Solving Complex Equations: gx x/(1-x) and hx x/(1-x)

Algebra provides a powerful tool to explore complex equations and functions, enabling us to uncover the underlying patterns and relationships that govern mathematical expressions. In this article, we will delve into the intricacies of solving equations of the form gx x/(1-x) and hx x/(1-x), exploring various methods and transformations to find the solutions.

Solving for x in gx x/(1-x)

Firstly, let's consider the equation gx x/(1-x). Our goal is to solve for x under this constraint. To begin, we need to understand the behavior of the function defined by g(x).

Initial Simplification

Assume gx 0. This implies that the numerator of the right-hand side must be zero, or the denominator must be non-zero and undefined, which isn't feasible in this context. Thus, we focus on the numerator:

[ 1x 0 ] [ x 0 ]

So, one potential solution is x 0. To check if this solution is valid, we substitute it back into the original function:

[ g(0) frac{1 cdot 0}{1 - 0} 0 ] This confirms that x 0 is indeed a solution to the equation gx 0.

Solving for x in hx x/(1-x)

Nexly, let's consider the equation hx x/(1-x). Similar to the previous case, we want to find the values of x that satisfy this equation. Assuming hx 0, we again have:

[ x 0 ]

We can verify this by substituting x 0 back into the original function:

[ h(0) frac{0}{1 - 0} 0 ] This confirms that x 0 is a solution to the equation hx 0.

Comparing gx and hx

To find the relationship between g(x) and h(x) when both are defined as x/(1-x), we solve for fgx hx:

[ frac{1x}{1-x} frac{x}{1-x} ] Since the denominators are the same, we can equate the numerators: [ 1x x ] This equation holds true for all x except x 1 (to avoid division by zero). Therefore, there is no unique solution for x in the context of the equation fgx hx.

Revisiting the Relationship: g(x) in terms of h(x)

We now explore the relationship between g(x) and h(x) by expressing g(x) as a function of h(x). Given that both functions are defined as x/(1-x), we can rewrite g(x) in terms of h(x):

[ g(x) frac{1x}{1-x} frac{1}{1-x}h(x) ] [ Rightarrow g(x) - h(x) frac{1}{1-x} ] [ Rightarrow g(x) h(x) frac{1}{1-x} ]

To find x in terms of h(x) and g(x), we rearrange the equation:

[ g(x) - h(x) frac{1}{1-x} ] [ Rightarrow 1-x frac{1}{g(x) - h(x)} ] [ Rightarrow x 1 - frac{1}{g(x) - h(x)} ] [ Rightarrow x frac{g(x) - h(x) - 1}{g(x) - h(x)} ]

Transforming Equations: Solving for x

Now, let's solve for x in terms of g(x) and h(x) using the equations x (g(x) - 1)/(g(x) 1) and x h(x)/(h(x) - 1):

[ frac{g(x) - 1}{g(x) 1} frac{h(x)}{h(x) - 1} ] Cross-multiplying, we get: [ (g(x) - 1)(h(x) - 1) (g(x) 1)h(x) ] [ Rightarrow g(x)h(x) - g(x) - h(x) 1 g(x)h(x) h(x) ] [ Rightarrow -g(x) - h(x) 1 h(x) ] [ Rightarrow -g(x) 2h(x) - 1 ] [ Rightarrow g(x) 2 - 2h(x) ] [ Rightarrow h(x) frac{g(x) - 1}{2} ]

Conclusion

In conclusion, we have explored the equations gx x/(1-x) and hx x/(1-x) and found that the solutions for x in these equations are x 0 when g(x) 0 and h(x) 0, respectively. We also found the relationship between g(x) and h(x) and derived a transformation that allows us to express g(x) in terms of h(x) and vice versa. Understanding these transformations and their implications is crucial for solving complex algebraic equations and functions.