Question

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question id: 504302
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match each compound inequality on the left to the graph that represents its solution on the right.
4x + 3 > 15 or -6x ≥ 12
-8x > -24 and -10 ≤ 2x - 6
-29 ≤ 9x - 2 < 16
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Explanation:

Step1: Solve \(4x + 3>15\) or \(-6x\geq12\)

• Solve \(4x + 3>15\):

Subtract 3 from both sides: \(4x>15 - 3=12\)
Divide by 4: \(x > 3\)

• Solve \(-6x\geq12\):

Divide by - 6 (reverse inequality): \(x\leq - 2\)
So the solution is \(x>3\) or \(x\leq - 2\)

Step2: Solve \(-8x>-24\) and \(-10\leq2x - 6\)

• Solve \(-8x>-24\):

Divide by - 8 (reverse inequality): \(x < 3\)

• Solve \(-10\leq2x - 6\):

Add 6 to both sides: \(-4\leq2x\)
Divide by 2: \(-2\leq x\)
So the solution is \(-2\leq x<3\)

Step3: Solve \(-29\leq9x - 2<16\)

• Solve \(-29\leq9x - 2\):

Add 2 to both sides: \(-27\leq9x\)
Divide by 9: \(-3\leq x\)

• Solve \(9x - 2<16\):

Add 2 to both sides: \(9x<18\)
Divide by 9: \(x < 2\)
So the solution is \(-3\leq x<2\)

Now, match the solutions to the graphs:

• \(4x + 3>15\) or \(-6x\geq12\) (solution \(x>3\) or \(x\leq - 2\)) should match the graph with open circle at 3 (for \(x > 3\)) and closed circle at - 2 (for \(x\leq - 2\))
• \(-8x>-24\) and \(-10\leq2x - 6\) (solution \(-2\leq x<3\)) should match the graph with closed circle at - 2 and open circle at 3
• \(-29\leq9x - 2<16\) (solution \(-3\leq x<2\)) should match the graph with closed circle at - 3 and open circle at 2

Answer:

• \(4x + 3>15\) or \(-6x\geq12\) matches the graph with open circle at 3 and closed circle at - 2.
• \(-8x>-24\) and \(-10\leq2x - 6\) matches the graph with closed circle at - 2 and open circle at 3.
• \(-29\leq9x - 2<16\) matches the graph with closed circle at - 3 and open circle at 2.