Solving the Equation yx-1 z for x

Solving the Equation yx-1 z for x

This article will guide you through the process of solving the equation yx - 1 z for the variable x. Understanding how to manipulate and solve such equations is crucial in various fields, including mathematics, physics, and engineering. We will explore the step-by-step solution while also addressing common confusions and clarifying some nuances.

Step-by-Step Solution

Given the equation:

yx - 1 z

Isolate the term with x:

Starting with the equation:

yx - 1 z

Divide both sides by y, assuming y ≠ 0:

This step simplifies the equation by isolating the term with x.

x - 1 z/y

Add 1 to both sides to solve for x:

This final step yields the value of x in terms of the constants y and z.

x z/y 1

Therefore, the solution for x is:

x (z/y) 1

Understanding the Inverse Operations

In mathematics, inverse operations are crucial for solving equations. For example, if you have an equation like:

x - 1 z/y

The inverse operation would be to add 1 to both sides:

x - 1 1 z/y 1

This simplifies to:

x z/y 1

Remember, if y 0, the equation becomes undefined because division by zero is not allowed in mathematics. In such a case, z would also have to be 0 for the equation to be valid, and x could be any value.

Special Case: y 0

If y 0, then the equation yx - 1 z is undefined because of division by zero. However, if the function is defined as y(x - 1) z, then it simplifies to:

y(x - 1) z

Dividing both sides by y:

x - 1 z/y

Then adding 1 to both sides:

x z/y 1

Again, if y 0, the equation becomes undefined, and x can take any value.

Verification Using WolframAlpha

WolframAlpha, a powerful computational knowledge engine, also supports this solution:

Assuming y ≠ 0, yx - 1 z implies x - 1 z/y, which implies:

x 1 z/y

This is the same result we obtained by manual calculation.

Moreover, if y 0, then x can be any value.

Conclusion

In summary, solving the equation yx - 1 z for x involves isolating the term with x and using inverse operations to find the solution. The process is straightforward and can be applied to similar algebraic equations.