Let's solve the equation step by step:

The given equation is:

\[
0 = (2a - 1)e^{a^2 - a} (x - a) + e^{a^2 - a}
\]

First, notice that \(e^{a^2 - a}\) appears as a common factor in both terms. We can factor it out:

\[
0 = e^{a^2 - a} \left( (2a - 1)(x - a) + 1 \right)
\]

Since \(e^{a^2 - a}\) is never zero (because the exponential function is never zero for any real number), we can divide both sides of the equation by \(e^{a^2 - a}\):

\[
0 = (2a - 1)(x - a) + 1
\]

Now, expand the terms on the right-hand side:

\[
0 = (2a - 1)(x - a) + 1 = (2a - 1)x - (2a - 1)a + 1
\]

Simplifying:

\[
0 = (2a - 1)x - (2a - 1)a + 1
\]

Group the constant terms together:

\[
0 = (2a - 1)x + \left( 1 - (2a - 1)a \right)
\]

Simplifying the constant term:

\[
1 - (2a - 1)a = 1 - 2a^2 + a
\]

Thus, the equation becomes:

\[
0 = (2a - 1)x + (1 - 2a^2 + a)
\]

Now solve for \(x\):

\[
(2a - 1)x = - (1 - 2a^2 + a)
\]

\[
x = \frac{-(1 - 2a^2 + a)}{2a - 1}
\]

So the solution for \(x\) is:

\[
x = \frac{2a^2 - a - 1}{2a - 1}
\]

### Final Answer:

\[
x = \frac{2a^2 - a - 1}{2a - 1}
\]

Would you like more details on any step or have any further questions?

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Here are 5 related questions to explore:

1. How would you solve this equation if \(a = 1\)?
2. What happens to the solution for \(x\) if \(a = 0.5\)?
3. Can this equation be solved graphically? How would the graph look?
4. What is the significance of \(e^{a^2 - a}\) in the original equation?
5. How would the solution change if the original equation had another exponential term?

**Tip**: Always check if a common factor can be factored out, as it simplifies solving equations significantly.

0 = (2a - 1)e^(a^2 - a) (x - a) + e^(a^2 - a)

Learn how to solve the equation 0 = (2a - 1)e^(a^2 - a) (x - a) + e^(a^2 - a) by factoring out the exponential term and solving for x. This problem involves exponential functions, algebraic manipulation, and factored linear equations. Suitable for students in Grades 10-12.

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