prelim - inequalities and modulus questions (1 Viewer)

[charisloke]

Would appreciate it if anyone could help me with these sums; I tried doing them and came up with different answers from the ones in the book. So I'm not sure now if I did them wrong, or there are mistakes in the book's answers:

(y² - 6)/y < y


|x-3|/(3-x) = x


x² > a²


|x - 3| + |x - 4| = |x - 2|

Thanks a lot!

 

[tommykins]

Would appreciate it if anyone could help me with these sums; I tried doing them and came up with different answers from the ones in the book. So I'm not sure now if I did them wrong, or there are mistakes in the book's answers:

(y² - 6)/y < y


|x-3|/(3-x) = x


x² > a²


|x - 3| + |x - 4| = |x - 2|

Thanks a lot!

1 . (y² - 6)/y < y
y(y²-6) < y³
y³-6y < y³
-6y > 0

y > 0


2. |x-3|/(3-x) = x

|x-3| = x(3-x) -> don't need to square denominator as it is an = sign.
|x-3| = 3x - x²
x² - 3x + |x-3| = 0

Positive case

x² - 3x + x - 3 = 0 -> x² - 2x - 3 = 0
(x-3)(x+1) = 0
.:. x = 3 and -1

Negative Case
x² -3x - x + 3 = 0 -> x² - 4x + 3 = 0
(x-3)(x-1) = 0
.:. x = 3 and +1

Simply sub this back into the original |x-3|/(3-x) = x to get your x values that satisfy it.

3. x² > a² -> x²-a² > 0
(x-a)(x+a) > 0
From this, if we think logically, x² > a² means the number x squared has to be greater than the number a², thus the magnitude of x has to be greater than a (since a number squared is always positive)

x>a
-x < -a


4.|x - 3| + |x - 4| = |x - 2| -> |x - 3| + |x - 4| - |x - 2| = 0
Positive case

x-3 + x-4 - x + 2 = 0
x - 5 = 0
x = 5

negative case
still leasd to x = 5


I think for this you need the +- case for each of the aboslute values, but I'm too busy to be able to do that at the moment.

If I've done anything incorrect, I apologise.

 

[charisloke]

3. It makes sense to me now *phew*.

4. I'll do the individual +- cases myself, no worries bout that.

Thanks for your help.

 

[tommykins]

3. It makes sense to me now *phew*.

4. I'll do the individual +- cases myself, no worries bout that.

Thanks for your help.

Please correct one of the solutions to 3.

x > a and x < -a

Sorry about that.

 

[sicmacao]

Would appreciate it if anyone could help me with these sums; I tried doing them and came up with different answers from the ones in the book. So I'm not sure now if I did them wrong, or there are mistakes in the book's answers:

(y² - 6)/y < y


|x-3|/(3-x) = x


x² > a²


|x - 3| + |x - 4| = |x - 2|

Thanks a lot!

The answers to |x - 3| + |x - 4| = |x - 2| are x = 3 or 5

Need to check 4 cases
if x=>4
x = 5

if 3<=x<4
x = 3

if 2<=x<3
x = 3 (impossible)

if x<2
x = 5 (impossible)