"and" and "or"

Question

-11≤ -5 +6x < 13

I also need this one

 

 

 

3x+ 2 ≤ -1 or 11 -x ≤ 5

Answer 1

What you want to do is solve for x. 

Let's take on the 1st problem.

You want to find out what numbers can x be in order for (-5+6x) to be less than 13 but more or equal to -11.

 

-11 ≤ -5 +6x < 13

Add 5 to all sides (because whatever you do to one side, you must do the same to the other. Just like with regular equation.)

-11 +5 ≤ -5 +6x +5< 13 +5

-6 ≤ 6x < 18

Now, divide by 6 to get x by itself

-1≤x<3

Therefore, x can be any between -1 (including -1) and 3(excluding 3).

 

For example, -1.00000 is included. So is 2.9 or 2.9999. However 3 will make (-5 + 6x) = (-5 +6x3)=13.

 

 

 

Second example:

3x+2≤-1

Subtract 2 from both sides

3x≤-1-2

3x≤-3

Divide by 3 to find x

x≤-1

So, x can be (-∞, -1]. From negative infinity (you do not include infinity because, well, it's infinity) to -1, including -1 (indicated by squared brackets "]" ).

 

You might ask, "well, what if I want to divide by 3 first".

Let's see

LEFT side is a SUM. therefore, if you divide a sum by a number (3), then you must divide each of the terms by that number.

3x+2≤-1

(3x+2)/3 ≤ -1/3

3x/3 +2/3 ≤ -1/3

x + 2/3 ≤-1/3

Now, subtract 2/3 from both sides

x +2/3 - 2/3 ≤ -1/3 - 2/3

x ≤ -3/3

x ≤ -1

You got the same answer.

 

I think this will explain the last problem.

 

Answer 2

3x+2≤ -1

3x+2-2≤-1-2 

3x≤-3

3x/3≤-3/3

x≤-1