Solving For C When 2(3-5c) Is One Less Than 4(1-c)
Introduction
In this article, we will delve into the realm of algebraic expressions and equations to determine the specific value of a variable that satisfies a given condition. Our core objective is to find the value of the variable 'c' for which the expression 2(3-5c) is exactly one less than the expression 4(1-c). This involves a step-by-step process of simplifying expressions, setting up an equation, and then employing algebraic techniques to isolate and solve for the unknown variable. Understanding how to manipulate expressions and solve equations is a fundamental skill in mathematics, forming the bedrock for more advanced concepts in algebra and calculus. This exploration will not only provide the solution to the problem at hand but also reinforce the importance of meticulous attention to detail and the application of mathematical principles.
Algebraic expressions are the building blocks of equations, and the ability to simplify them is crucial for problem-solving. In our case, we have two expressions, each involving the variable 'c'. The first step is to distribute the constants outside the parentheses into the terms within, a process that eliminates the parentheses and allows us to combine like terms. This simplification reduces the complexity of the expressions and paves the way for setting up a meaningful equation. Furthermore, the process of solving for a variable often involves performing inverse operations on both sides of the equation to maintain balance. For instance, if we have a term added to the variable, we subtract it from both sides; if a variable is multiplied by a constant, we divide both sides by that constant. This methodical approach ensures that we isolate the variable on one side of the equation, revealing its value. By mastering these techniques, readers will gain confidence in tackling similar algebraic challenges and deepen their understanding of mathematical problem-solving.
Setting Up the Equation
The heart of solving this problem lies in translating the given statement into a mathematical equation. The problem states that the value of the expression 2(3-5c) is "one less than" the value of the expression 4(1-c). This phrase is a crucial indicator of the mathematical operation we need to employ. "One less than" implies subtraction, specifically, we need to subtract 1 from the expression 4(1-c) to equate it to 2(3-5c). Thus, we can construct the equation as follows:
2(3-5c) = 4(1-c) - 1
This equation forms the foundation of our solution. It encapsulates the relationship described in the problem statement, setting the stage for the subsequent algebraic manipulations needed to solve for 'c'. The process of converting verbal statements into mathematical equations is a cornerstone of problem-solving in mathematics. It demands a careful understanding of the language used and the mathematical concepts it represents. Words like "less than," "more than," "times," and "divided by" all have specific mathematical connotations, and correctly interpreting them is paramount to formulating the correct equation. Moreover, the equal sign (=) in an equation signifies a balance between the expressions on either side. Any operation performed on one side must also be performed on the other to maintain this balance, a principle that guides the entire process of solving equations. The ability to set up equations accurately is a valuable skill that extends beyond the realm of algebra, finding applications in various fields such as physics, engineering, and economics.
Simplifying the Expressions
Before we can isolate the variable 'c', we must first simplify both sides of the equation. This involves applying the distributive property, which states that a(b + c) = ab + ac. Let's apply this property to each expression:
For the left side of the equation, 2(3-5c), we distribute the 2:
2 * 3 - 2 * 5c = 6 - 10c
For the right side of the equation, 4(1-c) - 1, we first distribute the 4:
4 * 1 - 4 * c - 1 = 4 - 4c - 1
Now, we can combine like terms on the right side:
4 - 4c - 1 = 3 - 4c
Our equation now looks like this:
6 - 10c = 3 - 4c
By simplifying the expressions, we've reduced the complexity of the equation, making it easier to manipulate and solve. The distributive property is a fundamental tool in algebra, enabling us to remove parentheses and combine terms effectively. It's important to remember the order of operations (PEMDAS/BODMAS) when simplifying expressions: Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right). Applying these rules consistently ensures that we arrive at the correct simplified form. Furthermore, combining like terms involves identifying terms with the same variable and exponent and then adding or subtracting their coefficients. This process streamlines the equation and brings us closer to isolating the variable we're trying to solve for. Mastering the art of simplification is crucial for success in algebra and beyond, laying the groundwork for tackling more complex mathematical problems.
Isolating the Variable
Now that we've simplified the equation to 6 - 10c = 3 - 4c, our next goal is to isolate the variable 'c' on one side of the equation. To achieve this, we'll employ a series of algebraic manipulations, ensuring that we perform the same operations on both sides to maintain the equation's balance.
First, let's add 10c to both sides of the equation. This will eliminate the '-10c' term on the left side:
6 - 10c + 10c = 3 - 4c + 10c
This simplifies to:
6 = 3 + 6c
Next, we subtract 3 from both sides to isolate the term with 'c':
6 - 3 = 3 + 6c - 3
This simplifies to:
3 = 6c
Finally, to solve for 'c', we divide both sides by 6:
3 / 6 = 6c / 6
This gives us:
c = 1/2
Therefore, the value of the variable 'c' that satisfies the given condition is 1/2.
The process of isolating a variable involves strategically applying inverse operations to both sides of the equation. The key is to identify the operations that are currently applied to the variable and then perform their opposites. For instance, if a variable is being multiplied by a number, we divide both sides by that number. If a number is being added to the variable, we subtract that number from both sides. This methodical approach ensures that we gradually peel away the layers surrounding the variable until it stands alone on one side of the equation. Throughout this process, it's essential to maintain the balance of the equation, meaning that any operation performed on one side must be mirrored on the other. This principle underpins all algebraic manipulations and guarantees that the solution we arrive at is valid. Isolating variables is a fundamental skill in algebra, and proficiency in this area opens doors to solving a wide range of mathematical problems.
Verification of the Solution
To ensure the accuracy of our solution, it's crucial to verify that the value we found for 'c', which is 1/2, indeed satisfies the original equation. This involves substituting c = 1/2 back into the original equation and checking if both sides of the equation are equal. Let's substitute c = 1/2 into the original equation:
2(3 - 5c) = 4(1 - c) - 1
Substituting c = 1/2, we get:
2(3 - 5 * 1/2) = 4(1 - 1/2) - 1
Now, let's simplify both sides of the equation. For the left side:
2(3 - 5/2) = 2(6/2 - 5/2) = 2(1/2) = 1
For the right side:
4(1 - 1/2) - 1 = 4(1/2) - 1 = 2 - 1 = 1
Since both sides of the equation equal 1 when c = 1/2, our solution is verified.
The verification step is a critical component of the problem-solving process in mathematics. It serves as a safeguard against errors that may have occurred during the algebraic manipulations. By substituting the solution back into the original equation, we confirm that the value we found for the variable truly satisfies the given condition. If the two sides of the equation do not match after substitution, it indicates that there was likely an error in our calculations, prompting us to revisit our steps and identify the mistake. This process of checking our work reinforces the importance of accuracy and attention to detail in mathematical problem-solving. Furthermore, it deepens our understanding of the relationship between equations and their solutions, solidifying our grasp of algebraic concepts. The habit of verifying solutions is a valuable practice that enhances both our confidence in our answers and our overall mathematical proficiency.
Conclusion
In this article, we successfully determined the value of the variable 'c' for which the expression 2(3-5c) is one less than the expression 4(1-c). Through a systematic approach involving simplifying expressions, setting up an equation, isolating the variable, and verifying the solution, we found that c = 1/2. This exercise underscores the importance of algebraic manipulation skills and the value of careful, step-by-step problem-solving techniques. By mastering these skills, individuals can confidently tackle a wide range of mathematical challenges.
