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Answer :
We begin with the equation
$$
\frac{1}{2}(x-14)+11=\frac{1}{2}x-(x-4).
$$
**Step 1: Expand both sides.**
On the left-hand side, distribute $\frac{1}{2}$ over $(x-14)$:
$$
\frac{1}{2}x - \frac{1}{2}\cdot 14 + 11 = \frac{1}{2}x - 7 + 11.
$$
Combining $-7$ and $11$, we have:
$$
\frac{1}{2}x + 4.
$$
On the right-hand side, distribute the negative sign over $(x-4)$:
$$
\frac{1}{2}x - (x-4) = \frac{1}{2}x - x + 4.
$$
Since $\frac{1}{2}x - x = -\frac{1}{2}x$, the right-hand side becomes:
$$
-\frac{1}{2}x + 4.
$$
Now, the equation is:
$$
\frac{1}{2}x + 4 = -\frac{1}{2}x + 4.
$$
**Step 2: Eliminate the constant.**
Subtract $4$ from both sides to isolate the terms with $x$:
$$
\frac{1}{2}x+4-4 = -\frac{1}{2}x+4-4,
$$
which simplifies to:
$$
\frac{1}{2}x = -\frac{1}{2}x.
$$
**Step 3: Combine like terms to solve for $x$.**
Add $\frac{1}{2}x$ to both sides to collect all $x$ terms on one side:
$$
\frac{1}{2}x + \frac{1}{2}x = 0,
$$
thus,
$$
x = 0.
$$
**Conclusion:**
The value of $x$ that satisfies the original equation is
$$
\boxed{0}.
$$
$$
\frac{1}{2}(x-14)+11=\frac{1}{2}x-(x-4).
$$
**Step 1: Expand both sides.**
On the left-hand side, distribute $\frac{1}{2}$ over $(x-14)$:
$$
\frac{1}{2}x - \frac{1}{2}\cdot 14 + 11 = \frac{1}{2}x - 7 + 11.
$$
Combining $-7$ and $11$, we have:
$$
\frac{1}{2}x + 4.
$$
On the right-hand side, distribute the negative sign over $(x-4)$:
$$
\frac{1}{2}x - (x-4) = \frac{1}{2}x - x + 4.
$$
Since $\frac{1}{2}x - x = -\frac{1}{2}x$, the right-hand side becomes:
$$
-\frac{1}{2}x + 4.
$$
Now, the equation is:
$$
\frac{1}{2}x + 4 = -\frac{1}{2}x + 4.
$$
**Step 2: Eliminate the constant.**
Subtract $4$ from both sides to isolate the terms with $x$:
$$
\frac{1}{2}x+4-4 = -\frac{1}{2}x+4-4,
$$
which simplifies to:
$$
\frac{1}{2}x = -\frac{1}{2}x.
$$
**Step 3: Combine like terms to solve for $x$.**
Add $\frac{1}{2}x$ to both sides to collect all $x$ terms on one side:
$$
\frac{1}{2}x + \frac{1}{2}x = 0,
$$
thus,
$$
x = 0.
$$
**Conclusion:**
The value of $x$ that satisfies the original equation is
$$
\boxed{0}.
$$
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