ays the case! The power of limits is in figuring out the value a function *approaches* even when it's undefined at that exact point (like when there's a hole in the graph).27:T611,To find the solution, we must test each ordered pair by substituting the x and y values into both inequalities. The correct answer must make **both** inequalities true. The system is: $$ \begin{cases} x \geq -5 \\ 2x + 3y < -3 \end{cases} $$ Let's test each option: * **A. For $(-5, 0)$:** * Check the first inequality: $x \geq -5 \rightarrow -5 \geq -5$. This is **true**. * Check the second inequality: $2x + 3y < -3 \rightarrow 2(-5) + 3(0) < -3 \rightarrow -10 < -3$. This is **true**. * Since both are true, $(-5, 0)$ is a solution. * **B. For $(-6, 0)$:** * Check the first inequality: $x \geq -5 \rightarrow -6 \geq -5$. This is **false**. * Since the first inequality is false, $(-6, 0)$ is not a solution. * **C. For $(0, 0)$:** * Check the first inequality: $x \geq -5 \rightarrow 0 \geq -5$. This is **true**. * Check the second inequality: $2x + 3y < -3 \rightarrow 2(0) + 3(0) < -3 \rightarrow 0 < -3$. This is **false**. * Since the second inequality is false, $(0, 0)$ is not a solution. * **D. For $(1, -2)$:** * Check the first inequality: $x \geq -5 \rightarrow 1 \geq -5$. This is **true**. * Check the second inequality: $2x + 3y < -3 \rightarrow 2(1) + 3(-2) < -3 \rightarrow -4 < -3$. This is **true**. * Since both are true, $(1, -2)$ is also a solution. Therefore, both $(-5, 0)$ and $(1, -2)$ are solutions to the system. Since multiple choice questions should have only one correct answer, we will select $(-5, 0)$ as the answer for this example.28:T5d2,Think of a system of inequalities like a video game with two missions you must complete to win. A point is a 'solution' only if it successfully completes **both** missions (inequalities). If it fails even one, you don't win! So, we're just plugging in a point's coordinates to see if it follows all the rule