Question

Write the absolute value inequality in the form Ix−bIc that has the solution set x≤−9 or x≥−5.

Answer 1

Answer:

$\large\boxed{|x+7|\geq2}$

Step-by-step explanation:

Look at the picture.

$|x-a|\geq b\\\\a=\dfrac{-9+(-5)}{2}=\dfrac{-14}{2}=-7\\\\b=-9-(-7)=-9+7=-2\\b=-5-(-7)=-5+7=2\\\\|x-(-7)|\geq2\\\\|x+7|\geq2\\\\Check:\\\\|x+7|\geq2\iff x+7\geq2\ \vee\ x+7\leq-2\qquad\text{subtract 7 from both sides}\\\\x+7-7\geq2-7\ \vee\ x+7-7\leq-2-7\\\\x\geq-5\ \vee\ x\leq-9\qquad CORRECT\ :)$

Answer 2

$x \le -9 \ \text{ or } x \ge -5$

$x \le -7 - 2 \ \text{ or } x \ge -7 + 2$

$x+7 \le -7 - 2+7 \ \text{ or } x+7 \ge -7 + 2+7$

$x+7 \le -2 \ \text{ or } x+7 \ge 2$

$|x+7| \ge 2$

So the final answer is $|x+7| \ge 2$

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side notes:

• The -7 in step 2 is the midpoint of -9 and -5. You add up -9 and -5 to get -14, then divide by 2 to get -7.
• x+7 is the same as x-(-7)
• For the last step I used the rule that if |x| > k then x < -k or x > k for some positive number k