Solving (1/2)a + (2/3)b = 50 With B = 30 A Step-by-Step Guide
Solving equations is a fundamental skill in mathematics, and this article delves into the process of finding the correct solution for an equation involving fractions and two variables. In this specific problem, we are given the equation $\frac{1}{2} a+\frac{2}{3} b=50$ and the value of one variable, $b=30$. Our goal is to determine the correct solution for the other variable, a, by substituting the given value and simplifying the equation. This involves a step-by-step approach, starting with the substitution, followed by basic arithmetic operations to isolate the variable a. Understanding this process is crucial for solving various algebraic problems and real-world applications where equations with multiple variables are encountered. This article will provide a detailed explanation of each step, ensuring a clear understanding of the solution. We'll break down the equation, perform the necessary calculations, and arrive at the final answer. This will not only help in solving this particular problem but also build a solid foundation for tackling more complex equations in the future. So, let's embark on this mathematical journey to find the correct solution and enhance our problem-solving skills.
Substituting the Value of b
The initial step in solving the equation is to substitute the given value of b into the equation. We are given that $b=30$, so we replace b with 30 in the equation $\frac{1}{2} a+\frac{2}{3} b=50$. This substitution yields the equation $\frac{1}{2} a+\frac{2}{3}(30)=50$. This is a critical step because it transforms the equation from one with two variables to one with a single variable, a, which we can then solve. The accuracy of this substitution is paramount as any error here will propagate through the rest of the solution. Once we have made the substitution, the next step involves simplifying the equation by performing the arithmetic operation. In this case, we need to multiply $ rac{2}{3}$ by 30. This simplification is crucial to isolate the term containing the variable a and move closer to finding its value. The process of substitution is widely used in algebra and is a fundamental technique for solving equations with multiple variables. It allows us to reduce the complexity of the equation and make it easier to solve. By carefully substituting the given value, we pave the way for further simplification and ultimately, the solution of the equation.
Simplifying the Equation
After substituting the value of b, we simplify the equation. The equation now reads $\frac{1}{2} a+\frac{2}{3}(30)=50$. We need to perform the multiplication $ rac{2}{3}(30)$. To do this, we multiply the fraction $ rac{2}{3}$ by 30, which can be written as $ rac{2}{3} \times 30$. This calculation simplifies to $ rac{2 \times 30}{3} = \frac{60}{3}$, which further simplifies to 20. So, the equation becomes $\frac{1}{2} a+20=50$. This simplification is a crucial step in isolating the term with a. By performing this multiplication, we have reduced the equation to a simpler form that is easier to manipulate. This step demonstrates the importance of order of operations in mathematics, where multiplication is performed before addition. Simplifying the equation step by step ensures that we maintain accuracy and avoid errors in the solution. The simplified equation $\frac{1}{2} a+20=50$ is now in a form where we can easily isolate the variable a by performing further algebraic manipulations. This simplification process is a key skill in solving algebraic equations and is essential for progressing towards the final solution.
Isolating the Variable a
To isolate the variable a, we need to get the term $ rac{1}{2} a$ by itself on one side of the equation. Currently, we have $\frac{1}{2} a+20=50$. To isolate $\frac{1}{2} a$, we subtract 20 from both sides of the equation. This is a fundamental algebraic principle – whatever operation you perform on one side of the equation, you must perform on the other side to maintain equality. Subtracting 20 from both sides gives us $\frac{1}{2} a+20-20=50-20$, which simplifies to $\frac{1}{2} a=30$. This step is crucial because it moves us closer to finding the value of a. By subtracting 20, we have effectively eliminated the constant term on the left side of the equation, leaving only the term containing a. The equation $\frac{1}{2} a=30$ is now in a simple form where we can easily solve for a. Isolating the variable is a key technique in solving equations, and this step demonstrates how to use subtraction to achieve this. The next step involves eliminating the fraction multiplying a to find the final solution.
Solving for a
Now that we have the equation $\frac{1}{2} a=30$, we need to solve for a. The variable a is being multiplied by $\frac{1}{2}$, so to isolate a, we need to perform the inverse operation, which is multiplication by the reciprocal of $\frac{1}{2}$. The reciprocal of $\frac{1}{2}$ is 2. Therefore, we multiply both sides of the equation by 2. This gives us $2 \times \frac{1}{2} a=2 \times 30$, which simplifies to $a=60$. This step is the final step in solving for a. By multiplying both sides of the equation by 2, we have effectively canceled out the fraction, leaving a by itself. This demonstrates the importance of using inverse operations to solve equations. The solution $a=60$ is the value of a that satisfies the original equation when $b=30$. This means that if we substitute a = 60 and b = 30 back into the original equation, $\frac{1}{2} a+\frac{2}{3} b=50$, the equation will hold true. This final step confirms that we have correctly solved the equation. The process of solving for a variable involves a series of steps, including substitution, simplification, isolation, and using inverse operations. Mastering these steps is crucial for success in algebra and other areas of mathematics.
Verification
To ensure that our solution is correct, it's always a good practice to verify it. We found that $a=60$ when $b=30$. Let's substitute these values back into the original equation: $\frac{1}{2} a+\frac{2}{3} b=50$. Substituting a = 60 and b = 30, we get $\frac{1}{2}(60)+\frac{2}{3}(30)=50$. Now, we simplify each term. $\frac{1}{2}(60)$ equals 30, and $\frac{2}{3}(30)$ equals 20. So, the equation becomes $30+20=50$, which simplifies to $50=50$. This confirms that our solution is correct. Verification is a critical step in problem-solving as it helps to identify any errors that may have occurred during the solution process. By substituting the values back into the original equation, we can ensure that the equation holds true. This step provides confidence in our solution and demonstrates a thorough understanding of the problem-solving process. In this case, the verification step confirms that $a=60$ is indeed the correct solution when $b=30$. This practice of verification is highly recommended for all mathematical problems to ensure accuracy and build strong problem-solving skills.
Conclusion
In conclusion, we have successfully solved the equation $\frac{1}{2} a+\frac{2}{3} b=50$ when $b=30$. The steps involved substitution, simplification, isolating the variable, and solving for a. We first substituted the value of b into the equation, then simplified the equation by performing arithmetic operations. Next, we isolated the variable a by subtracting 20 from both sides of the equation. Finally, we solved for a by multiplying both sides of the equation by 2, which gave us the solution $a=60$. We also verified our solution by substituting the values of a and b back into the original equation, confirming that the equation holds true. This process demonstrates a systematic approach to solving algebraic equations with fractions and multiple variables. By following these steps, we can confidently solve similar problems and develop strong problem-solving skills in mathematics. The ability to solve equations is a fundamental skill that is essential for various applications in mathematics, science, and engineering. This article provides a clear and detailed explanation of the steps involved in solving such equations, making it a valuable resource for students and anyone looking to improve their algebraic skills. The key takeaways from this article include the importance of substitution, simplification, isolating variables, and verification in the problem-solving process.
