We want to know which ordered pair is a solution of the system of inequalities shown:
[tex]\begin{cases}x-4y\ge0 \\ x-y<-1\end{cases}[/tex]For doing so, we will try to solve both inequalities for one variable, in this case, we will use y.
On the first equation:
[tex]\begin{gathered} x-4y\ge0 \\ x\ge4y \\ y\le\frac{x}{4} \end{gathered}[/tex]On the second equation:
[tex]\begin{gathered} x-y<-1 \\ x+1-y<0 \\ x+1And joining those two results we get:[tex]x+1Now we check each of the ordered pairs, if they hold the condition above:For (0, 2)
We have that x=0, and y=2. Thus,
[tex]\begin{gathered} x+1=1 \\ \frac{x}{4}=0 \\ \text{And as }2>0,\text{ (0, 2) is NOT a solution of the system.} \end{gathered}[/tex]For (-3, 8)
In this case, x=-3 and y=8.
[tex]\begin{gathered} x+1=-2 \\ \frac{x}{4}=-\frac{3}{4} \\ \text{As }8>-\frac{3}{4},\text{ this means that (-3, 8) is NOT a solution of the system.} \end{gathered}[/tex]For (2,5)
In this case, x=2 and y=5.
[tex]\begin{gathered} x+1=3 \\ \frac{x}{4}=\frac{2}{4}=\frac{1}{2} \\ \text{As }5>\frac{1}{2}\text{ this means that (2, 5) is NOT a solution of the system.} \end{gathered}[/tex]For (-7, -4)
In this case, x=-7 and y=-4.
[tex]\begin{gathered} x+1=-6 \\ \frac{x}{4}=-\frac{7}{4} \\ \text{As }-6<-4\le-\frac{7}{4},\text{ (-7, -4) is a SOLUTION of the system.} \end{gathered}[/tex]For (6, -1)
We have that x=6 and y=-1.
[tex]\begin{gathered} x+1=7 \\ \frac{x}{4}=\frac{6}{4}=\frac{3}{2} \\ \text{As }7>-1,\text{ (6, -1) is NOT a solution of the system. } \end{gathered}[/tex]