cdot \frac{2}{3}$. This gives $3x + 4 = 9x - 8$. Choice B uses 100 as multiplier unnecessarily. Choice C incorrectly multiplies the $x$ terms. Choice D uses 300 as multiplier and makes coefficient errors.
8. Recipe scaling: solve $\frac{x}{3}-\frac{1}{6}=\frac{1}{2}$ for $x$ cups.
   Answer: $x=2$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{3} - \frac{1}{6} = \frac{1}{2}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=2$. A common error, as seen in Choice A, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
9. Dividing items: solve $\frac{x}{9}-\frac{1}{9}=\frac{2}{3}$ for $x$.
   Answer: $x=7$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{9} - \frac{1}{9} = \frac{2}{3}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=7$. A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
10. Mixing: solve $\frac{x}{9}+\frac{2}{3}=\frac{8}{9}$ for $x$ mL.
    Answer: $x=2$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{9} + \frac{2}{3} = \frac{8}{9}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=2$. A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
11. In cost sharing, solve $\frac{x}{4}+\frac{3}{8}=\frac{7}{8}$ dollars for $x$.
    Answer: $x=2$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{4} + \frac{3}{8} = \frac{7}{8}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=2$. A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
12. Cost sharing: solve $\frac{x}{12}+\frac{1}{6}=\frac{1}{3}$ dollars for $x$.
    Answer: $x=2$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{12} + \frac{1}{6} = \frac{1}{3}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=2$. A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
13. Recipe adjustment: solve $\frac{x}{4}-\frac{1}{8}=\frac{3}{8}$ for $x$ cups.
    Answer: $x=2$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{4} - \frac{1}{8} = \frac{3}{8}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=2$. A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
14. Mixing: solve $\frac{x}{20}+\frac{1}{4}=\frac{2}{5}$ for $x$ mL.
    Answer: $x=3$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{20} + \frac{1}{4} = \frac{2}{5}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=3$. A common error, as seen in Choice D, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
15. Distance-speed: solve $\frac{x}{9}-\frac{1}{3}=\frac{1}{6}$ for $x$ miles.
    Answer: $x=\frac{9}{2}$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{9} - \frac{1}{3} = \frac{1}{6}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=\frac{9}{2}$. A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
16. Recipe scaling: solve $\frac{x}{3}+\frac{1}{4}=\frac{11}{12}$ for $x$ cups.
    Answer: $x=2$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{3} + \frac{1}{4} = \frac{11}{12}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=2$. A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
17. Mixing solutions: solve $\frac{x}{15}+\frac{2}{5}=\frac{3}{5}$ for $x$ mL.
    Answer: $x=3$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{15} + \frac{2}{5} = \frac{3}{5}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=3$. A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
18. Mixing: solve $\frac{x}{6}+\frac{1}{2}=\frac{5}{6}$ for $x$ mL.
    Answer: $x=2$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{6} + \frac{1}{2} = \frac{5}{6}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=2$. A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
19. Divide snacks: solve $\frac{x}{6}+\frac{1}{3}=\frac{5}{6}$ for $x$.
    Answer: $x=3$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{6} + \frac{1}{3} = \frac{5}{6}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=3$. A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.
20. Cost sharing: solve $\frac{x}{6}+\frac{1}{12}=\frac{5}{12}$ dollars for $x$.
    Answer: $x=2$ This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation $\frac{x}{6} + \frac{1}{12} = \frac{5}{12}$ involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution $x=2$. A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

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