Question

the movement of the progress bar may be uneven because questions can be worth more or less
match each compound inequality on the left to the graph that represents its solution on the right.
$4x - 3 > 9$ or $-8x \geq 16$
$8x \geq -24$ and $10 > 2x + 6$
$-10 \leq 6x + 2 < 20$

Explanation:

Step1: Solve \(4x - 3>9\) or \(-8x\geq16\)

• Solve \(4x - 3>9\):

Add 3 to both sides: \(4x>9 + 3=12\)
Divide by 4: \(x > 3\)

• Solve \(-8x\geq16\):

Divide by - 8 (reverse inequality): \(x\leq - 2\)
So the solution is \(x>3\) or \(x\leq - 2\). The graph with an open circle at 3 (for \(x > 3\)) and a closed circle at - 2 (for \(x\leq - 2\)) and lines extending is the second graph (with open at 3, closed at - 2).

Step2: Solve \(8x\geq - 24\) and \(10>2x + 6\)

• Solve \(8x\geq - 24\):

Divide by 8: \(x\geq - 3\)

• Solve \(10>2x + 6\):

Subtract 6: \(4>2x\)
Divide by 2: \(2>x\) or \(x < 2\)
So the solution is \(-3\leq x<2\). The graph with a closed circle at - 3 and an open circle at 2 is the first graph (closed at - 3, open at 2).

Step3: Solve \(-10\leq6x + 2<20\)

• Subtract 2 from all parts: \(-10-2\leq6x+2 - 2<20 - 2\)

\(-12\leq6x<18\)

• Divide by 6: \(\frac{-12}{6}\leq\frac{6x}{6}<\frac{18}{6}\)

\(-2\leq x<3\)
So the solution is \(-2\leq x<3\). The graph with a closed circle at - 2 and an open circle at 3 is the third graph (closed at - 2, open at 3).

Answer:

• \(4x - 3>9\) or \(-8x\geq16\) matches the second graph (with open circle at 3, closed at - 2).
• \(8x\geq - 24\) and \(10>2x + 6\) matches the first graph (closed at - 3, open at 2).
• \(-10\leq6x + 2<20\) matches the third graph (closed at - 2, open at 3).