Expert: Sherman D. - 7/21/2004

Question
Hi I am a high school student and I wanted to solve these problems. I am having trouble with the one's below.....

The question was -

Solve each of the following inequalities. Write solutions in interval notation.

1)-7<(2x-5)/2<1
2)2x^3-18x<=0
3)|x/x-1|<1

I also wanted to know is there was an answer to this.
the question was -

factor as completly as possible

4) k^15p-8

I could'nt get this one either.
the question was-

Simplify the following.

5)(7-i)/(3+i)+(4-3i)/(3i)

waiting for your reply,
Avinash

Answer
Solve each of the following inequalities. Write solutions in interval notation.

1.)
-7 < ((2x-5)/2) < 1
multiply -7 and 1 by 2
-14 < 2x - 5 < 2
add 5 to -14 and add 5 to 2
-9 < 2x < 7
divide -9 and 7 by 2
(-9/2) < x < (7/2)
Using interval notation, this would be
((-9/2),(7/2))
or you can write it as (-4.5,3.5) if you like.

-----------------------------------------------------------
2.)
2x^3 - 18x <= 0
2x(x^2 - 9) <= 0
2x(x - 3)(x + 3) <= 0
This gives you
x <= -3 or 0 <= x <= 3
Using interval notation you get
(-oo,-3] U [0,3]

To let you know, what i meant by -oo its suppose to be the -infinity sign.

-----------------------------------------------------------

3.)
|x/(x-1)| < 1
Can be written as
±(x/(x - 1)) < 1
divide both sides by ± and you get
(x/(x - 1)) < 1 or (x/(x - 1)) > -1
Multiply 1 and -1 by (x - 1)
x < x - 1 or x > -(x - 1)
x < x - 1 or x > -x + 1
subtract x to both sides in x < x - 1
0x < -1 or  x > -x + 1
add x to both sides in x > -x + 1
0x < -1 or 2x > 1
divide both sides by 0 in 0x < -1
Undefined or 2x > 1
divide both sides by 2 in 2x > 1
Since the 0x < -1 can't be used, this leaves you with an answer of
x > .5
I went to www.quickmath.com and they got an answer of
x < .5

-----------------------------------------------------------

4.)
k^15p-8
If by this one you mean
k^(15p - 8)

This may or may not be able to factor, i'm not totally certain.

-----------------------------------------------------------

5.)
((7 - i)/(3 + i)) + ((4 - 3i)/(3i))
Multiply everything by (3i)(3 + i)
(((7 - i)(3 + i)) + ((3i)(4 - 3i)))/((3i)(3 + i))
(((21 + 7i - 3i - (i^2)) + ((12i - (9i^2)))/(9i + 3i^2)
Keep in mind that i = sqrt(-1)
(((21 + (7 - 3)i - (-1)) + ((12i - (9*-1)))/(9i + (3*-1))
(((21 + 4i + 1) + (12i + 9)))/(9i - 3)
((22 + 4i) + (12i + 9))/(9i - 3)
(22 + 4i + 12i + 9)/(9i - 3)
((22 + 9) + (4 + 12)i)/(9i - 3)
(31 + 16i)/(-3 + 9i)
-(31 + 16i)/(3 - 9i)

Make sure to check over my work in case i messed up somewhere.