algebra ... :S

[ADZ]

I am posting this on behalf of a friend.

I looked at it and couldnt figure it out, so here goes...

State the solution set for each of the following inequalities:

a) |3x+2|>4


b) (x+1)
-------------------------------- <0
(x+2)(x-3)


ANY help would be muchly appreciated!

 

[Emily]

a.) I can't remember how to write this "officially", but I got x is greater than 2/3 or less than -2.

b.) x < -1 (it might be x < -2, since if x=-2 then you come out with division by 0, but x < -1 is what you get when you solve it properly (I think))

Probably wrong, but what the heck.

 

[ADZ]

dont spose you have the steps in which you took to get those answers.

 

[nak-1]

a) |3x+2|>4

Question means, the positive value of 3x+2 must be greater than 4.

Now you have to find the boundry conditions, ie. whats the minimum value of x. both positive and negative, that gives the result 4:

solve

3x + 2 = 4

and

3x + 2 = -4

In this case

x = 2/3 and x = -2

________________________________________

b) (x+1)
-------------- <0
(x+2)(x-3)

This is looks tougher at first but all you need to realise is that the term has to end up as a negative number. This means that either the top or the bottom has to be negative, but not both, as a negative divided by a negative is a positive value and hence >0

the top half is only negative when x<-1

the bottom half is negative when x<3 and x>-2

therfore -2 < x < -1 is the range of possible solutions to this inequality.

The only thing that matters about the divison by zero is that it is an undefined number but that it is greater than 0.

 

[bla!!]

For a,
1. set 3x+2 = 4 and then solve for x. It comes out to 2/3
2. Then set 3x+2 = -4 and solve for x. It comes out to -2

You're answer then is x > 2 or x < -2
The technical way to write this is {x: x > 2/3 ; x < -2}


For b,

To find the limits of the equation (where it cannot be sloved) set each part of the denominator equal to 0 and solve for x.

x+2 = 0 you get x=-2
x-3 = 0 you get x=3

I'm not sure on the steps for this, but the range winds up being this.

{x: x<3 ; x!=-2}

 

[bla!!]

Hey, no fair posting while I'm typing!!

Also, the range of the 2nd one extends further negative than -2.

Evaluated at -3, the answer is -1/3 and at -4, you get -3/14

It goes on to -infinity. You just can't hit -2.

 

[Emily]

a.) When you have an absolute value in an inequality, you solve it out two ways:
3x+2 > 4 is one way; subtract 2 from both sides
3x > 2 - divide both sides by 3
x > 2/3
The second way you write it is by reversing the sign and negating the 4. You do this because |x| is the same value as |-x|, and you expressed |x| in the first equation, so now you're doing |-x|. But you divide by -1 to get the negative sign on the right, and when you divide by a negative number you reverse the sign of an inequality.
3x+2 < -4 (subtract 2 from both sides)
3x < -6 (divide by 3)
x < -2

b.) Ok... My first answer for this is wrong, but I think I've got the steps now:
For any fraction to be less than 0, it has to be negative, and for a fraction to be negative, either the numerator or the denominator has to be negative. So if we make the numerator negative, than x+1 < 0, which simplifies to x < -1. So that's one condition for one solution. For the next condition for this solution, the denominator has to be positive. This means that neither of the terms can be negative, so x > -2 and x > 3. But we've got a problem because x can't be less than -1 and greater than -2 and greater than 3. So then can try making the denominator positive by multiplying two negative terms. Now, in addition to x < -1, we add the conditions x < -2 and x < 2. Since these are all "less than" conditions, we only need the one condition x < -2, which we already have. So one solution is x < -2.

So now we do the solution where the numerator is positive and the denominator is negative. To make the numerator positive, x > -1. To make the denominator negative, one (but not both) of the two terms has to be negative. To make (x+2) a negative number, x < -2. But x < -2 is negated by x > -1, so we can't use that condition. To make (x-3) negative, x < 3. Therefore, the two conditions for the second solution are x > -1 and x < 3.

So... um... lots of writing. Final solutions: x < -2 and -1 < x < 3.

P.S. I'm sure there is a much, much easier, more efficient way of doing this problem but this is just a very, er, thorough way.