If you can please explain to me how to solve these problems I'd be so greatful. Thanks a lot.

Printable View

- May 9th 2007, 07:19 PMalwaysalillostplease help me solve inequalities
If you can please explain to me how to solve these problems I'd be so greatful. Thanks a lot.

- May 9th 2007, 07:57 PMJhevon
1)

-2 <= a + 3 < 8

so with these, all we do is work on both sides at once. just like how when there is an equal sign, if you change something on one side, you do the same thing on the other, this is the same principle. if we change something in the center, we must do the same thing on both ends, if i do something on one end, i must do the same in the center and the other end and so on. the thing to look out for when dealing with inequalities are these:

- if we multiply by a negative number, we turn the signs around.

- if we take the inverse of everything, we must turn the signs around.

now to get to your question

-2 <= a + 3 < 8 ..........let's get a by itself. subtract 3 throughout the system

=> -2 - 3 <= a < 8 - 3

=> -5 <= a < 5

we wouldn't graph this as a two dimensional graph, but as a one-dimensional graph

basically we draw a number line, where a line on the number line represents a, and circles represent the end points f our solution. if we can be equal to a value, we shade the circle, if we can't, we leave it unshaded - May 9th 2007, 08:19 PMJhevon
2)

-6 <= 3 + x < 4

the same principle applies

-6 - 3 <= x < 4 - 3 ..............subtracted 3 throughout the system

-9 <= x < 1

so x is in the interval [-9,1)

i don't suppose we have to graph this, if so, you should try. i'd do it the same way as i did above, not in 2-D

3)

3x + 1 <= -2 or 2x + 3 >= 5

let's deal with each separately, then state a unified solution

3x + 1 <= -2

=> 3x <= -3

=> x <= -1

2x + 3 >= 5

=> 2x >= 2

=> x >= 1

so we have x <= -1 or x >=1, in interval form, that solution would be:

(-infinity, -1]U[1, infinity)

try graphing that on a number line

4)

|x - 4|> 3

those bars you see on the left are called absolute value signs. they basically force us to always output a positive result. so if we plug in a value that makes x - 4 negative, it turns it positive. any way, how do we do this one? we split it in two, and consider two parts, one inequality will have the values as you see them here, the other will have one side negative.

|x - 4| > 3

=> x - 4 > 3 or -(x - 4)> 3

=> x > 7 or x - 4 < -3 ............notice that i turned around the inequality since i just multiplied by -1

=> x > 7 or x < 1

so in interval form, that would be:

(-infinity, 1)U(7, infinity)

another way to do it is to keep it in one system, this is the usual way to do it in higher mathematics.

|x - 4| > 3

=> -3 > x - 4 > 3 ........keep the signs in the same direction, make one end the negative of the other

=> 1 > x > 7

=> 1 > x or x > 7 since x can't be both, and we get the same thing - May 9th 2007, 10:13 PMearboth
- May 11th 2007, 08:06 PMalwaysalillost
Thanks a lot Jhevon for your very thorough explaination and for all your help.

- May 11th 2007, 08:17 PMJhevon
- May 12th 2007, 06:12 AMalwaysalillost
hmm... for number one I got a different answer.

Shouldn't it be:

-2≤a+3<8

-2-3≤a and a<8-3

-5≤a and a<5

That's what I got. Which one would be the right answer? - May 12th 2007, 06:36 AMearboth
Hello,

I can't see Jhevon around here so I'm going to answer:

Your answer is correct and Jhevon's answer too. He only wrote the answer in one expression ("chain of inequalities"?) while you use two inequalities connected by an AND operation. Both answers - your's and Jehvon's - are the same. - May 12th 2007, 06:40 AMalwaysalillost
Thanks earboth!