how to Combined Inequalities problems?

Question

1. 2x > -6 and x - 4 < 3
2. x + 3 > 2x + 1 and -4x < -8
3. -6 < x + 3 < 6
4. -3x > -6 or x - 5 < 2
5.x - 2 > 2x + 1 or 10 > -2x + 2

Answer 1

Solve them like you would a regular equation first, and then look at the two together. For the first one, we have

2x > -6
x > -3

x - 4 < 3
x < 7

The "and" in between the two inequalities means that whatever x is, it has to satisfy both inequalities. So what can x be? We found that x has to be larger than -3 and x has to be less than 7. Thus, -3 < x < 7.

If you were to graph this, start with a number line and put open circles at -3 and 7. (The open circles represent the fact that x cannot equal -3 or 7.) Then just draw a line between them - the line represents the fact that your answer can be anywhere between the two numbers. As a check, pick something between -3 and 7 and plug it in to both inequalities. If it works, then your answer is correct.

Likewise for #2:

x + 3 > 2x + 1
2 > x

-4x < -8
x > 2 -- you have to change the direction of the inequality when you multiply or divide by a negative number

This one turns out to be x <> 2 (x can't equal 2). x can be anything else except 2 - plug in numbers to see for yourself.

The third one is already a compound problem, and in some cases it may be easier to separate it into two problems (i.e. -6 < x + 3 and x + 3 < 6). But since the expression with the x is relatively simple, just subtract 3 from both sides of the inequality, so that you have -9 < x < 3.

The last two are tricky because of the "or" situation. "and" problems usually involve answers that include a closed range - for the first one, for example, the answer was anything between -3 and 7. For "or" problems, the answer will be something like "anything less than -3 or anything greater than 5." Graphically, "and" problems include a small part of the number line, while "or" problems include an infinite part (or parts) of the number line.

Solve the fourth problem as usual:

-3x > 6
x < 3

x - 5 < 2
x < 7

If you graphed these, you would have two rays pointing left - one ray starts at 3 and the other at 7. In this case, the answer can satisfy either one - it doesn't have to satisfy both. So if you said that x = 6 and plugged it in, it would work for the second inequality but not the first. That's perfectly fine - since the problem said "or." So your answer would be x < 7.

Finally, let's look at the last problem:

x - 2 > 2x + 1
-3 > x

10 > -2x + 2
8 > -2x
4 < x

Here, you have x being less than -3 or greater than 4. Graphically, you would be omitting a line segment (versus including only a line segment if this were an "and" problem). And if you choose anything less than -3 or greater than 4 to test your answer, you'll find that it will work for one and not the other - which, again, is perfectly fine since there is the "or" statement between the two. So this answer would be x < 3 or x > 4.

Answer 2

First, come across the answer set of the first set of inequalities. ok-5<=x-6<=3 => ok - 5 <= x - 6 and x - 6 <= 3 => ok + a million <= x and x <= 9 => x >= ok + a million and x <= 9 i.e. x E [ok + a million, 9] .................................(a million) Now, the second one set of inequalities are: 3x-7<=2(a million+x) and 5x-7>=23 => 3x - 7 <= 2 + 2x and 5x - 7 >= 23 => x <= 9 and 5x >= 30 => x <= 9 and x >= 6 i.e. x E [6, 9] ......................................(2... Now from (a million) and (2) ok + a million <= x <= 9 6 <= x <= 9 For x to be a answer for each of the inequalities given contained in the challenge, we are going to ought to set: ok + a million = 6 => ok = 5 answer: For ok = 5, the answer set of each and every of the given inequalities are an same.