Question

In my math book, it tells me to use some method with +'s and -'s to solve inequalities. Can somebody explain this to me?

Answer 1

help me and that's ur 500th medal.

Answer 2

Solve for x: $$\frac{1}{1-x} \leq 0$$

Answer 3

oops I meant $\frac{1}{1-x}\geq 0$

Answer 4

$$\frac{1}{1-x}\geq 0$$

Answer 5

the book teaches me like so:
+/+=yes
-/+=no
-/-=yes
what does it mean?

Answer 6

That looks like the NXOR operation

Answer 7

what is the NXOR operation?
Can you explain it to me?

Answer 8

NOT exclusive OR

Answer 9

explain?

Answer 10

when does the XOR operation evaluate to true?

Answer 11

and what does it have to do with inequalities :-P

Answer 12

i don't know

Answer 13

Do you know how to solve an inequality?

Answer 14

sort of, but the book's way of doing it confuses me because every time I solve it the book's way, I get the exact opposite of the answer in the book

Answer 15

Let's do one now:\[\text{Find all numbers x for which:}\]\[4-x < 3 - 2x\]

Answer 16

How would you solve it?

Answer 17

\[4-x<3-2x\]
\[4-3<x\]
\[1<x\]

Answer 18

is it correct?

Answer 19

oh, it's a negative x
-1<x

Answer 20

actually it is \[x < 1\]

Answer 21

what?

Answer 22

oh, ok

Answer 23

Another one:\[5 - x^2 < 8\]

Answer 24

5-x^2<8
5-x^2-8<0
-3-x^2<0
I got this far

Answer 25

i think i need to complete the square

Answer 26

There's no need to complete the square. Just move X to one side of the inequality :-D

Answer 27

ok

Answer 28

5-x^2<8
5-x^2-8<0
-3-x^2<0
-x^2<3
x^2>-3
|x|>-3sqrt

Answer 29

\[|x|>\sqrt{-3}\]

Answer 30

i'm stuck

Answer 31

Instead of going to \[|x|>\sqrt{-3}\] try to conclude something from \[x^2 > -3\]

Answer 32

One more \[x^2-4x+3 > 0\]

Answer 33

do I split it into two different equations?

Answer 34

You can actually stop once you've reached the step \[x^2 > -3\]since \[x^2 \geq 0\] is true for all x.

Answer 35

(x-3)(x-1)>0
the critical points are 3 and 1
x>3 and x<0

Answer 36

in this situation, my book tells me to do some sort of +/- and +/+ and -/- thing

Answer 37

When is the product of two numbers positive?

Answer 38

when x>3

Answer 39

When you multiply two positive numbers, do you get a positive answer?

Answer 40

yes

Answer 41

or two negatives

Answer 42

but not if one number is positive and the other is negative.

Answer 43

yup, and that is not more than 0

Answer 44

so in our case, we should split the problem into two situations, each with two equations

Answer 45

When both numbers are positive:\[x -3 >0\ \ \ \ and\ \ \ \ x-1 >0\]
When both are negative\[x-3 < 0 \ \ \ \ and\ \ \ \ x-1 < 0\]

Answer 46

yup

Answer 47

what point are you trying to make?

Answer 48

There are two different solutions to the inequality

Answer 49

each accounting the fact that both expressions are either both positive or negative.

Answer 50

oh, ok

Answer 51

bbut the answer is positive, right?

Answer 52

not necessarily; the answer is two different inequalities

Answer 53

or a compound inequality :-D

Answer 54

x < 1 or x > 3

Answer 55

ok,well thanks!

Answer 56

that really helped me