Question

1. which system of linear inequalities has no solution?

$\

$$\begin{cases} y > 2x + 1 \\\\ y < 2x - 4 \\end{cases}$$

$
$\

$$\begin{cases} y > x - 5 \\\\ y - x > -5 \\end{cases}$$

$
$\

$$\begin{cases} y < 2x + 3 \\\\ -2x + y < 3 \\end{cases}$$

$
$\

$$\begin{cases} 2x + 3y \\geq 12 \\\\ x + 5y < 10 \\end{cases}$$

$

Explanation:

Step1: Analyze the first system

The first system is \(

$$\begin{cases}y > 2x + 1\\y < 2x - 4\end{cases}$$

\). The two lines \(y = 2x+1\) and \(y = 2x - 4\) are parallel (same slope \(m = 2\)). The region \(y>2x + 1\) is above \(y = 2x+1\) and \(y<2x - 4\) is below \(y = 2x - 4\). Since \(2x+1>2x - 4\) for all \(x\) (subtracting \(2x\) from both sides gives \(1>-4\), which is true), there is no \(y\) that can be both greater than \(2x + 1\) and less than \(2x - 4\).

Step2: Analyze the second system

Rewrite the second inequality \(y - x>-5\) as \(y>x - 5\). So the system is \(

$$\begin{cases}y > x - 5\\y > x - 5\end{cases}$$

\), which is just \(y>x - 5\), so there are solutions (all points above \(y=x - 5\)).

Step3: Analyze the third system

Rewrite \(-2x + y<3\) as \(y<2x + 3\). The system is \(

$$\begin{cases}y < 2x + 3\\y < 2x + 3\end{cases}$$

\), which is \(y<2x + 3\), so there are solutions (all points below \(y = 2x+3\)).

Step4: Analyze the fourth system

The first inequality \(2x+3y\geq12\) and the second \(x + 5y<10\). These are two non - parallel lines (different slopes, slope of first line \(y=-\frac{2}{3}x + 4\), slope of second line \(y=-\frac{1}{5}x+2\)), so there will be an intersection of the regions, meaning there are solutions.

Answer:

A. \(

$$\begin{cases}y > 2x + 1\\y < 2x - 4\end{cases}$$

\)