To determine the correct value of [tex]\( k \)[/tex], we need to solve the system of equations given by:
[tex]\[
\begin{cases}
(k+1)x + 2y = 7 \\
x - 4y = -7
\end{cases}
\][/tex]
We are given that [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex].
First, let's substitute these values into the second equation to confirm the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
x - 4y = -7
\][/tex]
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex]:
[tex]\[
2 - 4(1) = -7
\][/tex]
[tex]\[
2 - 4 = -2 \neq -7
\][/tex]
There seems to be an inconsistency here considering the standard approach. However, let's proceed with the first equation to find the value of [tex]\( k \)[/tex] given the equations match the constants provided.
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex] into the first equation:
[tex]\[
(k+1)x + 2y = 7
\][/tex]
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex]:
[tex]\[
(k+1)(2) + 2(1) = 7
\][/tex]
Simplify:
[tex]\[
2(k+1) + 2 = 7
\][/tex]
[tex]\[
2k + 2 + 2 = 7
\][/tex]
[tex]\[
2k + 4 = 7
\][/tex]
Subtract 4 from both sides:
[tex]\[
2k = 3
\][/tex]
Divide by 2:
[tex]\[
k = \frac{3}{2}
\][/tex]
Thus, there is no integer value of [tex]\( k \)[/tex] from the given options ([tex]\( k = 0 \)[/tex], [tex]\( k = 1 \)[/tex], [tex]\( k = 2 \)[/tex], [tex]\( k = 3 \)[/tex]) that satisfies these equations with [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex]. Hence, the correct option for [tex]\( k \)[/tex] does not exist within the provided choices.
But since we are to choose from the given options, we conclude that:
None of the given values of [tex]\( k \)[/tex] satisfy the equations with [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex].
Thus the correct answer is:
```
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