Question

exit card
wymerria mcneil ♥

1. circle all the values that are included in solution set of the inequality -2x + 1.5 ≥ 7.5

-2 -9.6 0 -3 1 3.5

1. solve the inequality and graph the solution: 5(x - 3) < -5
2. solve the compound inequality and graph the solution on a number line: 2x - 6 < -2 < x - 3

Explanation:

Question 1:

Step1: Solve the inequality \(-2x + 1.5 \geq 7.5\)

Subtract \(1.5\) from both sides: \(-2x \geq 7.5 - 1.5\) → \(-2x \geq 6\)

Step2: Divide by \(-2\) (reverse inequality)

\(x \leq \frac{6}{-2}\) → \(x \leq -3\)
Check values: \(-9.6 \leq -3\), \(-3 \leq -3\) (others: \(-2 > -3\), \(0 > -3\), \(1 > -3\), \(3.5 > -3\))

Question 2:

Step1: Divide by \(5\)

\(5(x - 3) < -5\) → \(x - 3 < \frac{-5}{5}\) → \(x - 3 < -1\)

Step2: Add \(3\)

\(x < -1 + 3\) → \(x < 2\)
(Graph: Open circle at \(2\), arrow left on number line)

Question 3:

Step1: Split compound inequality

\(2x - 6 < -2\) and \(-2 < x - 3\)

Step2: Solve \(2x - 6 < -2\)

Add \(6\): \(2x < 4\) → \(x < 2\)? Wait, no—wait, original is \(2x - 6 < -2 < x - 3\). Correct split: \(2x - 6 < -2\) and \(-2 < x - 3\)
Solve \(2x - 6 < -2\): \(2x < 4\) → \(x < 2\)? No, wait, no—wait, let's re-express:
\(2x - 6 < -2\) → \(2x < 4\) → \(x < 2\) (wrong, wait original is \(2x - 6 < -2 < x - 3\). Wait, no, the compound inequality is \(2x - 6 < -2\) and \(-2 < x - 3\). Wait, solving \(-2 < x - 3\): \(x - 3 > -2\) → \(x > 1\)? No, wait, let's do it properly:

Wait, the inequality is \(2x - 6 < -2 < x - 3\). So we have two parts:

1. \(2x - 6 < -2\)
2. \(-2 < x - 3\)

Solve part 1: \(2x - 6 < -2\) → \(2x < 4\) → \(x < 2\) (no, that can't be, wait maybe a typo? Wait the user wrote \(2x - 6 < -2 < x - 3\). Wait, maybe it's \(2x - 6 < -2\) and \(-2 < x - 3\). Wait, solving \(-2 < x - 3\): \(x - 3 > -2\) → \(x > 1\)? No, that doesn't make sense. Wait, maybe the inequality is \(2x - 6 < -2\) and \(-2 < x - 3\), but that would be \(x < 2\) and \(x > 1\), so \(1 < x < 2\)? No, that seems off. Wait, maybe the original is \(2x - 6 < -2\) and \(-2 < x - 3\), but let's check again.

Wait, the user's handwritten: "Solve the compound inequality and graph the solution on a number line: \(2x - 6 < -2 < x - 3\)". So we have to solve \(2x - 6 < -2\) and \(-2 < x - 3\) simultaneously.

First inequality: \(2x - 6 < -2\)
Add 6: \(2x < 4\) → \(x < 2\)

Second inequality: \(-2 < x - 3\)
Add 3: \(1 < x\) → \(x > 1\)

So combining both: \(1 < x < 2\)? Wait, but that seems conflicting. Wait, maybe the compound inequality is \(2x - 6 < -2\) or \(-2 < x - 3\)? No, the wording is "compound inequality" (usually "and" or "or"). Wait, maybe a typo in the problem. Alternatively, maybe the inequality is \(2x - 6 < -2\) and \(-2 < x - 3\), so solution is \(1 < x < 2\). But let's verify:

If \(x = 1.5\) (between 1 and 2):

\(2(1.5) - 6 = 3 - 6 = -3 < -2\) (true)

\(-2 < 1.5 - 3 = -1.5\) → \(-2 < -1.5\) (true). So yes, \(1 < x < 2\).

(Graph: Open circles at 1 and 2, line between them on number line)

Answer:

-9.6, -3