To solve the equation involving the ceiling function:

$\left\lceil 2x - 1 \right\rceil + \left\lceil x + 3 \right\rceil = 12$

we need to understand how the ceiling function $\lceil \cdot \rceil$ works. The ceiling function $\lceil y \rceil$ gives the smallest integer greater than or equal to $y$.

Let's denote: $\left\lceil 2x - 1 \right\rceil = a$ $\left\lceil x + 3 \right\rceil = b$

where $a$ and $b$ are integers, and we know that: $a + b = 12$

Analyzing $\left\lceil 2x - 1 \right\rceil = a$:

$a - 1 < 2x - 1 \leq a$ $a < 2x \leq a + 1$ $\frac{a}{2} < x \leq \frac{a + 1}{2}$

Analyzing $\left\lceil x + 3 \right\rceil = b$:

$b - 1 < x + 3 \leq b$ $b - 4 < x \leq b - 3$

Solving the Combined Inequalities:

We need $x$ to satisfy both sets of inequalities simultaneously.

Given $a + b = 12$, let's substitute $b = 12 - a$:

1. For $\left\lceil 2x - 1 \right\rceil = a$: $\frac{a}{2} < x \leq \frac{a + 1}{2}$

2. For $\left\lceil x + 3 \right\rceil = 12 - a$: $11 - a < x \leq 12 - a$

Therefore, we need to find $a$ such that the intervals: $\frac{a}{2} < x \leq \frac{a + 1}{2}$ and $11 - a < x \leq 12 - a$

Finding the Overlapping Intervals:

To find a common $x$ that fits both intervals:

1. $\frac{a}{2} < 12 - a$
2. $11 - a \leq \frac{a + 1}{2}$

Solving $\frac{a}{2} < 12 - a$:

$a < 24 - 2a$ $3a < 24$ $a < 8$

Solving $11 - a \leq \frac{a + 1}{2}$:

$22 - 2a \leq a + 1$ $21 \leq 3a$ $7 \leq a$

Thus, $a$ must be an integer such that: $7 \leq a < 8$

This means $a = 7$.

Verifying $a = 7$ and Finding $x$:

For $a = 7$: $\frac{7}{2} < x \leq \frac{8}{2}$ $3.5 < x \leq 4$

And for $b = 5$ (since $b = 12 - a = 12 - 7 = 5$): $11 - 7 < x \leq 12 - 7$ $4 < x \leq 5$

The intervals do not overlap since $3.5 < x \leq 4$ and $4 < x \leq 5$ are disjoint.

Thus, there are no real numbers $x$ that satisfy the given equation.

Conclusion:

There are no real numbers $x$ that satisfy the equation $\left\lceil 2x - 1 \right\rceil + \left\lceil x + 3 \right\rceil = 12$.

Would you like more details on the analysis or have any other questions?

Here are some related questions to expand on this topic:

1. What are the properties of the ceiling function?
2. How do the ceiling and floor functions differ?
3. Can ceiling functions ever produce non-integer values?
4. How would the solution change if we replaced the ceiling functions with floor functions?
5. What real-world applications make use of ceiling functions?

Tip: When dealing with piecewise functions like the ceiling function, always consider the intervals where the function changes value to identify possible solutions.