SOLUTION: Find a and b.
If | x + 2 | < 5, then a < x - 2 < b.
Must I solve left side inequality first?
If so, -5 < x + 2 < 5 is the set up.
Subteact 2 from all side
Algebra.Com
Question 1199370: Find a and b.
If | x + 2 | < 5, then a < x - 2 < b.
Must I solve left side inequality first?
If so, -5 < x + 2 < 5 is the set up.
Subteact 2 from all sides.
-5 - 2 < x < 5 - 2
-7 < x < 3
Is this correct? If correct, what is the next step?
Found 3 solutions by ikleyn, greenestamps, math_tutor2020:
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
Find a and b.
If | x + 2 | < 5, then a < x - 2 < b.
Must I solve left side inequality first?
No. As you do starting with your next line, because the absolute value inequality is "less than", you solve both inequalities (the "compound inequality") all at once.
If so, -5 < x + 2 < 5 is the set up.
Subtract 2 from all sides.
-5 - 2 < x < 5 - 2
-7 < x < 3
Is this correct? If correct, what is the next step?
That is all correct. The next step is to look at the inequality that shows you how a and b are defined in the problem:
a < x - 2 < b
You have taken the original inequality and solved for x; to answer the question that is asked, you need to solve the inequality for x-2:
-7 < x < 3
Subtract 2 from all sides:
-9 < x-2 < 1
ANSWER: a = -9; b=1
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
The jump from
|x+2| < 5
to
-5 < x + 2 < 5
is valid.
The same can be said when you went from
-5 < x + 2 < 5
to
-7 < x < 3
Then, as the tutor @greenestamps has pointed out, you subtract 2 from all sides to get x-2 in the middle.
-7 < x < 3
-7-2 < x-2 < 3-2
-9 < x-2 < 1
-----------------------------------------------
Or you can subtract 4 from all sides after the absolute value bars have been removed.
Why 4?
Because we want to go from x+2 to x-2
The gap from +2 to -2 on the number line is 4 units.
I.e.
x+2 to (x+2)-4 to x-2
So,
-5 < x + 2 < 5
-5-4 < (x + 2)-4 < 5-4
-9 < x-2 < 1
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