Solve For Z: Step-by-Step Solution

Hey there, math enthusiasts! Today, we're diving into an exciting equation that involves solving for the variable z. Equations like this might seem daunting at first, but don't worry, we'll break it down step by step and make it super easy to understand. So, grab your calculators and let's get started!

The Equation We're Tackling

The equation we're going to solve is:

2 + \frac{3}{z-1} = -\frac{3}{(z+3)(z-1)}

This looks a bit complicated, right? We've got fractions, variables in the denominators, and a whole lot of mathematical operations going on. But fear not! With a systematic approach, we can conquer this equation and find the values of z that make it true.

Step 1: Identifying the Domain and Restrictions

Before we jump into solving, it's crucial to identify the domain of the equation. The domain is the set of all possible values of z for which the equation is defined. In simpler terms, we need to figure out which values of z would cause any problems, like division by zero.

Looking at our equation, we have two fractions with denominators that involve z: z - 1 and (z + 3)(z - 1). Division by zero is a big no-no in mathematics, so we need to make sure these denominators are never equal to zero.

Let's analyze each denominator:

• z - 1 ≠ 0 implies that z cannot be equal to 1. If z were 1, we'd have division by zero in the first fraction.
• (z + 3)(z - 1) ≠ 0 implies that z cannot be equal to -3 or 1. If z were -3 or 1, the second fraction would involve division by zero.

Therefore, the restrictions on z are that z cannot be equal to 1 or -3. These values are excluded from the domain of the equation. This means that any solutions we find later must not be 1 or -3, or they will be considered extraneous solutions.

Step 2: Eliminating the Fractions

Fractions can make equations look messy and complicated. Our goal now is to get rid of them by finding a common denominator and multiplying both sides of the equation by it. This will clear the fractions and give us a simpler equation to work with.

Looking at our equation again:

2 + \frac{3}{z-1} = -\frac{3}{(z+3)(z-1)}

The denominators we have are z - 1 and (z + 3)(z - 1). The least common denominator (LCD) is the smallest expression that is divisible by both denominators. In this case, the LCD is (z + 3)(z - 1).

To eliminate the fractions, we'll multiply both sides of the equation by the LCD:

(z + 3)(z - 1) \left( 2 + \frac{3}{z-1} \right) = (z + 3)(z - 1) \left( -\frac{3}{(z+3)(z-1)} \right)

Now, we distribute the LCD on the left side of the equation:

2(z + 3)(z - 1) + \frac{3(z + 3)(z - 1)}{z - 1} = -\frac{3(z + 3)(z - 1)}{(z+3)(z-1)}

Notice how some terms cancel out beautifully:

2(z + 3)(z - 1) + 3(z + 3) = -3

We've successfully eliminated the fractions! The equation now looks much cleaner and easier to handle.

Step 3: Simplifying and Rearranging the Equation

Our next step is to simplify the equation by expanding the expressions and combining like terms. This will help us get the equation into a standard form that we can easily solve.

Let's start by expanding the products on the left side of the equation:

2(z^2 + 3z - z - 3) + 3(z + 3) = -3

Simplify further:

2(z^2 + 2z - 3) + 3z + 9 = -3

Distribute the 2:

2z^2 + 4z - 6 + 3z + 9 = -3

Now, combine like terms:

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

To solve for z, we want to get the equation into the standard quadratic form, which is ax^2 + bx + c = 0. To do this, we add 3 to both sides of the equation:

2z^2 + 7z + 6 = 0

Great! We now have a quadratic equation in standard form.

Step 4: Solving the Quadratic Equation

We have a quadratic equation: 2z^2 + 7z + 6 = 0. There are several ways to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. For this equation, factoring seems like a good approach.

To factor the quadratic, we need to find two numbers that multiply to give 2 * 6 = 12 and add up to 7. Those numbers are 3 and 4.

Now we rewrite the middle term using these numbers:

2z^2 + 4z + 3z + 6 = 0

Next, we factor by grouping:

2z(z + 2) + 3(z + 2) = 0

Notice that (z + 2) is a common factor. We can factor it out:

(2z + 3)(z + 2) = 0

Now, we apply the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So, we set each factor equal to zero and solve for z:

2z + 3 = 0 \quad \text{or} \quad z + 2 = 0

Solving the first equation:

2z = -3
z = -\frac{3}{2}

Solving the second equation:

z = -2

So, we have two potential solutions: z = -\frac{3}{2} and z = -2.

Step 5: Checking for Extraneous Solutions

Remember those restrictions we identified in Step 1? We need to check if our solutions satisfy those restrictions. The restrictions were that z cannot be equal to 1 or -3.

Our solutions are z = -\frac{3}{2} and z = -2. Neither of these values is equal to 1 or -3, so they are both valid solutions.

The Final Answer

We've done it! We've successfully solved the equation and found the values of z that make it true. The solutions are:

z = -\frac{3}{2}, -2

So, if you were asked to provide the solutions separated by commas, the answer would be:

-\frac{3}{2}, -2

Wrapping Up

Solving equations with fractions and variables in the denominators might seem tricky at first, but by following a systematic approach, you can conquer them with ease. Remember to identify the domain and restrictions, eliminate the fractions, simplify the equation, solve for the variable, and check for extraneous solutions.

Keep practicing, and you'll become a pro at solving all sorts of equations! And remember, math can be fun when you break it down step by step. Keep exploring, keep learning, and keep solving!

If you guys have any questions, feel free to ask in the comments below. Until next time, happy solving!