Problem 56
Question
MULTIPLE CHOICE Which equations are equivalent? I. \(\frac{3}{5} x=3\) II. \(\frac{x}{5}=2\) III. \(2 x=10\) VI. \(-x=-5\) A. I and II B. II and IV C. I, II, and III D. I, III, and IV
Step-by-Step Solution
Verified Answer
The correct answer is D. Equations I, III, and IV are equivalent.
1Step 1: Solve each equation individually
Solve each equation for its value of x. For equation I: Multiply both sides by \(\frac{5}{3}\) to get \(x=5\). For equation II: Multiply both sides by 5 to get \(x=10\). For equation III: Divide both sides by 2 to get \(x=5\). For equation IV: Multiply both sides by -1 to get \(x=5\).
2Step 2: Identifying equivalent equations
From the solutions obtained in step one, equations I, III, and IV yield the same value of \(x=5\). Hence, these equations are equivalent.
3Step 3: Matching with the choices
The result from step two corresponds to the choice D (I, III, and IV are equivalent).
Key Concepts
Understanding Solving EquationsMastering Algebraic SolutionsNavigating Multiple Choice Problems
Understanding Solving Equations
Solving equations is a fundamental part of algebra that helps us find the value of unknown variables. When we solve an equation, we try to isolate the variable on one side of the equation to discover its value. Think of it as unraveling a puzzle where each step you take gets you closer to the solution.
For instance, let's consider the equations from the multiple choice problem. To solve such equations, we employ basic algebraic operations like addition, subtraction, multiplication, or division. These operations help us simplify the equation and zero in on the variable we want to solve for. Take equation I: \(\frac{3}{5} x = 3\). By multiplying both sides by \(\frac{5}{3}\), we isolate \(x\), giving us \(x = 5\).
This same process of isolating \(x\) helps in simplifying other equations too, as seen with equations II, III, and IV in our example. The ability to solve equations is crucial and forms the backbone of understanding algebra.
For instance, let's consider the equations from the multiple choice problem. To solve such equations, we employ basic algebraic operations like addition, subtraction, multiplication, or division. These operations help us simplify the equation and zero in on the variable we want to solve for. Take equation I: \(\frac{3}{5} x = 3\). By multiplying both sides by \(\frac{5}{3}\), we isolate \(x\), giving us \(x = 5\).
This same process of isolating \(x\) helps in simplifying other equations too, as seen with equations II, III, and IV in our example. The ability to solve equations is crucial and forms the backbone of understanding algebra.
Mastering Algebraic Solutions
Algebraic solutions involve using various techniques and methods to solve equations for unknown variables. These solutions can be reached by manipulating the equations through different mathematical operations.
One effective technique is to perform the same operation on both sides of the equation. This maintains the balance of the equation and helps in simplifying it further. In our exercise, we used operations like multiplication and division to find solutions for equations I, II, III, and IV:
One effective technique is to perform the same operation on both sides of the equation. This maintains the balance of the equation and helps in simplifying it further. In our exercise, we used operations like multiplication and division to find solutions for equations I, II, III, and IV:
- In equation I: Multiplication was used to both isolate and find the value of \(x\).
- In equation II: Multiplying both sides by 5 unveiled \(x = 10\).
- With equation III, division of both sides by 2 disclosed \(x = 5\).
- Equation IV required multiplying by -1 to reach \(x = 5\).
Navigating Multiple Choice Problems
Multiple choice problems can be tricky, especially ones that involve equations. However, with a clear understanding of solving and analyzing equations, they can become easier.
The strategy for tackling such problems involves first solving each given equation to find its solution. Once the solutions have been determined, as we did in the exercise, we can then compare them to ascertain equivalency. This approach helps in narrowing down the correct choice by identifying which solutions match.
For our problem, once the solutions for equations I, II, III, and IV were calculated, it became evident that equations I, III, and IV were equivalent as they all resulted in \(x = 5\). Therefore, option D was the correct multiple choice answer. By simplifying each equation separately and comparing their results, navigating through multiple choice problems becomes more manageable and less intimidating.
The strategy for tackling such problems involves first solving each given equation to find its solution. Once the solutions have been determined, as we did in the exercise, we can then compare them to ascertain equivalency. This approach helps in narrowing down the correct choice by identifying which solutions match.
For our problem, once the solutions for equations I, II, III, and IV were calculated, it became evident that equations I, III, and IV were equivalent as they all resulted in \(x = 5\). Therefore, option D was the correct multiple choice answer. By simplifying each equation separately and comparing their results, navigating through multiple choice problems becomes more manageable and less intimidating.
Other exercises in this chapter
Problem 56
Find the product. $$ -8(-5) $$
View solution Problem 56
Find the opposite of the number. $$ 0.2 $$
View solution Problem 56
Solve \(15 x+6-x=16 x+6-2 x\) $$ \begin{array}{\underline{\phantom{xx}}}{ (A) \text { \(x=0\) }} \quad {(B) \text { \(x=6\) }} \quad{(C) \text { No solution }}\quad {(D) \text { Identi
View solution Problem 56
In Exercises 54–56, match the real-life problem with an equation. Then solve the problem. A. \(x+15=7\) B. \(15-x=7\) C. \(15+7=x\) D. \(x+15=-7\) The temperatu
View solution