Algebra help? Quick and easy!?

Question

I have already answered these questions, and i'm just wondering if i have the correct answer. I would GREATLY appreciate it if you could just look at them. You can do as many or as little as you feel like, any you have time for would be wonderful :)

Question:

2x-4 < 6-7x < 3x+6

my answer was: 5<x<0 but i don't think that's possible...

Question:

(x+5)(x+2)squared(2x-1) > 0

my answer was: x to the fourth power > 4

Question:

-3 < 4x-9 < 11

my answer: -3 < x < -4 again i don't know if that's possible...

thank you for help with ANY of these :) :)

and if you want tell me the correct answer as well as what i did wrong. but you don't have to again :)

THANKS!!!!

Answer 1

OH MY GOSHHH i hateeeeeeeeee algebra. im so glad ive left school now. i got a c in maths which is likeee awesome for me lololol

Answer 2

Sorry! Math and I got a divorce! We are not on speaking terms.

Answer 3

my answer for the last one is -3&lt;4x-9=-3+9&lt;4x =6&lt;4x ,4x&gt;6 so 2x&gt;3

x&gt;2over3theis is no.1

4x-9&lt;11 = 4x&lt;11+9 = x&lt;5 theis is no.2

so the answer is x = 4                                                                                            

that all what i can solve.sorry

Answer 4

Thanks for answering my question but I have nooooo idea. im so sorry. :/

Answer 5

sorry, i took algebra last year and learned this but i cant understand my notes. all i can tell you is that 1 and 3 are wrong because in 1. a number can&#039;t be greater than five and less than zero at the same time, and in 2. a number cant be greater than -3 but less than -4. if you cant get the answers on here you should ask your parents, an older sibling, or call someone in ur class. sorry this wasn&#039;t that helpful

Answer 6

When a compound inequality has x&#039;s in all three parts, I find it easier to work with them in pieces.

2x - 4 &lt; 6 - 7x &lt; 3x+ 6 means

2x - 4 &lt; 6 - 7x and 6 - 7x &lt; 3x+ 6

2x - 4 &lt; 6 - 7x   Add 7x to both sides:

9x - 4 &lt; 6   Add 4 to both sides:

9x &lt; 10   Divide both sides by 9 (which is positive, so the sense of the inequality is unchanged)

x &lt; 10/9

6 - 7x &lt; 3x+ 6

6 &lt; 10x + 6

0 &lt; 10x

0 &lt; x, or equivalently, x &gt; 0

So we must have x simultaneously greater than 0 and less than 10/9

0 &lt; x &lt; 10/9

(x+5)(x+2)²(2x-1) &gt; 0

If a product like the expression on the left changes sign, it has to be zero at the value(s) where it changes sign. So first determine where the expression is zero:

(x+5)(x+2)²(2x-1) = 0

x+5 = 0 or x+2 = 0 or 2x-1 = 0

x = -5 or x = -2 or x = 1/2

Divide a number line using these values, like this:

__|___|___|__

..-5...-2.....½

Pick a number in each of the pieces and evaluate the expression. The numerical value doesn&#039;t matter, only its sign. If it&#039;s positive, then the expression is positive throughout that piece; if it&#039;s negative, then the expression is negative throughout that piece.

x = -6: (-6+5)(-6+2)²(2(-6)-1) &gt; 0

x = -3: (-3+5)(-3+2)²(2(-3)-1) &lt; 0

x = 0: (0+5)(0+2)²(2(0)-1) &lt; 0

x = 1: (1+5)(1+2)²(2(1)-1) &gt; 0

So (x+5)(x+2)²(2x-1) &gt; 0 for x &lt; -6 or x &gt; 1

Since x appears in only one part of the compound inequality, let&#039;s work with it as is:

-3 &lt; 4x-9 &lt; 11   Add 9 to all three parts:

6 &lt; 4x &lt; 20   Divide all three parts by 4:

3/2 &lt; x &lt; 5

(Again, since I&#039;m dividing by a positive number, the sense of the inequalities is preserved)

Answer 7

dont tell me we have to do this in high school!?!?!

d**n, and i thought what i was doing now was hard......

proved me wrong....