https://file.aiquickdraw.com/mathbot/file/17267793260308nph77du.jpg

Let's work through both equations step by step.

### First Equation:
\[
4(u - 1) - 1 = 2(2u - 3)
\]

1. Expand both sides:
\[
4(u - 1) - 1 = 4u - 4 - 1
\]
\[
2(2u - 3) = 4u - 6
\]

Now the equation is:
\[
4u - 5 = 4u - 6
\]

2. Subtract \( 4u \) from both sides:
\[
-5 = -6
\]

This is a false statement, meaning there is **no solution** to this equation.

### Second Equation:
\[
3(x + 2) + 4 = 5(x - 1) + x
\]

1. Expand both sides:
\[
3(x + 2) + 4 = 3x + 6 + 4
\]
\[
5(x - 1) + x = 5x - 5 + x
\]

Now the equation is:
\[
3x + 10 = 6x - 5
\]

2. Move all terms involving \( x \) to one side and constant terms to the other side:
\[
3x - 6x = -5 - 10
\]
\[
-3x = -15
\]

3. Divide by \( -3 \):
\[
x = 5
\]

So, the solution to the second equation is \( x = 5 \).

### Final Answers:
- **First equation**: No solution.
- **Second equation**: \( x = 5 \).

Would you like more details or clarifications? Here are 5 related questions to expand your understanding:

1. How can you identify if an equation has no solution?
2. What is the difference between an equation having no solution and infinite solutions?
3. Can equations have multiple solutions, and how would you solve them?
4. How do you deal with equations involving fractions?
5. How would the approach change for systems of equations?

**Tip**: Always simplify both sides of an equation fully before solving; this makes it easier to spot inconsistencies or special cases like no solution or infinite solutions.

Solve the following equations: 4(u - 1) - 1 = 2(2u - 3) and 3(x + 2) + 4 = 5(x - 1) + x

Learn how to solve the linear equations 4(u - 1) - 1 = 2(2u - 3) and 3(x + 2) + 4 = 5(x - 1) + x. This guide explains why the first equation has no solution and provides a step-by-step solution for x = 5 in the second equation. Suitable for Grades 7-9 students.

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