Absolute value inequality q (1 Viewer)

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[ExtremelyBoredUser]

View attachment 34814

pretty sure its x >= 1.5 just from seeing the Q? 2.5 - 0.5 = 2 and hence x = 1.5 is the min/starting value.

If you're looking for a way to do it, I think just substituting reasonable values should work but if you want some form of reasoning besides algebra of abs values (which seems like a pain/not worth it) then u can try graphing both |x+1| and |x-2| and seeing where the difference is exactly 2 and hence values greater than that will also solve the inequality.

 

[=)(=]

pretty sure its x >= 1.5 just from seeing the Q? 2.5 - 0.5 = 2 and hence x = 1.5 is the min/starting value.

If you're looking for a way to do it, I think just substituting reasonable values should work but if you want some form of reasoning besides algebra of abs values (which seems like a pain/not worth it) then u can try graphing both |x+1| and |x-2| and seeing where the difference is exactly 2 and hence values greater than that will also solve the inequality.

oohhh okay yea with trial and error its pretty quick, Thanks!

 

[5uckerberg]

View attachment 34814

This is for anything between -1 and 2 less than -1 is -3 and greater than 2 is simply 3. Justifies my working.

 

[=)(=]

ohh okay thank you!

 

[5uckerberg]

pretty sure its x >= 1.5 just from seeing the Q? 2.5 - 0.5 = 2 and hence x = 1.5 is the min/starting value.

If you're looking for a way to do it, I think just substituting reasonable values should work but if you want some form of reasoning besides algebra of abs values (which seems like a pain/not worth it) then u can try graphing both |x+1| and |x-2| and seeing where the difference is exactly 2 and hence values greater than that will also solve the inequality.

Some people love doing guesswork huh. Very well, The technique is to find the value where the absolute value is 0 and then find what happens before it and then find the value where the absolute value is 0 on the other side. Afterwards if there two absolute value functions find what happens in between

 

[ExtremelyBoredUser]

Some people love doing guesswork huh. Very well, The technique is to find the value where the absolute value is 0 and then find what happens before it and then find the value where the absolute value is 0 on the other side. Afterwards if there two absolute value functions find what happens in between

work smarter not harder haha but yes i wouldn't put it on my exam paper "hi sir i just put x = 1.5 and ye the inequality works for x>=1.5 ". I would probably reason something similar.

 

[Drongoski]

I have a simpler geometric way to do this that I investigated 2 to 3 years ago. Let me work this one out.

 

[=)(=]

Because I have a rather wordy explanation of the method, it would appear this approach is more complicated. It is not. I may post another similar type of question I did sometime ago.

ohh so you basically just find the equations of the lines in each different region and since the 3rd region is a line of equation y=3 then y=2 will be in that middle bridging section which you then set equal to 2 and solve for x?

 

[Drongoski]

Because I have a rather wordy explanation of the method, it would appear this approach is more complicated. It is not. I may post another similar type of question I did sometime ago.

I did a similar type of equality a few years ago: |x+1| + |x-2| = 3
If you look at region 2: every point x where -1 <= x <= 2 satisfies this equality BUT NOT the points OUTSIDE this region.

 

[Drongoski]

oohhh okay yea with trial and error its pretty quick, Thanks!

Not really trial-&-error; there is a logical approach to this.