Solving Xxvii. 1/3(4x-1) + 2/5(2x+5) - 5 14/15 = 0 A Step-by-Step Guide

Introduction to Linear Equations

In the realm of mathematics, particularly in algebra, solving linear equations is a fundamental skill. Linear equations are algebraic expressions where the highest power of the variable is one. These equations can be represented in the form ax + b = 0, where x is the variable, and a and b are constants. Mastering the techniques to solve these equations is crucial for understanding more complex mathematical concepts. This article delves into solving a specific linear equation, providing a step-by-step guide to ensure clarity and comprehension.

The equation we will tackle is: $\frac{1}{3}(4 x-1)+\frac{2}{5}(2 x+5)-5 \frac{14}{15}=0$. This equation may seem daunting at first glance due to the fractions and mixed numbers involved. However, by systematically applying the principles of algebra, we can simplify it and find the value of x that satisfies the equation. The process involves several key steps, including distributing coefficients, combining like terms, eliminating fractions, and isolating the variable. Each step is carefully explained to facilitate a thorough understanding.

Solving linear equations is not just an academic exercise; it has practical applications in various fields, including engineering, physics, economics, and computer science. For instance, in engineering, linear equations are used to model electrical circuits and structural mechanics. In economics, they are used to analyze supply and demand curves. Therefore, a strong grasp of solving linear equations is invaluable for anyone pursuing these fields. By the end of this article, you will not only be able to solve the given equation but also gain a deeper appreciation for the importance and versatility of linear equations in problem-solving.

Step-by-Step Solution

1. Simplify the Mixed Number

The first step in solving the equation $rac{1}{3}(4 x-1)+\frac{2}{5}(2 x+5)-5 \frac{14}{15}=0$ is to convert the mixed number into an improper fraction. This simplifies the equation and makes it easier to work with. The mixed number is $5 \frac{14}{15}$, which can be converted to an improper fraction by multiplying the whole number (5) by the denominator (15) and adding the numerator (14). This gives us (5 * 15) + 14 = 75 + 14 = 89. The improper fraction is then $\frac{89}{15}$. So, the equation becomes:

$$\frac{1}{3}(4 x-1)+\frac{2}{5}(2 x+5)-\frac{89}{15}=0$$

Converting mixed numbers to improper fractions is a crucial step in simplifying equations. It allows us to perform arithmetic operations, such as addition and subtraction, more easily. Improper fractions have a numerator that is greater than or equal to the denominator, which can sometimes make them seem more complex than mixed numbers. However, in the context of algebraic equations, improper fractions are often more convenient to work with because they eliminate the need to deal with whole numbers separately.

In this specific case, converting $5 \frac{14}{15}$ to $\frac{89}{15}$ allows us to combine it with other fractions in the equation more seamlessly. This is because we can now find a common denominator and perform addition and subtraction operations without having to worry about the whole number part of the mixed number. The ability to convert between mixed numbers and improper fractions is a fundamental skill in algebra, and mastering it is essential for solving a wide range of equations and mathematical problems. The simplification achieved in this initial step sets the stage for the subsequent steps, making the overall solution process more manageable and less prone to errors.

2. Distribute the Coefficients

The next step involves distributing the coefficients outside the parentheses to the terms inside. This means multiplying $\frac{1}{3}$ by both terms in $(4x - 1)$ and $\frac{2}{5}$ by both terms in $(2x + 5)$.

For the first term, $\frac{1}{3}(4x - 1)$, we multiply $\frac{1}{3}$ by $4x$ and $\frac{1}{3}$ by $-1$. This gives us $\frac{4x}{3} - \frac{1}{3}$.

For the second term, $\frac{2}{5}(2x + 5)$, we multiply $\frac{2}{5}$ by $2x$ and $\frac{2}{5}$ by $5$. This gives us $\frac{4x}{5} + \frac{10}{5}$, which simplifies to $\frac{4x}{5} + 2$.

Now, substituting these back into the equation, we get:

$$\frac{4x}{3} - \frac{1}{3} + \frac{4x}{5} + 2 - \frac{89}{15} = 0$$

Distributing coefficients is a key technique in simplifying algebraic expressions. It allows us to remove parentheses and combine like terms, which is essential for solving equations. The distributive property, which states that a(b + c) = ab + ac, is the foundation of this step. By applying this property, we can break down complex expressions into simpler components, making them easier to manipulate.

In the context of this equation, distributing the fractions $\frac{1}{3}$ and $\frac{2}{5}$ ensures that all terms are expressed individually, which is a prerequisite for combining like terms. This step is particularly important when dealing with fractions, as it allows us to work with numerators and denominators separately, reducing the risk of errors. The result of this step is a more expanded form of the equation, but it is also a more manageable form, as it sets the stage for the next step: combining like terms.

3. Combine Like Terms

After distributing the coefficients, the equation is: $\frac{4x}{3} - \frac{1}{3} + \frac{4x}{5} + 2 - \frac{89}{15} = 0$. Now, we need to combine like terms. Like terms are terms that have the same variable raised to the same power. In this case, we have two terms with x ($\frac{4x}{3}$ and $\frac{4x}{5}$) and several constant terms ($-\frac{1}{3}$, $2$, and $-\frac{89}{15}$).

First, let's combine the x terms. To add $\frac{4x}{3}$ and $\frac{4x}{5}$, we need a common denominator, which is 15. So, we convert the fractions:

$$\frac{4x}{3} = \frac{4x * 5}{3 * 5} = \frac{20x}{15}$$

$$\frac{4x}{5} = \frac{4x * 3}{5 * 3} = \frac{12x}{15}$$

Now, we can add them: $\frac{20x}{15} + \frac{12x}{15} = \frac{32x}{15}$.

Next, let's combine the constant terms. We have $-\frac{1}{3}$, $2$, and $-\frac{89}{15}$. Again, we need a common denominator, which is 15. So, we convert the fractions:

$$-\frac{1}{3} = -\frac{1 * 5}{3 * 5} = -\frac{5}{15}$$

$$2 = \frac{2 * 15}{1 * 15} = \frac{30}{15}$$

Now, we can combine them: $-\frac{5}{15} + \frac{30}{15} - \frac{89}{15} = \frac{-5 + 30 - 89}{15} = \frac{-64}{15}$.

Substituting these combined terms back into the equation, we get:

$$\frac{32x}{15} - \frac{64}{15} = 0$$

Combining like terms is a fundamental algebraic skill that simplifies equations and makes them easier to solve. It involves identifying terms with the same variable and exponent and then adding or subtracting their coefficients. This process reduces the number of terms in the equation, making it more manageable.

In this step, we combined the x terms and the constant terms separately. This approach is crucial because it ensures that we are only adding or subtracting terms that are mathematically compatible. The process of finding a common denominator for fractions is an essential part of this step, as it allows us to perform addition and subtraction operations on fractions with different denominators. The result of combining like terms is a more concise equation that retains the same mathematical meaning as the original equation but is significantly easier to solve. This simplification is a key step in isolating the variable and finding the solution to the equation.

4. Eliminate Fractions

To eliminate fractions from the equation $\frac{32x}{15} - \frac{64}{15} = 0$, we can multiply both sides of the equation by the common denominator, which in this case is 15. This will clear the fractions and make the equation easier to solve.

Multiplying both sides by 15, we get:

$$15 * (\frac{32x}{15} - \frac{64}{15}) = 15 * 0$$

This simplifies to:

$$32x - 64 = 0$$

Eliminating fractions is a common technique used in solving equations that involve fractional coefficients. It simplifies the equation by transforming it into an equivalent equation without fractions. This is achieved by multiplying both sides of the equation by the least common multiple (LCM) of the denominators. In this case, the LCM of the denominators is 15, so multiplying both sides by 15 effectively cancels out the denominators.

The rationale behind this technique is based on the principle that if we perform the same operation on both sides of an equation, the equality remains valid. By multiplying both sides by the LCM, we are ensuring that the equation remains balanced while simultaneously eliminating the fractions. This step is particularly useful when dealing with equations that have multiple fractions, as it can significantly reduce the complexity of the equation. The resulting equation, in this case, $32x - 64 = 0$, is a much simpler linear equation that can be solved using basic algebraic techniques.

5. Isolate the Variable

Now that we have the equation $32x - 64 = 0$, we need to isolate the variable x. This means getting x by itself on one side of the equation. To do this, we first add 64 to both sides of the equation:

$$32x - 64 + 64 = 0 + 64$$

This simplifies to:

$$32x = 64$$

Next, we divide both sides by 32:

$$\frac{32x}{32} = \frac{64}{32}$$

This gives us:

$$x = 2$$

Isolating the variable is the final and crucial step in solving an equation. It involves performing operations on both sides of the equation to get the variable by itself. The goal is to manipulate the equation in such a way that the variable is the only term on one side, and the constant value is on the other side. This process typically involves using inverse operations, such as addition and subtraction or multiplication and division.

In this case, we first added 64 to both sides to eliminate the constant term on the left side. This is based on the principle that adding the same value to both sides of an equation preserves the equality. Then, we divided both sides by 32 to eliminate the coefficient of x. This is based on the principle that dividing both sides of an equation by the same non-zero value preserves the equality. The result of these operations is the solution to the equation, which is the value of x that satisfies the original equation. In this case, we found that x = 2.

Conclusion

In conclusion, the solution to the equation $\frac{1}{3}(4 x-1)+\frac{2}{5}(2 x+5)-5 \frac{14}{15}=0$ is x = 2. We arrived at this solution by following a series of logical steps, including simplifying the mixed number, distributing coefficients, combining like terms, eliminating fractions, and isolating the variable. Each step is crucial in transforming the original equation into a simpler form that can be easily solved.

Solving linear equations is a fundamental skill in mathematics, and the techniques used in this example are applicable to a wide range of algebraic problems. By mastering these techniques, students can gain confidence in their ability to solve complex equations and develop a deeper understanding of algebraic principles. The step-by-step approach outlined in this article provides a clear and concise method for solving linear equations, making it a valuable resource for students and anyone interested in mathematics.

Furthermore, the process of solving this equation highlights the importance of precision and attention to detail in mathematics. Each step must be performed carefully to avoid errors, and the order of operations must be followed correctly. This reinforces the idea that mathematics is not just about finding the right answer but also about understanding the process and reasoning behind the solution. The skills learned in solving linear equations are transferable to other areas of mathematics and science, making it a foundational topic in education.

By understanding and applying these principles, anyone can confidently tackle linear equations and appreciate the elegance and power of algebra. The journey from the initial complex equation to the final simple solution demonstrates the beauty of mathematical problem-solving and the satisfaction of arriving at the correct answer.