Solving The Equation 9/z + 9/(z-2) = 12

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Welcome, math enthusiasts! Today, we're diving deep into the fascinating world of algebraic equations. Specifically, we'll be tackling a rational equation: $\frac{9}{z}+\frac{9}{z-2}=12$. This type of equation involves fractions with variables in the denominator, and solving them often requires a systematic approach to eliminate those pesky denominators and arrive at a simpler polynomial equation. Our goal is to find the value(s) of $z$ that make this equation true. We'll explore the steps involved in isolating the variable and uncover the potential solutions, which could be a set of numbers, a single value, or even no solution at all. Get ready to flex those mathematical muscles as we unravel this intriguing problem!

Understanding Rational Equations

Let's first set the stage by understanding what a rational equation is. In essence, it's an equation that contains one or more rational expressions. A rational expression is simply a fraction where the numerator and the denominator are polynomials. The key characteristic, and often the source of complexity, is the presence of variables in the denominator. This means we have to be mindful of certain values of the variable that would make the denominator zero, as division by zero is undefined. For the equation $\frac{9}{z}+\frac{9}{z-2}=12$, the denominators are $z$ and $z-2$. Therefore, we must acknowledge that $z \neq 0$ and $z \neq 2$. These are called extraneous solutions – values that might appear as solutions to our simplified equation but are not valid for the original equation because they would lead to division by zero. Keeping these restrictions in mind is crucial as we proceed with solving.

The Strategy: Clearing the Denominators

The most common and effective strategy for solving rational equations is to eliminate the denominators. We achieve this by multiplying every term in the equation by the least common denominator (LCD). The LCD is the smallest expression that is a multiple of all the denominators in the equation. In our case, the denominators are $z$ and $z-2$. Since these two expressions have no common factors, their LCD is simply their product: $z(z-2)$. By multiplying each term by $z(z-2)$, we will cancel out the denominators, transforming our rational equation into a more manageable polynomial equation. This step is pivotal because it simplifies the problem significantly, allowing us to use standard algebraic techniques to isolate the variable. Remember, whatever operation we perform on one side of the equation, we must perform on the other to maintain equality. This principle guides us as we systematically eliminate the fractional components and move closer to finding the solution(s).

Step-by-Step Solution

Now, let's roll up our sleeves and work through the equation $\frac{9}{z}+\frac{9}{z-2}=12$ step-by-step. Our first move, as discussed, is to multiply every term by the LCD, which is $z(z-2)$.

$$z(z-2) \left( \frac{9}{z} \right) + z(z-2) \left( \frac{9}{z-2} \right) = z(z-2)(12)$$

Notice how in the first term, the $z$ in the denominator cancels out with the $z$ from the LCD, leaving us with $(z-2) \times 9$. Similarly, in the second term, the $(z-2)$ in the denominator cancels out with the $(z-2)$ from the LCD, leaving us with $z \times 9$. On the right side of the equation, we distribute the $12$ across $z(z-2)$, resulting in $12z(z-2)$.

This gives us:

$$9(z-2) + 9z = 12z(z-2)$$

Next, we distribute and simplify:

$$9z - 18 + 9z = 12z^2 - 24z$$

Combine like terms on the left side:

$$18z - 18 = 12z^2 - 24z$$

Now, we want to rearrange this into a standard quadratic equation form, $ax^2 + bx + c = 0$. To do this, we move all terms to one side. Let's move the terms from the left to the right side to keep the $z^2$ term positive:

$$0 = 12z^2 - 24z - 18z + 18$$

Combine the $z$ terms:

$$0 = 12z^2 - 42z + 18$$

This is our quadratic equation. We can simplify it further by dividing all terms by their greatest common divisor, which is 6:

$$0 = 2z^2 - 7z + 3$$

We have successfully transformed the original rational equation into a quadratic equation. The next step is to solve this quadratic equation to find the possible values of $z$. Remember, these are potential solutions, and we'll need to check them against our initial restrictions ($z \neq 0$ and $z \neq 2$) later.

Solving the Quadratic Equation

We are now faced with the quadratic equation $2z^2 - 7z + 3 = 0$. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. Let's try factoring first, as it's often the quickest if the equation is factorable.

We are looking for two numbers that multiply to $(2 \times 3 = 6)$ and add up to $-7$. These numbers are $-1$ and $-6$. We can use these numbers to rewrite the middle term $-7z$ as $-z - 6z$:

$$2z^2 - z - 6z + 3 = 0$$

Now, we can factor by grouping. Group the first two terms and the last two terms:

$$(2z^2 - z) + (-6z + 3) = 0$$

Factor out the greatest common factor from each group:

$$z(2z - 1) - 3(2z - 1) = 0$$

Notice that we have a common binomial factor $(2z - 1)$. We can factor this out:

$$(2z - 1)(z - 3) = 0$$

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

Case 1: $2z - 1 = 0$

$$2z = 1$$

$$z = \frac{1}{2}$$

Case 2: $z - 3 = 0$

$$z = 3$$

So, our potential solutions are $z = \frac{1}{2}$ and $z = 3$. These are the values that satisfy the quadratic equation $2z^2 - 7z + 3 = 0$. The next crucial step is to verify if these solutions are valid for the original rational equation by checking them against the restrictions we identified earlier: $z \neq 0$ and $z \neq 2$.

Checking for Extraneous Solutions

This is a critical phase in solving rational equations. We found two potential solutions: $z = \frac{1}{2}$ and $z = 3$. Now, we must compare these with the values that would make the original denominators zero. Our restrictions were $z \neq 0$ and $z \neq 2$.

Let's examine our first potential solution, $z = \frac{1}{2}$. Is $\frac{1}{2}$ equal to 0? No. Is $\frac{1}{2}$ equal to 2? No. Therefore, $z = \frac{1}{2}$ is a valid solution.

Now, let's look at our second potential solution, $z = 3$. Is 3 equal to 0? No. Is 3 equal to 2? No. Therefore, $z = 3$ is also a valid solution.

Since both potential solutions do not violate the initial restrictions, they are both actual solutions to the original equation $\frac{9}{z}+\frac{9}{z-2}=12$. This means that when $z$ is $\frac{1}{2}$ or $3$, the equation holds true. It's always a good practice to plug these values back into the original equation to double-check, but since they passed the restriction test, we can be confident in our answer.

Conclusion: The Solutions Revealed

After carefully navigating the steps of clearing denominators, simplifying, and solving the resulting quadratic equation, we have arrived at our answer. The equation $\frac{9}{z}+\frac{9}{z-2}=12$ has two distinct solutions. These solutions are $z = \frac{1}{2}$ and $z = 3$. Both of these values are valid because they do not cause any denominator in the original equation to become zero. This meticulous process ensures that we have found all possible values for $z$ that satisfy the given condition.

Understanding how to solve rational equations is a fundamental skill in algebra, opening doors to solving more complex problems in various fields, including engineering, physics, and economics. The key is to be systematic, mindful of restrictions, and confident in your algebraic manipulation.

For further exploration and practice with algebraic equations, you might find the resources at Khan Academy to be incredibly helpful. They offer a wide range of lessons and practice problems on various mathematical topics, including solving rational and quadratic equations.

Options Recap:

A. no solution B. 3 C. $\frac{1}{2}$ and 3 D. $\frac{1}{2}$

Based on our analysis, the correct option is C. $\frac{1}{2}$ and 3.