Question

question 3a
use inequality notation.

1. solve:

a. $|x + 4| > 11$
b. $5|x - 2| - 3 < 62$
c. $5|x + 7| + 2 \geq 27$
d. $-4|1 - 2x| - 5 \geq -25$

Explanation:

Step1: Split absolute value inequality

For \(|x+4|>11\), split into two cases:
\(x+4 > 11\) or \(x+4 < -11\)

Step2: Solve first inequality

Isolate \(x\) by subtracting 4:
\(x > 11 - 4\)
\(x > 7\)

Step3: Solve second inequality

Isolate \(x\) by subtracting 4:
\(x < -11 - 4\)
\(x < -15\)

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Step1: Isolate absolute value term

For \(5|x-2|-3 < 62\), add 3 to both sides:
\(5|x-2| < 62 + 3\)
\(5|x-2| < 65\)

Step2: Divide to simplify inequality

Divide both sides by 5:
\(|x-2| < 13\)

Step3: Split and solve inequalities

Split into \(-13 < x-2 < 13\), add 2 to all parts:
\(-13 + 2 < x < 13 + 2\)
\(-11 < x < 15\)

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Step1: Isolate absolute value term

For \(5|x+7|+2 \geq 27\), subtract 2 from both sides:
\(5|x+7| \geq 27 - 2\)
\(5|x+7| \geq 25\)

Step2: Divide to simplify inequality

Divide both sides by 5:
\(|x+7| \geq 5\)

Step3: Split and solve inequalities

Split into \(x+7 \geq 5\) or \(x+7 \leq -5\):
For \(x+7 \geq 5\): \(x \geq 5 - 7\) → \(x \geq -2\)
For \(x+7 \leq -5\): \(x \leq -5 - 7\) → \(x \leq -12\)

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Step1: Isolate absolute value term

For \(-4|1-2x|-5 \geq -25\), add 5 to both sides:
\(-4|1-2x| \geq -25 + 5\)
\(-4|1-2x| \geq -20\)

Step2: Reverse inequality on division

Divide by -4 (reverse inequality):
\(|1-2x| \leq 5\)

Step3: Split and solve inequalities

Split into \(-5 \leq 1-2x \leq 5\). Subtract 1 from all parts:
\(-5 - 1 \leq -2x \leq 5 - 1\)
\(-6 \leq -2x \leq 4\)

Step4: Divide and reverse inequality

Divide by -2 (reverse inequalities):
\(3 \geq x \geq -2\)
\(-2 \leq x \leq 3\)

Answer:

a. \(x < -15\) or \(x > 7\)
b. \(-11 < x < 15\)
c. \(x \leq -12\) or \(x \geq -2\)
d. \(-2 \leq x \leq 3\)