SOLUTION: How will {x ∈ R|x ≥ −2 or 1 < x} be represented on the number line

Algebra ->  Number-Line -> SOLUTION: How will {x ∈ R|x ≥ −2 or 1 < x} be represented on the number line      Log On


   



Question 1180082: How will {x ∈ R|x ≥ −2 or 1 < x} be represented on the number line
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
here's a reference

https://the-world-is-my-classroom.weebly.com/unit-63---introduction-to-linear-inequalities.html

here's a picture of another reference.



here's a picture of my solution to your problem.



x >= -2 has a solid circle on -2 witgh an arrow going to the right indefinitely.

1 < x is the same as x > 1 so that has an open cicle on the 1 with an arrow going to the right indefinitely.

if you put them together, that becomes a little more tricky.

the arrow goes to the right in both cases.

the OR means you can have either the first inequality or the second inequality or both.

if you have the first inequality, then the top inequality on my graph is used.
if you have the second inequality, then the bottom inequality on my graph is used.

if you have both, then x has to be greater than or equal to -2 AND it has to be greater than 1.
any value of x >= -2 and <= 1 would not be valid because, while it satisfied x >= -2, it doesn't satisfy x > 1 at the same time.

on the other hand, any value of x > 1 will satisfy both inequalities because it is >= -2 and it is also > 1.

so, the AND portion of x >= -2 OR x > 1 is only satisfied when x > 1.

i don't think you can graph all of this in one representation on the number line.

it is either x >= -2 OR x > 1.
in the x >= -2 part, x = 1 is valid.
in the x > 1 part, x <= 1 is not valid.
in the x >= -2 AND x > 1 part, any value of x >= -2 and <= 1 is not valid.

i tried to find a reference online but was not successful, so i'll go with my best guess which is what i told you above.

it's either x >= -2 by itself or it's x > 1 by itself or it's x > 1 by itself if both conditions have to be satisfied at the same time.

that's my best guess, anyway.