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Answer :
We start with the equation
[tex]$$
\frac{1}{2}(x-14)+11=\frac{1}{2} x-(x-4).
$$[/tex]
First, we simplify both sides by applying the distributive property:
1. On the left-hand side, distribute [tex]$\frac{1}{2}$[/tex] over [tex]$(x-14)$[/tex]:
[tex]$$
\frac{1}{2}(x-14) = \frac{1}{2}x - 7.
$$[/tex]
Then add 11:
[tex]$$
\frac{1}{2}x - 7 + 11 = \frac{1}{2}x + 4.
$$[/tex]
2. On the right-hand side, distribute the negative over [tex]$(x-4)$[/tex]:
[tex]$$
\frac{1}{2}x - (x-4) = \frac{1}{2}x - x + 4.
$$[/tex]
Notice that
[tex]$$
\frac{1}{2}x - x = -\frac{1}{2}x,
$$[/tex]
so the right-hand side becomes:
[tex]$$
-\frac{1}{2}x + 4.
$$[/tex]
The simplified equation is now
[tex]$$
\frac{1}{2}x + 4 = -\frac{1}{2}x + 4.
$$[/tex]
Next, subtract 4 from both sides to remove the constant term:
[tex]$$
\frac{1}{2}x + 4 - 4 = -\frac{1}{2}x + 4 - 4,
$$[/tex]
which simplifies to
[tex]$$
\frac{1}{2}x = -\frac{1}{2}x.
$$[/tex]
To solve for [tex]$x$[/tex], add [tex]$\frac{1}{2}x$[/tex] to both sides:
[tex]$$
\frac{1}{2}x + \frac{1}{2}x = 0,
$$[/tex]
which gives
[tex]$$
x = 0.
$$[/tex]
Thus, the value of [tex]$x$[/tex] is [tex]$\boxed{0}$[/tex].
[tex]$$
\frac{1}{2}(x-14)+11=\frac{1}{2} x-(x-4).
$$[/tex]
First, we simplify both sides by applying the distributive property:
1. On the left-hand side, distribute [tex]$\frac{1}{2}$[/tex] over [tex]$(x-14)$[/tex]:
[tex]$$
\frac{1}{2}(x-14) = \frac{1}{2}x - 7.
$$[/tex]
Then add 11:
[tex]$$
\frac{1}{2}x - 7 + 11 = \frac{1}{2}x + 4.
$$[/tex]
2. On the right-hand side, distribute the negative over [tex]$(x-4)$[/tex]:
[tex]$$
\frac{1}{2}x - (x-4) = \frac{1}{2}x - x + 4.
$$[/tex]
Notice that
[tex]$$
\frac{1}{2}x - x = -\frac{1}{2}x,
$$[/tex]
so the right-hand side becomes:
[tex]$$
-\frac{1}{2}x + 4.
$$[/tex]
The simplified equation is now
[tex]$$
\frac{1}{2}x + 4 = -\frac{1}{2}x + 4.
$$[/tex]
Next, subtract 4 from both sides to remove the constant term:
[tex]$$
\frac{1}{2}x + 4 - 4 = -\frac{1}{2}x + 4 - 4,
$$[/tex]
which simplifies to
[tex]$$
\frac{1}{2}x = -\frac{1}{2}x.
$$[/tex]
To solve for [tex]$x$[/tex], add [tex]$\frac{1}{2}x$[/tex] to both sides:
[tex]$$
\frac{1}{2}x + \frac{1}{2}x = 0,
$$[/tex]
which gives
[tex]$$
x = 0.
$$[/tex]
Thus, the value of [tex]$x$[/tex] is [tex]$\boxed{0}$[/tex].
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