Question

solve the following inequation: x+3≤(1-x^2)/(2-x)

Answer 1

"inequation"

Haha!

Well, why don't you try solving\[x+3=\frac{1-x^2}{2-x}\]instead?

Answer 2

\[2<x\leq 5 \]

Answer 3

\[x+3 \le \frac{1-x^2}{2-x}\]
\[x+3-\frac{1-x^2}{2-x} \le 0\]

\[\frac{(x+3)(2-x)-(1-x^2)}{2-x} \le 0\]

\[\frac{(2x-x^2+6-3x)-(1-x^2)}{2-x} \le 0\]

\[\frac{-x^2+x^2+2x-3x+6-1}{2-x} \le 0\]
\[\frac{-x+5}{2-x} \le 0\]

so the expression (-x+5)/(2-x) is zero when x=5
the expression (-x+5)/(2-x) is undefined when x=2

so


--------|-----|----
2 5
-------------------
-x+5| + | + | -
|
2-x | + | - | -
==============
+ - +

so the expression (-x+5)/(2-x) is less than zero when 2<x<5

-----------------------------------------------
final answer
(-x+5)/(2-x) is less than or equal to when
\[2 \le x <5\]

Answer 4

oops

Answer 5

\[2 < x \le 5\]

Answer 6

Good work myininaya. I cheated and used the Mathematica "Reduce" function.

Answer 7

lol

Answer 8

thanks