Question

Solve the following inequalities; that is, find the range(s) of values of X for which the inequality is satisfied. (Picture Included)

Answer 1

I have a bunch that I have worked on but these 3 are giving me trouble.

Conviniently labelled a,b,c. :) Thanks in advance anyone that can help!

Answer 2

Let's work on letter (a) for now.
\[\frac{x^{2}-6x+3}{x-3}\le1\]
For these types of problems, particularly when quadratic expressions involved, see if they can be factored. Incidentally...

Answer 3

Incidentally, the numerator can be factored, like so
\[\frac{\left( x-3\right)^{2}}{x-3}\le1\]

Answer 4

once its factored as such, can I cancel 1 set of (x-3) and thus make it x-3 < 1?

Answer 5

I was wondering when you'd show up :D
lol Just kidding...
Yes, indeed, you can cancel it out, and make it much simpler as an inequality :)

Answer 6

had to go toilet lol. So from here could I do +3 to both sides to make it x < 4 ?

Answer 7

You're incredibly thorough :)
And you're RIGHT ;)

Answer 8

Thanks for that, I am new to inequalities and still learning all the tricks. :) How would I tackle b) ?

Answer 9

Well, can you factor it?

Answer 10

Right, I think it can't be factored in any manner that I know of. How about you?

Answer 11

Not unless it got really complex with decimals but I can't see how to factor it either.

Answer 12

Ok, before anything else, what's your final answer to (a) ?

Answer 13

It should be in the form of a set (interval), since it's a range you're looking for.

Answer 14

I thought x<4 was the final answer for a) :S

Answer 15

Is that how you normally do it? If it is, I won't contest it :)

Answer 16

From what I was taught, as long as there is a single x on one side and a variable on the other the problem is solved, is it possible to manipulate further for a range?

Answer 17

TYPO
lol
Work from this one instead
\[\frac{5x^{2}+5x-2}{2x+1}=3x-5\]
my bad, sorry :/

Answer 18

5x^2 + 5x -2 = 0, thus quad formula for +/- etc. Is what I'd do if it wasn't an inequality.

Answer 19

Yeah, granted, but
5x^2 + 5x - 2 = 0 is not what I'm asking you, I'm asking about the one more similar to your (b) problem:
\[\frac{5x^{2}+5x-2}{2x+1}=3x-5\]

Answer 20

You multiply 2x + 1 on both sides, first, right?

Answer 21

hmm, yes making the right a fraction as well.

Answer 22

Thus ending up with
\[5x^{2}+5x-2=(3x-5)(2x+1)\]

Answer 23

No fraction, making the right a *polynomial* as well :)

Answer 24

ahh lol a poly rather.

Answer 25

Ok, so that's good and well with equalities. You can almost do the same thing with inequalities, following a few rules. You realise that what we did was multiply both sides of the equality by the same expression, right?

Answer 26

correct and eliminating the fraction.

Answer 27

Well, allow me to present a scenario of sorts :)

5 < 7

This inequality holds, undoubtedly

What if you multiply 2 to both sides of the inequality, does the "<" still hold?

Answer 28

Multiplying 2 to both sides, you get
10 < 14
And indeed, the inequality holds.

However, what if instead of 2, we multiply -2 on both sides? Does the "<" still hold?

Answer 29

No, the sign would reverse as -10 is > -14.

Answer 30

That's right :D
Behold the multiplication property of inequality :D
If
a < b
Then
ka < kb
if k > 0

AND

ka > kb
if k < 0

Answer 31

Now, back to our dilemma
The problem is we don't have
\[\frac{5x^{2}+5x-2}{2x+1}=3x-5\]
instead, we have
\[\frac{5x^{2}+5x-2}{2x+1}<3x-5\]
What do we do now?

Answer 32

Well, following the Multiplication Property of Inequality, first, let's consider the case
where
\[x>-\frac{1}{2}\]or\[2x+1>0\]
Then...?

Answer 33

would that thus infer the x value is >0?

Answer 34

Well, technically, no, but that's not the reason we consider this case.
If indeed,
\[2x+1 > 0\]
then\[5x^{2}+5x-2 \ \ \ \left\{ <,>,= \right\} \ \ \ (3x-5)(2x+1)\]
Which of these three signs do you use?

Answer 35

Remember, we multiplied both sides by 2x + 1
And
2x + 1 > 0
Therefore...?

Answer 36

I think it would now be < as you mentioned earlier if ka < kb if k >0, if our k here is 2x+1 which is >0 and it has multiplied both sides then would < hold true?

Answer 37

You're right :D
In other words, since we multiplied both sides by something *positive*, the inequality "<" is preserved
Now then, do everything you can, evaluate, simplify, until you make it so that you have 0 on one side of the inequality.

Answer 38

If you get stuck, just give me a holler ;)

Answer 39

sure thing, I'll work on it now.

Answer 40

Ok, I got as far as this...

Answer 41

5x^2+5x-2 < (3x-5)(2x+1)
-3x -3x
5x^2+2x-2 < (-5)(2x+1)
-2x -2x
5x^2-2 < (-5)(1)
+2 +2
5x^2<(-3)*(1)
5x^2< -3

Answer 42

Ok, there's something not quite right about your first step.
You ought to evaluate (3x - 5)(2x + 1) before you do any of your addition property tricks :)
You know FOIL method, right? ;)

Answer 43

yep, I'll foil now.

Answer 44

6x^2+3x-10x-5 ?

Answer 45

Yes, good, so far, now simplify 3x - 10x, and then simplify the rest of the inequality

Answer 46

Final result was 12x+3>x^2

Answer 47

err <

Answer 48

I don't know how to reduce the x^2 further from here :\

Answer 49

Ok, wait, are you sure that the the equation has
5x^2 + 5x -2
And not
5x^2 - 5x - 2
?

Answer 50

yep it's +5x, I screenshotted the question sheet.

Answer 51

Oh, well, there is a solution, but it's not pretty :(
I hope you're ready for this...

Answer 52

Hit me! :)

Answer 53

Ok, I think I better take you through this :)
We have
5x^2 + 5x - 2 < (3x - 5)(2x + 1)
Evaluating through FOIL, we get
5x^2 + 5x - 2 < 6x^2 -7x - 5
Subtracting 5x^2 from both sides, we get
5x - 2 < x^2 - 7x - 5
Subtract 5x from both sides, you get
-2 < x^2 - 12x - 5
Add 2 to both sides, you get
0 < x^2 - 12x - 3

Getting it? :)

Answer 54

yep, this way everything lines up on one side.

Answer 55

But now that we have 0<x^2-12x-3, How do we determine the range?

Answer 56

Well, if it were, say,
0 < x^2 - 2x + 3, for instance
First, we determine the results of
0 = x^2 - 2x - 3
0 = (x - 3)(x + 1)
And the results happen to be -1 and 3.
We'll call these "critical numbers"

So, for x <-1, (x+1) < 0 and (x-3) < 0
So (x+1)(x-3) > 0
For x > -1 and x < 3
x + 1 > 0
x - 3 < 0
(x+1)(x-3) < 0
For x > 3
x + 1 > 0
x - 3 > 0
(x+1)(x-3) > 0

So the range for the inequality x^2 - 2x - 3 is
x < -1 UNION x > 3
Get it?

Answer 57

Yep, I've learned them as critical points or inflection points I assume this is the same thing? but the term "UNION" eludes me, what does that mean, are you able to post a symbol from the equation generator?

Answer 58

Union, uhh OR
\[x < -1 \cup x > 3\]

Answer 59

ahh yep, I know that symbol as "set of elements in"

Answer 60

Ok, well, you do something similar, except your expression
x^2 -12x -3 is not factorable. I'm sorry, it looks like you have to use your quadratic formula o.O

Answer 61

lol no worries. To err and err then to err less and less :)

Answer 62

Thank you for all your help, I however must go to bed, I will be on later tomorrow. Cheers for everything mate.

Answer 63

oh
I was about to tell you the same thing :D
We must be from adjacent or similar timezones :D

Answer 64

I would assume such although a fantastic userbase, 118 on at the moment is quite small. The night owls are out.

I partially quoted one of my favourites:

”The road to wisdom? — Well, it’s plain
and simple to express:
Err
and err
and err again
but less
and less
and less.”
— Piet Hein (1905–1996)

Have a good night bud. :)

Answer 65

Need help with b) and c) :|

Answer 66

i didn't read all of the above, but for the first two you have to set the expression less than zero or greater than zero, you cannot solve as it is written
and you cannot multiply both sides by any expression containing the variable, as you do not know if it is positive or negative

Answer 67

\[\frac{5x^2+5x-2}{2x+1}-(3x-5)<0\] is the first step of number 2
then you have to actually do the subtraction

Answer 68

kk I'll attempt now.

Answer 69

Should I multiply both sides by 2x+1 to make it 5x^2+5x-2 - (3x-5)(2x+1) < 0 ?

Answer 70

no you cannot do that because you don't know if \(2x+1\) is positive or negative
you have to change the sense of the inequality when you multiply by a negative number, that is why you cannot multiply both sides of an inequality by any expression containing a variable

Answer 71

ahh ok

Answer 72

you just have to do the subtraction, get a numerator and a denominator

Answer 73

you should get
\[\frac{-x^2+12x+3}{2x+1}<0\]

Answer 74

So would that mean, 5x^2+2x+3/2x+1<0

5x-3x, -2-(-5), is what I did here with the like terms.

Answer 75

How did you get -x^2+12x+3, I got the +3 but not the others?

Answer 76

you need to subtract. i am pretty sure what i wrote is correct

Answer 77

I just mean can you show me the syntax of your subtraction?

Answer 78

\[\frac{5x^2+5x-2}{2x+1}-(3x-5)<0\]
\[\frac{5x^2+5x-2-(3x-5)(2x+1)}{2x+1}<0\] and then algebra to clean up th enumerator

Answer 79

ahh foiled the right hand side then did subtractions.

Answer 80

kk I'm caught up now, from here with -x^2+12x+3/2x+1 < 0 How should I proceed?

Answer 81

@satellite73 Need help on proceeding. :S