ANSWER
(5, -2) and (0, -3)
EXPLANATION
We want to find which of the ordered pairs is a solution to the system of inequalities.
Ordered pairs are written in the form (x, y), this means, whichever ordered pair is a solution, when inserted into the system of inequalities, should be true.
This means that the values of x and y must be true for both inequalities in the system.
The system of inequalities is:
[tex]\begin{cases}-2x-3\leq\text{ y} \\ y-1<\text{ }\frac{1}{2}x\end{cases}[/tex]A. (5, -2)
[tex]\begin{gathered} -2(5)\text{ - 3 }\leq-2\Rightarrow\text{ -10 - 3}\leq-2\Rightarrow\text{ -13 }\leq-2 \\ -2\text{ - 1 < }\frac{1}{2}(5)\Rightarrow\text{ -3 < }\frac{5}{2} \end{gathered}[/tex]Since both inequalities are correct, this is a solution.
B. (-3, -4)
[tex]-2(-3)\text{ - 3 }\leq-4\Rightarrow\text{ 6 - 3 }\leq-4\Rightarrow\text{ 3}\leq-4[/tex]Since the first inequality is already incorrect, we do not need to go further.
It is not a solution
C. (0, -3)
[tex]\begin{gathered} -2(0)\text{ - 3 }\leq\text{ -3 }\Rightarrow\text{ -3 }\leq\text{ -3} \\ -3\text{ - 1 < }\frac{1}{2}(0)\Rightarrow\text{ -4 < 0} \end{gathered}[/tex]Since both inequalities are correct, this is a solution.
D. (0, 1)
[tex]\begin{gathered} -2(0)\text{ - 3 }\leq\text{ 1 }\Rightarrow\text{ -3 }\leq\text{ 1} \\ 1\text{ - 1 < }\frac{1}{2}(0)\Rightarrow\text{ 0 < 0} \end{gathered}[/tex]Since 0 is not less than 0, this is not a solution.
E. (-4, 1)
[tex]-2(-4)\text{ - 3 }\leq\text{ 1}\Rightarrow\text{ 8 - 3 }\leq1\Rightarrow\text{ 5 }\leq1[/tex]Since 5 is not less than 1, this is not a solution.
Therefore, the solutions are (5, -2) and (0, -3)