Asked by anonymous
Solve the inequality.
|2x+3|< 26
So there are two answers to this problem? Do you switch the inequality sign for the second answer?
This is the absolute value of some quantity I suppose. We have
|2x+3|< 26
Thus we want numbers such that
-26 < 2x+3 < 26
You can split this into two equations requiring x to satisfy both. You could split them like this
-26 < 2x+3
2x+3 < 26 and solve each expression, then take the intersection of the solutions, or add -3 to
-26 < 2x+3 < 26 to get
-29 < 2x < 23 now divide by 2
-26/2 < x < 23/2 or
-13 < x < 11.5 should be the final solution. This is an open interval.
For this problem I used the fact that
|x|< a means -a < x < a.
After reviewing, I see I didn't finsish the problem correctly. We started with |2x+3|< 26
We then have -26 < 2x+3 < 26
Followed by -29 < 2x < 23
The next line has a typo. This is what I had: -26/2 < x < 23/2
It should be: -29/2 < x < 23/2, and the final line should be
-14.5 < x < 11.5
This is the open interval (-14.5,11.5). It has a radius of 13 and is centered at -1.5.
|2x+3|< 26
So there are two answers to this problem? Do you switch the inequality sign for the second answer?
This is the absolute value of some quantity I suppose. We have
|2x+3|< 26
Thus we want numbers such that
-26 < 2x+3 < 26
You can split this into two equations requiring x to satisfy both. You could split them like this
-26 < 2x+3
2x+3 < 26 and solve each expression, then take the intersection of the solutions, or add -3 to
-26 < 2x+3 < 26 to get
-29 < 2x < 23 now divide by 2
-26/2 < x < 23/2 or
-13 < x < 11.5 should be the final solution. This is an open interval.
For this problem I used the fact that
|x|< a means -a < x < a.
After reviewing, I see I didn't finsish the problem correctly. We started with |2x+3|< 26
We then have -26 < 2x+3 < 26
Followed by -29 < 2x < 23
The next line has a typo. This is what I had: -26/2 < x < 23/2
It should be: -29/2 < x < 23/2, and the final line should be
-14.5 < x < 11.5
This is the open interval (-14.5,11.5). It has a radius of 13 and is centered at -1.5.
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