Question

1. which system of linear inequalities has no solution?

$\

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

$
$\

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

$
$\

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

$
$\

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

$

Explanation:

Step1: Analyze the slope of each inequality

For a system of linear inequalities \(y > mx + b\) and \(y < mx + c\) (where \(m\) is the slope), if the slopes are equal and the regions they define do not overlap, there is no solution. Let's rewrite each system in slope - intercept form (\(y=mx + b\)):

System 1: \(

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

\)
Rewrite the second inequality: \(y<2x + 3\). Both inequalities are the same. So the solution set is the region defined by \(y < 2x+3\), there are solutions.

System 2: \(

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

\)
The first inequality is \(y>2x + 1\) (slope \(m = 2\), y - intercept \(b_1=1\)) and the second is \(y<2x-4\) (slope \(m = 2\), y - intercept \(b_2=- 4\)). The region above \(y = 2x + 1\) and below \(y=2x - 4\) does not exist because for any \(x\), \(2x+1>2x - 4\) (since \(1>-4\)). So there is no \(y\) that can satisfy both inequalities.

System 3: \(

$$\begin{cases}2x + 3y\geq12\\x + 5y<10\end{cases}$$

\)
Rewrite the first inequality: \(3y\geq - 2x + 12\Rightarrow y\geq-\frac{2}{3}x + 4\). The second inequality: \(5y<-x + 10\Rightarrow y<-\frac{1}{5}x+2\). The slopes are different (\(-\frac{2}{3}
eq-\frac{1}{5}\)), so the lines will intersect and there will be a region of overlap (solution set).

System 4: \(

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

\)
Rewrite the second inequality: \(y>x - 5\). Both inequalities are the same. So the solution set is the region defined by \(y>x - 5\), there are solutions.

Answer:

The system \(\boldsymbol{

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

}\) (the second option)