Solving Rational Equations: A Step-by-Step Guide

Welcome, math enthusiasts! Today, we're diving into the fascinating world of rational equations. These equations, which involve fractions with variables in the numerator or denominator, can sometimes seem a bit tricky. However, with a clear understanding of the steps involved, you'll find them quite manageable. Our main goal is to solve the rational equation for z and to identify any potential extraneous solutions, ensuring our final answers are valid. We'll walk through the process systematically, breaking down each step to make it as easy to follow as possible.

Understanding Rational Equations and Extraneous Solutions

Before we jump into solving, let's clarify what rational equations are and why we need to be mindful of extraneous solutions. A rational equation is essentially an equation that contains one or more rational expressions (fractions with polynomials). The presence of variables in the denominators is what introduces the possibility of extraneous solutions. An extraneous solution is a value that we obtain during the solving process that, when substituted back into the original equation, makes a denominator equal to zero. Since division by zero is undefined, any solution that causes this is not a true solution to the original equation.

Why are extraneous solutions important? Imagine you're trying to solve a problem where a variable represents a physical quantity, like the number of items produced or a distance. If you get a solution that leads to an undefined situation, it means that specific scenario is impossible within the context of the problem. Therefore, always double-checking your solutions against the original equation is a crucial step. For the specific equation we'll be tackling, $\frac{3}{z}-\frac{z}{z+6}=\frac{18}{z^2+6 z}$, we need to be particularly attentive to values of $z$ that would make any denominator zero.

Let's identify the denominators in our equation: $z$, $z+6$, and $z^2+6z$. For these denominators to be non-zero, we must have $z \neq 0$ and $z+6 \neq 0$ (which means $z \neq -6$). The third denominator, $z^2+6z$, can be factored as $z(z+6)$. So, the restrictions we identified from the first two denominators are sufficient. This means that if we find $z=0$ or $z=-6$ as potential solutions during our calculations, we must discard them as they are extraneous solutions. Recognizing these restrictions upfront is a powerful strategy for efficiently solving rational equations.

Step-by-Step Solution Process

Now, let's get down to the business of solving our rational equation. The most effective strategy for tackling equations like $\frac{3}{z}-\frac{z}{z+6}=\frac{18}{z^2+6 z}$ is to eliminate the denominators by multiplying both sides of the equation by the Least Common Denominator (LCD). This simplifies the equation into a polynomial form that is much easier to solve.

Step 1: Find the Least Common Denominator (LCD).

First, we need to factor all the denominators. We already did this for $z^2+6z$, which factors into $z(z+6)$. Our denominators are $z$, $z+6$, and $z(z+6)$. The LCD is the smallest expression that is a multiple of all these denominators. In this case, the LCD is simply $z(z+6)$.

Step 2: Multiply every term by the LCD.

We will multiply each term on both sides of the equation by $z(z+6)$:

$$z(z+6) \left( \frac{3}{z} \right) - z(z+6) \left( \frac{z}{z+6} \right) = z(z+6) \left( \frac{18}{z(z+6)} \right)$$

Now, let's simplify each term by canceling out common factors:

• For the first term: $z(z+6) \left( \frac{3}{z} \right) = (z+6) \cdot 3 = 3z + 18$
• For the second term: $z(z+6) \left( \frac{z}{z+6} \right) = z \cdot z = z^2$
• For the third term: $z(z+6) \left( \frac{18}{z(z+6)} \right) = 18$

Step 3: Simplify and solve the resulting equation.

After canceling the denominators, our equation transforms into:

$$3z + 18 - z^2 = 18$$

This is a quadratic equation. To solve it, we need to set it equal to zero and simplify. Let's move all terms to one side:

$$0 = z^2 + 18 - (3z + 18)$$

$$0 = z^2 + 18 - 3z - 18$$

$$0 = z^2 - 3z$$

Now we have a simplified quadratic equation: $z^2 - 3z = 0$. We can solve this by factoring. Notice that both terms have a common factor of $z$:

$$z(z - 3) = 0$$

For this product to be zero, at least one of the factors must be zero. So, we set each factor equal to zero:

• $z = 0$
• $z - 3 = 0 \implies z = 3$

Step 4: Check for extraneous solutions.

This is a critical step! We found two potential solutions: $z=0$ and $z=3$. Now we must check these against the restrictions we identified earlier: $z \neq 0$ and $z \neq -6$.

• Check $z=0$: When we substitute $z=0$ back into the original equation $\frac{3}{z}-\frac{z}{z+6}=\frac{18}{z^2+6 z}$, the first term $\frac{3}{z}$ becomes $\frac{3}{0}$, which is undefined. Therefore, $z=0$ is an extraneous solution.
• Check $z=3$: When we substitute $z=3$ into the original equation: Left side: $\frac{3}{3} - \frac{3}{3+6} = 1 - \frac{3}{9} = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}$ Right side: $\frac{18}{3^2+6(3)} = \frac{18}{9+18} = \frac{18}{27} = \frac{2 \cdot 9}{3 \cdot 9} = \frac{2}{3}$ Since the left side equals the right side ($\frac{2}{3} = \frac{2}{3}$), $z=3$ is a valid real solution.

Analyzing the Solutions

Based on our step-by-step process and the crucial check for extraneous solutions, we can now complete the table provided:

This means that out of the two potential solutions we found ($z=0$ and $z=3$), only one ($z=3$) is a true solution to the original rational equation. The other ($z=0$) was identified as an extraneous solution because it made the denominators in the original equation undefined.

Conclusion

Solving rational equations requires a methodical approach, and the identification of extraneous solutions is a non-negotiable step. By finding the LCD, multiplying to clear denominators, simplifying, and then rigorously checking our potential solutions against the original equation's restrictions, we can confidently arrive at the correct answers. The process for solving $\frac{3}{z}-\frac{z}{z+6}=\frac{18}{z^2+6 z}$ yielded one valid real solution, $z=3$, and one extraneous solution, $z=0$. Remember, always be aware of the values that would make any denominator zero and exclude them from your final set of solutions.

For further exploration into algebraic equations and problem-solving techniques, you might find the resources at Khan Academy incredibly helpful. They offer a vast array of free courses and exercises covering many mathematical topics, including rational equations and functions.