Solving for Z in the Equation X YZ - Y/Z

When dealing with mathematical equations, understanding how to solve for a specific variable can be a challenging but rewarding skill.

In this article, we will walk you through the steps to find the value of Z in the equation X YZ - Y/Z when X YZ is utilized. This process involves substitution, algebraic manipulation, and solving a quadratic equation. Let's break down the steps in detail.

Step 1: Substitute X YZ

The given equation is X YZ - Y/Z. The first step to solving for Z is to substitute X YZ. This simplifies the equation and makes it easier to manipulate algebraically:

YZ YZ - Y/Z

Multiplying both sides of the equation by Z to eliminate the fraction, we get:

Step 2: Multiply by Z

Multiplying both sides of the equation YZ YZ - Y/Z by Z, we obtain:

YZ^2 YZ - Y

Now, to isolate terms involving Z, divide both sides of the equation by Y:

Step 3: Divide by Y

Dividing by Y, we get:

Z^2 Z - 1/Y

Finally, we move all terms to one side of the equation to form a standard quadratic equation:

Step 4: Move all to the left side

Moving all terms to one side, we get:

Z^2 - Z 1/Y 0

This is a standard quadratic equation of the form:

Z^2 bZ c 0

Using the quadratic formula, where a 1, b -1, and c 1/Y, we can solve for Z.

Step 5: Apply the Quadratic Formula

The quadratic formula is given by:

Z (-b ± V(b^2 - 4ac)) / 2a

Substituting the values of a, b, and c into this formula, we get:

Z (1 ± V(-1^2 - 4(1)(1/Y))) / 2(1)

Simplifying, we get:

Z (1 ± V(1 - 4/Y)) / 2

Since we are dealing with a real-world problem, the term under the square root should be non-negative. Thus, we are only interested in the real solutions where:

1 - 4/Y > 0

Therefore, the real values of Z are obtained as:

Z (1 - V(1 - 4/Y)) / 2

If Y 0, the equation simplifies to:

Z (1 - V(1)) / 2 Z 0

This gives us the complete solution set for Z.

Conclusion

Understanding how to solve for a variable in complex algebraic equations is a crucial skill. In this article, we utilized substitution, algebraic manipulation, and the quadratic formula to find the value of Z in the given equation. With these steps, you can now apply similar techniques to a variety of other algebraic problems.