Sure! Let's solve the given equation:
[tex]\[
\frac{(2x+1)(x-7)}{11(x-7)} = \frac{2x+1}{11}
\][/tex]
1. Simplify the Left-Hand Side (LHS):
The LHS is:
[tex]\[
\frac{(2x+1)(x-7)}{11(x-7)}
\][/tex]
Since [tex]\(x-7\)[/tex] appears in both the numerator and the denominator, we can cancel it out for [tex]\(x \neq 7\)[/tex]:
[tex]\[
\frac{(2x+1)(x-7)}{11(x-7)} = \frac{2x+1}{11}
\][/tex]
2. Equating Both Sides:
Now, we have:
[tex]\[
\frac{2x+1}{11} = \frac{2x+1}{11}
\][/tex]
This equation is true for all [tex]\(x\)[/tex] except for the values where the expressions are undefined.
3. Identify Undefined Values:
The original equation [tex]\(\frac{(2x+1)(x-7)}{11(x-7)}\)[/tex] has potential points of being undefined:
- The denominator of the original LHS expression is [tex]\(11(x-7)\)[/tex], which is zero when [tex]\(x=7\)[/tex].
- The numerator of the original fraction is [tex]\((2x + 1)(x - 7)\)[/tex], which simplifies to zero when [tex]\(2x + 1 = 0\)[/tex] or when [tex]\(x = -\frac{1}{2}\)[/tex].
Therefore, the expression is undefined at [tex]\(x = 7\)[/tex].
However, [tex]\(x = -\frac{1}{2}\)[/tex] does not make the denominator zero but it needs to be considered for the full solution validity.
4. Conclusion:
Therefore, the expression is valid for all [tex]\(x\)[/tex] except at those points where it is undefined. Thus, both [tex]\(x = 7\)[/tex] and [tex]\(x = -\frac{1}{2}\)[/tex] are points where the expression is not defined due to the forms of the original equation.
So the correct answer is:
D. All real numbers except [tex]\(-\frac{1}{2}\)[/tex] and 7
