To determine whether the point [tex]\((2,1)\)[/tex] is in the solution set of the given system of linear inequalities, we need to evaluate the inequalities with the coordinates [tex]\(x = 2\)[/tex] and [tex]\(y = 1\)[/tex].
1. First inequality: [tex]\( y < -x + 3 \)[/tex]
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex] into the inequality:
[tex]\[
1 < -2 + 3
\][/tex]
Simplify the right-hand side:
[tex]\[
1 < 1
\][/tex]
This statement is false. Hence, the point [tex]\((2,1)\)[/tex] does not satisfy the first inequality.
2. Second inequality: [tex]\( y \leq \frac{1}{2} x + 3 \)[/tex]
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex] into the inequality:
[tex]\[
1 \leq \frac{1}{2} \cdot 2 + 3
\][/tex]
Simplify the right-hand side:
[tex]\[
1 \leq 1 + 3
\][/tex]
[tex]\[
1 \leq 4
\][/tex]
This statement is true. Hence, the point [tex]\((2,1)\)[/tex] satisfies the second inequality.
In conclusion, the point [tex]\((2,1)\)[/tex] is not in the solution set of the inequality [tex]\( y < -x + 3 \)[/tex], but it is in the solution set of the inequality [tex]\( y \leq \frac{1}{2} x + 3 \)[/tex]. Therefore, for the system of inequalities:
[tex]\[
y < -x + 3
\][/tex]
[tex]\[
y \leq \frac{1}{2} x + 3
\][/tex]
The point [tex]\((2,1)\)[/tex] belongs to the solution set of the second inequality [tex]\( y \leq \frac{1}{2} x + 3 \)[/tex] but does not belong to the solution set of the first inequality [tex]\( y < -x + 3 \)[/tex].
