Equations with Variables on Both Sides - Pre Algebra
- When solving , a student gets , then . What should the student conclude? Answer: The equation has no solution because after subtracting from both sides When you encounter an equation that simplifies to something like , you're dealing with a special case where the variable terms are identical on both sides. The key is recognizing what happens when you isolate the variable terms from the constants. Starting from , subtract from both sides to eliminate the variable terms. This gives you , which is clearly a false statement. When solving an equation leads to a false statement (like ), the original equation has no solution. No value of can make the equation true because the equation is inherently contradictory. Choice A is incorrect because having the same coefficient for on both sides doesn't guarantee infinitely many solutions—you must also check what happens to the constant terms. Choice C is wrong because equal coefficients don't mean is the solution; you need to solve completely by eliminating the variable terms. Choice D makes an error by trying to manipulate the numbers and arithmetically, but this approach ignores the fundamental contradiction that . The correct answer is B because after properly eliminating the variable terms, you're left with the impossible statement . Remember this pattern: when solving linear equations, if you end up with identical variable terms on both sides but different constants, the equation has no solution. Always complete the solving process by isolating the constants to check for contradictions.
- For what value of will the equation have the solution ? Answer: If is the solution, then substituting into the equation: . This gives us , so . Therefore . Choice B gives the value of the right side before subtracting 12. Choice C incorrectly subtracts instead of adding to isolate . Choice D gives the value of when .
- The equation has the solution . If , what are the values of and ? Answer: When you encounter an equation with unknown coefficients and a given solution, you need to use the fact that the solution must satisfy the equation to find relationships between those unknowns. Since is the solution to , substituting into the equation must make it true: Now you have two equations with two unknowns: - (from substituting the solution) - (given condition) Adding these equations eliminates : Substituting back: , so . Let's check why the other answers fail. Choice A () gives us , violating our first equation. Choice B () produces , also incorrect. Choice D () yields , failing the substitution test. Only choice C satisfies both and . Study tip: When solving systems involving given solutions, always substitute the solution into the original equation first to create a new relationship. This converts an abstract problem into a concrete system of linear equations you can solve using elimination or substitution.
- Consider the equation . After finding a common denominator and clearing fractions, what is the coefficient of in the resulting equation? Answer: When you encounter an equation with fractions, your goal is to eliminate them by finding a common denominator and then clearing the fractions entirely. This transforms the equation into a simpler form where you can easily identify coefficients. Let's work through step by step. First, find the least common denominator of 4 and 6, which is 12. Rewrite each fraction: Now your equation becomes: Combine the fractions: To clear the fraction, multiply both sides by 12: The coefficient of is 1, making C correct. Choice A (3) likely comes from mistakenly using just the coefficient from the first fraction. Choice B (2) might result from using the coefficient from the second fraction. Choice D (-1) could occur if you incorrectly handled the subtraction when combining fractions. Remember: when clearing fractions, always multiply the entire equation by the common denominator. The resulting linear equation will have integer coefficients that are much easier to work with than the original fractions.
- A student solves by expanding both sides. After simplification, what conclusion should the student reach? Answer: The equation has no solution because it leads to a contradiction Expanding the left side: . Expanding the right side: . So we have . Subtracting from both sides gives , which is false. This contradiction means the equation has no solution.
- Marcus is solving the equation . After moving all terms with to one side and all constants to the other side, what equation should he have before solving for ? Answer: Starting with , subtract from both sides: . Then subtract from both sides (or add 9): . Finally, subtract from both sides to get , or equivalently . Choice B incorrectly adds coefficients instead of subtracting. Choice C incorrectly adds all terms. Choice D has incorrect signs from moving terms incorrectly.
- Sarah claims that the equation has infinitely many solutions. Jake argues it has no solution. Who is correct and why? Answer: Sarah is correct because the equation simplifies to , which is always true Expanding the left side: . This gives us . Subtracting from both sides yields , or , which is always true regardless of the value of . This means the equation is an identity with infinitely many solutions. Choice B incorrectly calculates the simplification. Choices C and D incorrectly assume there's a unique solution.
- The equation can be solved by first converting to integers. After multiplying by an appropriate power of 10, what equation results? Answer: When you encounter linear equations with decimals, converting them to integers by multiplying by an appropriate power of 10 makes solving much easier and reduces calculation errors. To eliminate all decimals in , you need to identify the decimal with the most places after the decimal point. Here, all decimals have one decimal place (0.5, 1.2, 0.3, 0.6), so multiply the entire equation by 10¹ = 10. Multiplying each term by 10: , which gives you . This matches answer choice D. Let's examine why the other choices are incorrect. Choice A multiplies only the first term by 10 while leaving the other decimals unchanged—you must multiply every term in the equation. Choice B multiplies by 100 instead of 10, which unnecessarily creates larger numbers: . While this equation is mathematically equivalent, the question asks for the result after multiplying by the "appropriate" power of 10, which is the smallest power that eliminates all decimals. Choice C multiplies only the x-terms by 10 but leaves the constants as decimals, failing to achieve the goal of converting to integers. Study tip: Always multiply the entire equation by the same power of 10. Find the decimal with the most places after the decimal point—that tells you which power of 10 to use. One decimal place means multiply by 10, two decimal places means multiply by 100, and so on.
- Lisa solves and gets . To check her answer, she substitutes back into the original equation. What should she get for the left side when ? Answer: When checking solutions to equations, you need to substitute your answer back into the original equation and evaluate both sides to see if they're equal. This verification step catches any algebraic errors you might have made. To find what Lisa gets for the left side when , substitute into : So the left side equals when . You can verify this is correct by checking the right side: . Since , Lisa's solution is actually incorrect, but the question only asks what she gets for the left side. Looking at the wrong answers: Choice A () might come from incorrectly calculating as instead of . Choice B () is what you get if you evaluate the right side instead of the left side, or if you forget to subtract the term entirely. Choice C () could result from mistakenly calculating by using just the coefficient inside the parentheses. Remember: when substituting into expressions with parentheses, always follow the order of operations carefully. First handle what's inside parentheses, then multiply, then subtract. Double-check by working through each step slowly rather than trying to do multiple operations at once.
- The equation can be solved by first clearing the fractions. What is the resulting equation after multiplying both sides by the least common denominator? Answer: The LCD of 3 and 2 is 6. Multiplying both sides by 6: . This simplifies to because . Choice B incorrectly calculates . Choice C incorrectly keeps the 6 with the first fraction. Choice D uses wrong coefficients from incorrect LCD application.