which system of linear inequalities has the point (2, 1) in its solution set?\n$y < -x + 3$\n$yleqslant\frac{…

Explanation:

Step1: Substitute x = 2 and y = 1 into the first inequality

For $y < -x + 3$, substitute $x = 2$ and $y = 1$: $1<-(2)+3$ $1 < 1$ (False)

Step2: Since the first - inequality is false for the given point, we don't need to check the second one for this option. Let's assume we have other options to check in a similar way. But since the problem only shows one option completely and it's incorrect, we need to re - evaluate the process.

Let's assume we have the correct system of inequalities and we check the point $(2,1)$ in the general way. For a system of inequalities $y < -x + 3$ and $y\leq\frac{1}{2}x+3$: For $y < -x + 3$, when $x = 2$ and $y = 1$, we have $1<-(2)+3=1$, which is false. For $y\leq\frac{1}{2}x + 3$, when $x = 2$ and $y = 1$, we have $1\leq\frac{1}{2}(2)+3=1 + 3=4$, which is true. But since the first inequality is not satisfied, this is not the correct system. If we assume the correct system: Let's check the point $(x = 2,y = 1)$ in a system. Suppose the system is $y\geq - x+1$ and $y<2x - 3$. For $y\geq - x + 1$, substitute $x = 2$ and $y = 1$: $1\geq-(2)+1$ $1\geq - 1$ (True) For $y<2x - 3$, substitute $x = 2$ and $y = 1$: $1<2(2)-3$ $1<1$ (False)

Let's assume the system $y < x$ and $y\geq - 2x+5$ For $y < x$, when $x = 2$ and $y = 1$, we have $1<2$ (True) For $y\geq - 2x+5$, when $x = 2$ and $y = 1$: $1\geq-2(2)+5$ $1\geq1$ (True)

Answer:

The system of inequalities $y < x$ and $y\geq - 2x + 5$ (assuming this is one of the options and is the correct one after checking all options in a similar substitution - based way as shown above. Since the provided option in the problem is incorrect, we show a general way of checking and a sample correct - like system).