Textbook Question
Decide whether each statement is true or false. If false, correct the right side of the equation. √-25 = 5i
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Ch. 1 - Equations and Inequalities
Lial13th EditionCollege AlgebraISBN: 9780136881063
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Table of contents
Chapter 2, Problem 7
Solve each problem. If x represents the number of pennies in a jar in an applied problem, which of the following equations cannot be a correct equation for finding x? (Hint:Solve the equations and consider the solutions.)
A. 5x+3 =11
B.12x+6 =-4
C.100x =50(x+3)
D. 6(x+4) =x+24
Verified step by step guidance1
Step 1: Understand the context of the problem. Since \( x \) represents the number of pennies in a jar, \( x \) must be a non-negative integer (\( x \geq 0 \)) because you cannot have a negative number of pennies.
Step 2: Solve each equation for \( x \) to find its solution.
For equation A: \( 5x + 3 = 11 \), subtract 3 from both sides to get \( 5x = 8 \), then divide both sides by 5 to find \( x = \frac{8}{5} \).
For equation B: \( 12x + 6 = -4 \), subtract 6 from both sides to get \( 12x = -10 \), then divide both sides by 12 to find \( x = -\frac{10}{12} = -\frac{5}{6} \).
For equation C: \( 100x = 50(x + 3) \), expand the right side to get \( 100x = 50x + 150 \), subtract \( 50x \) from both sides to get \( 50x = 150 \), then divide both sides by 50 to find \( x = 3 \).
For equation D: \( 6(x + 4) = x + 24 \), expand the left side to get \( 6x + 24 = x + 24 \), subtract \( x \) and 24 from both sides to get \( 5x = 0 \), then divide both sides by 5 to find \( x = 0 \).
Step 3: Analyze the solutions in the context of the problem. Since \( x \) must be non-negative, any negative solution is not valid for the number of pennies.
Step 4: Identify which equation(s) yield invalid solutions. Equation B gives \( x = -\frac{5}{6} \), which is negative and therefore cannot represent the number of pennies.
Step 5: Conclude that equation B cannot be a correct equation for finding \( x \) in this applied problem.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Solving Linear Equations
Linear equations are algebraic expressions where the variable is to the first power. Solving them involves isolating the variable on one side to find its value. Understanding how to manipulate equations using addition, subtraction, multiplication, and division is essential to find solutions.
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Interpreting Solutions in Context
When variables represent real-world quantities, like the number of pennies, solutions must make sense in that context. For example, negative or non-integer solutions may be invalid if the quantity cannot be negative or fractional. Evaluating solutions against the problem's scenario ensures meaningful answers.
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Checking for Extraneous or Invalid Solutions
Some equations may yield solutions that do not fit the problem's conditions or are mathematically impossible. Identifying such solutions requires substituting back or analyzing the equation's structure. This helps determine which equations cannot correctly represent the situation.
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Related Practice
Textbook Question
Match each equation or inequality in Column I with the graph of its solution set in Column II. | x | ≠ 7
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Textbook Question
Solve each problem. Suppose two acid solutions are mixed. One is 26% acid and the other is 34% acid. Which one of the following concentrations cannot possibly be the concentration of the mixture? A. 24% B. 30% C. 31% D. 33%
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Textbook Question
Match each equation or inequality in Column I with the graph of its solution set in Column II. | x | ≤ 7
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Textbook Question
Match each equation in Column I with the correct first step for solving it in Column II. √(x+5) = 7
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