Solving The Equation 3/(a+2) + 4/(a-1) = 0 A Step-by-Step Guide

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In this article, we will explore the process of finding the solution(s) to the equation $\frac{3}{a+2} + \frac{4}{a-1} = 0$. This equation involves fractions with algebraic expressions in the denominators, making it a valuable exercise in algebraic manipulation and equation-solving techniques. We will break down the steps involved in solving this equation, providing a clear and comprehensive guide for anyone looking to enhance their understanding of algebraic equations.

The equation we are tasked with solving is $\frac{3}{a+2} + \frac{4}{a-1} = 0$. This is a rational equation, which means it involves fractions where the numerators and denominators are polynomials. To solve this type of equation, our primary goal is to eliminate the fractions. This can be achieved by finding a common denominator and combining the fractions into a single expression. It's important to note that we must also consider any values of a that would make the denominators zero, as these values would make the fractions undefined.

Before we dive into the solution, let's consider the restrictions on the variable a. The denominators of the fractions are a + 2 and a - 1. If a = -2, the first denominator becomes zero, and if a = 1, the second denominator becomes zero. Therefore, a cannot be equal to -2 or 1. These restrictions are crucial to remember, as they will help us identify any extraneous solutions that may arise during the solving process.

  1. Find the Common Denominator: To combine the two fractions, we need to find a common denominator. In this case, the common denominator is the product of the two denominators, which is (a + 2)(a - 1).

  2. Rewrite the Fractions with the Common Denominator: We multiply the numerator and denominator of each fraction by the appropriate factor to obtain the common denominator:

    3a+2ā‹…aāˆ’1aāˆ’1+4aāˆ’1ā‹…a+2a+2=0\frac{3}{a+2} \cdot \frac{a-1}{a-1} + \frac{4}{a-1} \cdot \frac{a+2}{a+2} = 0

    This gives us:

    3(aāˆ’1)(a+2)(aāˆ’1)+4(a+2)(aāˆ’1)(a+2)=0\frac{3(a-1)}{(a+2)(a-1)} + \frac{4(a+2)}{(a-1)(a+2)} = 0

  3. Combine the Fractions: Now that the fractions have a common denominator, we can combine them:

    3(aāˆ’1)+4(a+2)(a+2)(aāˆ’1)=0\frac{3(a-1) + 4(a+2)}{(a+2)(a-1)} = 0

  4. Simplify the Numerator: Expand and simplify the numerator:

    3aāˆ’3+4a+8(a+2)(aāˆ’1)=0\frac{3a - 3 + 4a + 8}{(a+2)(a-1)} = 0

    7a+5(a+2)(aāˆ’1)=0\frac{7a + 5}{(a+2)(a-1)} = 0

  5. Solve for a: A fraction is equal to zero if and only if its numerator is equal to zero. Therefore, we set the numerator equal to zero and solve for a:

    7a+5=07a + 5 = 0

    7a=āˆ’57a = -5

    a=āˆ’57a = -\frac{5}{7}

  6. Check for Extraneous Solutions: We need to ensure that our solution does not make any of the original denominators equal to zero. We already identified that a cannot be -2 or 1. Our solution, a = -5/7, is not equal to either of these values, so it is a valid solution.

In this section, we will delve deeper into each step of the solution process, providing additional context and explanation. This will help solidify your understanding of the techniques used and the reasoning behind them.

1. Finding the Common Denominator

The first critical step in solving the equation $\frac{3}{a+2} + \frac{4}{a-1} = 0$ is finding the common denominator. The common denominator allows us to combine the two fractions into a single expression, which simplifies the equation and makes it easier to solve. When dealing with algebraic fractions, the common denominator is often the product of the individual denominators. In this case, the denominators are a + 2 and a - 1, so the common denominator is (a + 2)(a - 1). This method ensures that we have a denominator that is divisible by both original denominators.

Finding the common denominator is a fundamental concept in algebra, and it's not just limited to solving equations. It's also used in simplifying algebraic expressions, adding and subtracting fractions, and performing various other algebraic operations. Mastering this skill is essential for anyone looking to excel in algebra and related fields.

The importance of the common denominator lies in its ability to unify the fractions. By expressing both fractions with the same denominator, we can directly add or subtract their numerators, which simplifies the overall expression. This is analogous to adding fractions with numerical denominators, where we need a common denominator before we can combine the fractions.

2. Rewriting the Fractions with the Common Denominator

Once we've identified the common denominator as (a + 2)(a - 1), the next step is to rewrite each fraction with this common denominator. Rewriting fractions involves multiplying both the numerator and the denominator of each fraction by the appropriate factor. This ensures that the value of the fraction remains unchanged while expressing it with the desired denominator.

For the first fraction, $\frac{3}{a+2}$, we need to multiply both the numerator and the denominator by (a - 1) to obtain the common denominator. This gives us $\frac{3(a-1)}{(a+2)(a-1)}$. Similarly, for the second fraction, $\frac{4}{a-1}$, we multiply both the numerator and the denominator by (a + 2) to obtain the common denominator, resulting in $\frac{4(a+2)}{(a-1)(a+2)}$. These manipulations are crucial for combining the fractions and simplifying the equation.

This process highlights the importance of the fundamental principle of fractions, which states that multiplying both the numerator and the denominator of a fraction by the same non-zero quantity does not change the value of the fraction. This principle allows us to manipulate fractions without altering their underlying value, which is essential for solving equations and simplifying expressions.

3. Combining the Fractions

After rewriting the fractions with the common denominator, we can now combine the fractions into a single expression. This is achieved by adding the numerators while keeping the common denominator. In our case, we have:

3(aāˆ’1)(a+2)(aāˆ’1)+4(a+2)(aāˆ’1)(a+2)=3(aāˆ’1)+4(a+2)(a+2)(aāˆ’1)\frac{3(a-1)}{(a+2)(a-1)} + \frac{4(a+2)}{(a-1)(a+2)} = \frac{3(a-1) + 4(a+2)}{(a+2)(a-1)}

This step is a direct application of the rule for adding fractions with a common denominator. By combining the fractions, we've reduced the equation to a simpler form, which is easier to analyze and solve. The combined fraction now represents the entire left-hand side of the equation as a single expression.

The act of combining fractions is a powerful technique in algebra, allowing us to simplify complex expressions and equations. It's a skill that is frequently used in various mathematical contexts, including calculus, differential equations, and more. Understanding how to combine fractions effectively is crucial for success in these areas.

4. Simplifying the Numerator

The next step in solving the equation is simplifying the numerator of the combined fraction. This involves expanding any products and combining like terms. In our case, the numerator is 3(a - 1) + 4(a + 2). Expanding the products, we get 3a - 3 + 4a + 8. Combining the like terms (3a and 4a, and -3 and 8), we obtain 7a + 5. Therefore, the simplified numerator is 7a + 5.

Simplifying the numerator is an important step because it reduces the complexity of the equation and makes it easier to isolate the variable. This step often involves the distributive property and the combination of like terms, which are fundamental algebraic skills. A simplified numerator allows us to focus on the essential part of the equation that determines the solution.

Simplifying expressions is a cornerstone of algebra, and it's a skill that is used throughout mathematics. Whether you're solving equations, simplifying expressions, or working with more advanced concepts, the ability to simplify is crucial for success.

5. Solving for a

Now that we have the simplified fraction $\frac{7a + 5}{(a+2)(a-1)} = 0$, we can solve for a. A fraction is equal to zero if and only if its numerator is equal to zero. Therefore, we set the numerator equal to zero and solve the resulting equation: 7a + 5 = 0.

To solve this equation, we first subtract 5 from both sides, which gives us 7a = -5. Then, we divide both sides by 7, which yields a = -5/7. This is our potential solution for the equation. However, it's essential to check whether this solution is valid by considering any restrictions on the variable a.

The principle that a fraction is zero if and only if its numerator is zero is a key concept in solving rational equations. This principle allows us to transform a potentially complex equation into a simpler one by focusing solely on the numerator. Solving linear equations, such as 7a + 5 = 0, is a fundamental skill in algebra, and it's a skill that is applied in numerous mathematical contexts.

6. Checking for Extraneous Solutions

The final step in solving the equation is checking for extraneous solutions. Extraneous solutions are solutions that arise during the solving process but do not satisfy the original equation. These solutions often occur when dealing with rational equations, where certain values of the variable may make the denominator equal to zero, rendering the fraction undefined.

In our case, we identified at the beginning that a cannot be equal to -2 or 1 because these values would make the denominators of the original fractions equal to zero. Our solution, a = -5/7, is not equal to either -2 or 1, so it is a valid solution. If we had obtained a solution that was equal to -2 or 1, we would have had to discard it as an extraneous solution.

Checking for extraneous solutions is a crucial step in solving rational equations and other types of equations where restrictions on the variable may exist. Failing to check for extraneous solutions can lead to incorrect results and a misunderstanding of the solution set of the equation.

In conclusion, the solution to the equation $\frac{3}{a+2} + \frac{4}{a-1} = 0$ is a = -5/7. We arrived at this solution by finding a common denominator, combining the fractions, simplifying the numerator, solving for a, and checking for extraneous solutions. This process demonstrates the key steps involved in solving rational equations and highlights the importance of algebraic manipulation and careful consideration of restrictions on the variable.

Understanding how to solve rational equations is a valuable skill in algebra and beyond. By mastering these techniques, you'll be well-equipped to tackle more complex problems and deepen your understanding of mathematical concepts.

  • Solving rational equations
  • Common denominator
  • Extraneous solutions
  • Algebraic manipulation
  • Simplifying expressions
  • Solving for a
  • Fractions with algebraic expressions