Answer :
Sure! Let's solve the equation step by step to find the value of [tex]\( x \)[/tex]:
1. Start with the original equation:
[tex]\[
\frac{1}{2}(x-14) + 11 = \frac{1}{2}x - (x - 4)
\][/tex]
2. Distribute and simplify:
- On the left side: [tex]\(\frac{1}{2}(x-14) = \frac{1}{2}x - 7\)[/tex]
- On the right side: [tex]\(- (x - 4) = -x + 4\)[/tex]
So the equation now looks like:
[tex]\[
\frac{1}{2}x - 7 + 11 = \frac{1}{2}x - x + 4
\][/tex]
3. Combine like terms:
- On the left side: [tex]\(-7 + 11 = 4\)[/tex], so the equation becomes [tex]\(\frac{1}{2}x + 4\)[/tex]
- On the right side: [tex]\(\frac{1}{2}x - x + 4\)[/tex] is simplified to [tex]\(-\frac{1}{2}x + 4\)[/tex]
The equation now is:
[tex]\[
\frac{1}{2}x + 4 = -\frac{1}{2}x + 4
\][/tex]
4. Subtract 4 from both sides:
[tex]\[
\frac{1}{2}x = -\frac{1}{2}x
\][/tex]
5. Add [tex]\(\frac{1}{2}x\)[/tex] to both sides to combine the [tex]\( x \)[/tex] terms:
[tex]\[
\frac{1}{2}x + \frac{1}{2}x = 0
\][/tex]
This simplifies to:
[tex]\[
x = 0
\][/tex]
So, the value of [tex]\( x \)[/tex] is [tex]\( 0 \)[/tex].
1. Start with the original equation:
[tex]\[
\frac{1}{2}(x-14) + 11 = \frac{1}{2}x - (x - 4)
\][/tex]
2. Distribute and simplify:
- On the left side: [tex]\(\frac{1}{2}(x-14) = \frac{1}{2}x - 7\)[/tex]
- On the right side: [tex]\(- (x - 4) = -x + 4\)[/tex]
So the equation now looks like:
[tex]\[
\frac{1}{2}x - 7 + 11 = \frac{1}{2}x - x + 4
\][/tex]
3. Combine like terms:
- On the left side: [tex]\(-7 + 11 = 4\)[/tex], so the equation becomes [tex]\(\frac{1}{2}x + 4\)[/tex]
- On the right side: [tex]\(\frac{1}{2}x - x + 4\)[/tex] is simplified to [tex]\(-\frac{1}{2}x + 4\)[/tex]
The equation now is:
[tex]\[
\frac{1}{2}x + 4 = -\frac{1}{2}x + 4
\][/tex]
4. Subtract 4 from both sides:
[tex]\[
\frac{1}{2}x = -\frac{1}{2}x
\][/tex]
5. Add [tex]\(\frac{1}{2}x\)[/tex] to both sides to combine the [tex]\( x \)[/tex] terms:
[tex]\[
\frac{1}{2}x + \frac{1}{2}x = 0
\][/tex]
This simplifies to:
[tex]\[
x = 0
\][/tex]
So, the value of [tex]\( x \)[/tex] is [tex]\( 0 \)[/tex].