Question

match the following inequality with the graph of its solution set.

$xleq -2$ or $xgeq 2$

$x + 3lt 0$ and $x - 4lt -4$

$x + 3lt 0$ or $x - 4lt -4$

$xgeq -2$ and $xleq 2$

Explanation:

Step1: Solve \(x + 3\lt0\)

Subtract \(3\) from both sides: \(x\lt - 3\)

Step2: Solve \(x - 4\lt - 4\)

Add \(4\) to both sides: \(x\lt0\)

Step3: Analyze \(x + 3\lt0\) and \(x - 4\lt - 4\)

Since it's an "and" compound inequality, we take the intersection. The intersection of \(x\lt - 3\) and \(x\lt0\) is \(x\lt - 3\)

Step4: Analyze \(x + 3\lt0\) or \(x - 4\lt - 4\)

Since it's an "or" compound inequality, we take the union. The union of \(x\lt - 3\) and \(x\lt0\) is \(x\lt0\)

Step5: Analyze \(x\geq - 2\) and \(x\leq2\)

This is an "and" compound inequality. The solution is \(-2\leq x\leq2\)

Step6: Analyze \(x\leq - 2\) or \(x\geq2\)

This is an "or" compound inequality. The solution is two separate intervals \(x\leq - 2\) and \(x\geq2\)

Answer:

• \(x\leq - 2\) or \(x\geq2\): The graph with two separate closed - circle points (at \(-2\) and \(2\)) and lines extending to the left from \(-2\) and to the right from \(2\)
• \(x + 3\lt0\) and \(x - 4\lt - 4\): The graph with an open - circle at \(-3\) and a line extending to the left
• \(x + 3\lt0\) or \(x - 4\lt - 4\): The graph with an open - circle at \(0\) and a line extending to the left
• \(x\geq - 2\) and \(x\leq2\): The graph with two closed - circle points (at \(-2\) and \(2\)) and a line connecting them