How to Check your inequality problem

[Cindy Burgess]

I am suppose to solve and then check these problems: 1. 2x+5<8 When I worked it, I got x<3/2 I don't know how to check it 2. 3x+1>10 When I worked it, I got x>3 Again, I don't know how to check I don't know how to work this problem nor check it: 1<3x+2<12 I worked it like this: 1<3x+2 3x+2<12 First Answer -1/3<x x<3 1/3 If this is correct, I still don't know how to check it. Please Help!

 

[stapel]

1. 2x+5<8 When I worked it, I got x<3/2 I don't know how to check it

Plug 3/2 in. (This confirms equality, which isn't actually included, but will confirm your end-point.) Plug in a sample value from below 3/2. (This should work, confirming the "less than" inequality.) Plug in a sample value from above 3/2. (This should not work, again confirming the "less than" inequality.)

2. 3x+1>10 When I worked it, I got x>3 Again, I don't know how to check

Use the same method as above.

I don't know how to work this problem nor check it: 1<3x+2<12

I worked it like this: 1<3x+2 3x+2<12

You don't need to split it into various cases. Just work all three "sides" at the same time. Subtract 2 from all three sides, and then divide through by 3.

 

[Cindy Burgess]

Plug 3/2 in. (This confirms equality, which isn't actually included, but will confirm your end-point.) Plug in a sample value from below 3/2. (This should work, confirming the "less than" inequality.) Plug in a sample value from above 3/2. (This should not work, again confirming the "less than" inequality.)


Use the same method as above.


You don't need to split it into various cases. Just work all three "sides" at the same time. Subtract 2 from all three sides, and then divide through by 3.

THANK YOU! I couldn't find a definitive way to check these and you gave me one! I really appreciate it!!!

 

[HallsofIvy]

What stapel said is completely true and all you need to know for this problem. But there are some "hidden" technical matters you may need to know later. If you were to claim that "all numbers in the interval a< x< b" have some specific property, you could NOT prove it by showing that one number in that interval have it. Bit inequality is a special situation. For function f and g, if f(x)< g(x) for some specific value then f(y)< g(y) will be true as long as f and g are continuous and f(z) is NOT equal to g(z) for some z between x and y. So it is sufficient to find values at which f and g are equal or where one or the other is not continuous and find the intervals between those points.