Solving Rational Equations Find Values That Make Denominators Zero

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In the realm of algebra, rational equations present a unique challenge and opportunity for problem-solving. These equations, characterized by fractions with variables in the denominator, demand a meticulous approach to ensure accurate solutions. This article delves into the intricacies of solving the rational equation 4x+12−3x−12=5xx2−144\frac{4}{x+12}-\frac{3}{x-12}=\frac{5x}{x^2-144}, providing a step-by-step guide and insightful explanations to enhance your understanding. We will also address the crucial aspect of identifying values that make the denominator zero, as these values must be excluded from the solution set.

Understanding Rational Equations

Rational equations, at their core, are equations that contain at least one fraction whose numerator and/or denominator are polynomials. The equation we're tackling, 4x+12−3x−12=5xx2−144\frac{4}{x+12}-\frac{3}{x-12}=\frac{5x}{x^2-144}, perfectly exemplifies this type. The presence of variables in the denominators necessitates careful handling to avoid division by zero, an undefined operation in mathematics. Before diving into the solution, it's crucial to identify any values of x that would make any of the denominators zero. These values, known as excluded values, cannot be part of the solution set.

To find the excluded values, we set each denominator equal to zero and solve for x. In our equation, the denominators are x + 12, x - 12, and x² - 144. Setting x + 12 = 0 gives us x = -12, and setting x - 12 = 0 gives us x = 12. The denominator x² - 144 is a difference of squares, which can be factored as (x + 12)(x - 12). This confirms that x = -12 and x = 12 are the values that make this denominator zero as well. Therefore, x cannot be -12 or 12.

Identifying Values That Make a Denominator Zero

This is a critical first step in solving any rational equation. Failing to identify these values can lead to extraneous solutions – solutions that satisfy the transformed equation but not the original equation. We need to determine the values of x that will make any of the denominators in the equation equal to zero. These values are excluded from the solution set because division by zero is undefined.

In our equation, 4x+12−3x−12=5xx2−144\frac{4}{x+12}-\frac{3}{x-12}=\frac{5x}{x^2-144}, we have three denominators to consider: (x + 12), (x - 12), and (x² - 144). Let's analyze each one:

  • x + 12 = 0: Solving for x, we subtract 12 from both sides, giving us x = -12. This means that if x is -12, the denominator (x + 12) will be zero.
  • x - 12 = 0: Adding 12 to both sides, we find x = 12. So, if x is 12, the denominator (x - 12) becomes zero.
  • x² - 144 = 0: This is a difference of squares, which can be factored as (x + 12)(x - 12) = 0. As we already found, this equation is satisfied when x = -12 or x = 12. Alternatively, we can add 144 to both sides to get x² = 144. Taking the square root of both sides gives us x = ±12, confirming our previous findings.

Therefore, the values of x that make a denominator zero are -12 and 12. These values must be excluded from our final solution set. This process of identifying excluded values is not merely a preliminary step; it is a fundamental aspect of solving rational equations, ensuring the mathematical integrity of the solution.

Solving the Rational Equation

Now that we've identified the excluded values, we can proceed to solve the equation. The strategy here is to eliminate the fractions by multiplying both sides of the equation by the least common denominator (LCD). This transforms the equation into a more manageable form, typically a polynomial equation.

Finding the Least Common Denominator (LCD)

The LCD is the smallest expression that is divisible by all the denominators in the equation. In our case, the denominators are (x + 12), (x - 12), and (x² - 144). As we noted earlier, x² - 144 can be factored as (x + 12)(x - 12). Therefore, the LCD is simply (x + 12)(x - 12), as it encompasses all the factors present in the denominators. Recognizing this pattern significantly simplifies the process of finding the LCD and, consequently, solving the equation.

Multiplying by the LCD

Multiplying both sides of the equation 4x+12−3x−12=5xx2−144\frac{4}{x+12}-\frac{3}{x-12}=\frac{5x}{x^2-144} by the LCD, (x + 12)(x - 12), is the next crucial step. This action clears the fractions, transforming the rational equation into a polynomial equation that is easier to solve. It is essential to distribute the LCD correctly to each term in the equation to maintain equality and ensure an accurate solution. Here's how it unfolds:

(x + 12)(x - 12) * [4x+12−3x−12\frac{4}{x+12}-\frac{3}{x-12}] = (x + 12)(x - 12) * [5xx2−144\frac{5x}{x^2-144}]

Distributing the LCD on the left side, we get:

(x + 12)(x - 12) * 4x+12\frac{4}{x+12} - (x + 12)(x - 12) * 3x−12\frac{3}{x-12} = (x + 12)(x - 12) * 5xx2−144\frac{5x}{x^2-144}

Now, we simplify by canceling out common factors:

4(x - 12) - 3(x + 12) = 5x

Notice how the denominators have been eliminated, leaving us with a linear equation.

Simplifying and Solving the Equation

Now we expand and simplify the equation:

4x - 48 - 3x - 36 = 5x

Combining like terms on the left side:

x - 84 = 5x

Subtracting x from both sides:

-84 = 4x

Finally, dividing both sides by 4:

x = -21

Checking for Extraneous Solutions

The final step in solving rational equations is to check the solution against the excluded values we identified earlier. This is crucial because multiplying both sides of an equation by an expression containing a variable can sometimes introduce extraneous solutions. An extraneous solution is a value that satisfies the transformed equation but not the original equation. In our case, we found that x = -21 is a potential solution. We also determined that x cannot be -12 or 12.

Since -21 is not equal to -12 or 12, it is not an excluded value. Therefore, we can confidently accept x = -21 as a valid solution to the equation.

To be absolutely sure, it's always a good practice to substitute the solution back into the original equation to verify that it holds true. Let's substitute x = -21 into 4x+12−3x−12=5xx2−144\frac{4}{x+12}-\frac{3}{x-12}=\frac{5x}{x^2-144}:

4−21+12−3−21−12=5(−21)(−21)2−144\frac{4}{-21+12}-\frac{3}{-21-12}=\frac{5(-21)}{(-21)^2-144}

4−9−3−33=−105441−144\frac{4}{-9}-\frac{3}{-33}=\frac{-105}{441-144}

−49+111=−105297\frac{-4}{9}+\frac{1}{11}=\frac{-105}{297}

To check if the left side equals the right side, we need a common denominator for the fractions on the left. The LCD of 9 and 11 is 99:

−4(11)9(11)+1(9)11(9)=−4499+999=−3599\frac{-4(11)}{9(11)}+\frac{1(9)}{11(9)}=\frac{-44}{99}+\frac{9}{99}=\frac{-35}{99}

Now we simplify the right side:

−105297\frac{-105}{297}. Both -105 and 297 are divisible by 3: −105÷3297÷3\frac{-105 ÷ 3}{297 ÷ 3} = −3599\frac{-35}{99}

Thus, we have −3599=−3599\frac{-35}{99} = \frac{-35}{99}, which confirms that x = -21 is indeed a valid solution.

Conclusion

Solving rational equations requires a systematic approach, including identifying excluded values, finding the least common denominator, multiplying to eliminate fractions, solving the resulting equation, and checking for extraneous solutions. By carefully following these steps, you can confidently tackle even the most complex rational equations. Remember, the key is to maintain accuracy and attention to detail throughout the process. The solution to the equation 4x+12−3x−12=5xx2−144\frac{4}{x+12}-\frac{3}{x-12}=\frac{5x}{x^2-144} is x = -21. This exercise illustrates the importance of understanding the underlying principles of algebra and applying them methodically to arrive at the correct solution.

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