The problem shown is written in Kazakh and asks for solving an equation. Let’s focus on the given equation:

Given equation:

$\frac{3}{x} + \frac{4}{x - 1} = \frac{5 - x}{x^2 - x}$

Step 1: Simplify the right-hand side

Notice that the denominator $x^2 - x$ can be factored: $x^2 - x = x(x - 1)$ Thus, the equation becomes: $\frac{3}{x} + \frac{4}{x - 1} = \frac{5 - x}{x(x - 1)}$

Step 2: Find a common denominator

The left-hand side has two fractions: $\frac{3}{x}$ and $\frac{4}{x - 1}$. Their common denominator is $x(x - 1)$.

• Rewrite both fractions with this common denominator: $\frac{3}{x} = \frac{3(x - 1)}{x(x - 1)}, \quad \frac{4}{x - 1} = \frac{4x}{x(x - 1)}$

• Now, the left-hand side becomes: $\frac{3(x - 1) + 4x}{x(x - 1)}$

Step 3: Expand and combine terms in the numerator

$3(x - 1) + 4x = 3x - 3 + 4x = 7x - 3$ Thus, the equation becomes: $\frac{7x - 3}{x(x - 1)} = \frac{5 - x}{x(x - 1)}$

Step 4: Eliminate the denominator by multiplying both sides by $x(x - 1)$

$7x - 3 = 5 - x$

Step 5: Solve the linear equation

Add $x$ to both sides: $7x + x - 3 = 5$ $8x - 3 = 5$ Add 3 to both sides: $8x = 8$ Divide by 8: $x = 1$

Step 6: Check for restrictions

The denominators $x$ and $x - 1$ must not be 0. Thus: $x \neq 0 \quad \text{and} \quad x \neq 1$ Since $x = 1$ is a solution, but it’s restricted, there are no valid solutions.

Final Answer:

There are no solutions to the equation.

Would you like further clarification on any step?

Here are five related questions to practice:

1. How do you factor quadratic expressions like $x^2 - x$?
2. What are some techniques for solving equations with fractions?
3. How do you handle equations with no valid solutions due to restrictions?
4. What is the significance of finding a common denominator in equations?
5. How do you check for extraneous solutions when solving rational equations?

Tip: Always check for undefined values caused by the denominators before concluding a solution!