Solving Equations Find The Solution For (1/2)a + (2/3)b = 50

In the realm of mathematics, linear equations stand as fundamental tools for modeling and solving a myriad of real-world problems. Mastering the art of solving these equations is crucial for anyone venturing into fields like physics, engineering, economics, and computer science. This article delves into the intricacies of solving a specific linear equation, providing a step-by-step guide to ensure clarity and understanding. We will meticulously dissect the equation, unravel its components, and arrive at the correct solution, all while emphasizing the underlying principles that govern this mathematical process.

Before we plunge into the specifics, let's establish a solid grasp of what linear equations entail. A linear equation, at its core, is an algebraic statement that asserts the equality between two expressions, where each expression is a combination of constants and variables raised to the power of one. These equations can take various forms, but they all share the characteristic of representing a straight line when plotted on a graph. Understanding this graphical representation can provide valuable insights into the nature of the equation and its solutions.

• Key Components of a Linear Equation

  Linear equations comprise several key components that work in harmony to define the relationship between variables and constants. These include:

  • Variables: Symbols, typically letters like 'x' or 'y', that represent unknown quantities we seek to determine.
  • Constants: Numerical values that remain fixed throughout the equation, providing a stable foundation for the relationship.
  • Coefficients: Numbers that multiply variables, indicating the magnitude of their contribution to the equation.
  • Operators: Mathematical symbols like addition (+), subtraction (-), multiplication (*), and division (/) that dictate the operations performed on the terms.
  • Equality Sign: The symbol (=) that signifies the balance between the two expressions, establishing the core concept of an equation.

By recognizing and understanding these components, we can effectively navigate the process of solving linear equations.

Let's turn our attention to the specific equation at hand:

$\frac{1}{2}a + \frac{2}{3}b = 50$

This equation presents a linear relationship between two variables, 'a' and 'b'. Our objective is to find the value of 'a' when 'b' is given as 30. To achieve this, we will systematically manipulate the equation, isolating 'a' on one side and calculating its value based on the known value of 'b'.

• Breaking Down the Equation

  To effectively tackle the equation, let's break it down into its individual components:

  • $\frac{1}{2}a$: This term represents half the value of the variable 'a'.
  • $\frac{2}{3}b$: This term signifies two-thirds of the value of the variable 'b'.
  • 50: This constant represents the total value that the sum of the two terms must equal.

  By dissecting the equation in this manner, we gain a clearer understanding of the relationships between the variables and constants, paving the way for a methodical solution.

Now, let's embark on the journey of solving for 'a' when 'b' is given as 30. We will proceed step by step, explaining each operation and its purpose, ensuring a clear and comprehensible solution process.

• Step 1: Substitute the Value of 'b'

  The first step involves substituting the given value of 'b' (30) into the equation. This substitution transforms the equation into one with a single unknown, 'a', making it easier to solve.

  $\frac{1}{2}a + \frac{2}{3}(30) = 50$

  This substitution replaces the variable 'b' with its numerical value, simplifying the equation and bringing us closer to isolating 'a'.

• Step 2: Simplify the Equation

  Next, we simplify the equation by performing the multiplication operation within the second term:

  $\frac{1}{2}a + 20 = 50$

  This simplification combines the constant terms, further streamlining the equation and isolating 'a' within a smaller expression.

• Step 3: Isolate the Term with 'a'

  To isolate the term containing 'a', we subtract 20 from both sides of the equation. This operation maintains the equality while moving the constant term to the right side:

  $\frac{1}{2}a + 20 - 20 = 50 - 20$

  $\frac{1}{2}a = 30$

  By subtracting 20 from both sides, we effectively isolate the term with 'a', bringing us closer to the final solution.

• Step 4: Solve for 'a'

  Finally, to solve for 'a', we multiply both sides of the equation by 2. This operation cancels out the fraction multiplying 'a', leaving us with the value of 'a':

  $2 * \frac{1}{2}a = 30 * 2$

  $a = 60$

  Therefore, the solution to the equation when 'b' is 30 is a = 60.

To ensure the accuracy of our solution, it's always prudent to verify it by substituting the values of 'a' and 'b' back into the original equation. If the equation holds true, our solution is correct.

• Substituting 'a' = 60 and 'b' = 30 into the Original Equation

  $\frac{1}{2}(60) + \frac{2}{3}(30) = 50$

  $30 + 20 = 50$

  $50 = 50$

  As the equation holds true, our solution a = 60 is indeed correct.

In this article, we embarked on a journey to solve a linear equation, unraveling its intricacies and arriving at the correct solution. We began by establishing a firm understanding of linear equations and their components. We then dissected the given equation, identifying the variables, constants, and their relationships. Through a step-by-step approach, we substituted the value of 'b', simplified the equation, isolated the term with 'a', and ultimately solved for 'a'. Finally, we verified our solution, ensuring its accuracy.

Mastering the art of solving linear equations is a fundamental skill in mathematics and its applications. By understanding the underlying principles and following a systematic approach, you can confidently tackle a wide range of equations and unlock their solutions. Remember, practice is key to solidifying your understanding and honing your problem-solving skills. So, embrace the challenge, explore the world of linear equations, and watch your mathematical prowess flourish!