how do u solve this: | 6x-1 | + | 3x+2 | > 7-3x ?

Question

i know the solution by graphing it but i need another way to do the problem

Answer 1

First find out when the two sides are equal:

| 6x-1 | + | 3x+2 | = 7-3x

for x > 1/6, the equation becomes:

6x-1 + 3x+2 = 7-3x

9x + 1 = 7 - 3x

12x = 6

x = 1/2

For -2/3 <= x <= 1/6, the equation becomes:

1 - 6x + 3x + 2 = 7 - 3x

3 = 7 (no solution)

For x < -2/3, the equation becomes:

-6x + 1 - 3x - 2 = 7-3x

x = -4/3.


Thus the two expressions are equal for x = 1/2, and x = -1.

to find when the one on the left is larger, divide the number line into three regions: x < -4/3, -4/3 < x < 1/2, x > 1/2, and test a number from each region: we will test the following pts.:

Region----Test Point----Which side bigger?
x < -4/3------------ (-2) ------ left
-4/3 < x < 1/2---- 0 --------- right
x > 1/2---------- 1---------- left

Since we are solving LHS > RHS, the solution is:

x < -4/3 OR x > 1/2.

Answer 2

I'm going to assume that you are solving for real numbers rather than complex. The notation would be the same.

The formal method is to set up all four cases where you simply flip the signs on each part in that is in the absolute value bars and constrain the values.

You get
case 1) 6x-1 + 3x+2 > 7-3x where x>=1/6 (and x>=-2/3)
case 2) 1-6x+3x+2>7-3x where x<=1/6 and x>=-2/3
case 3) 6x-1-3x-2>7-3x where x<-2/3 and x>=1/6 (a null case)
case 4) 1-6x-3x-2>7-3x where (x<=1/6 and) x<=-2/3

Is this starting to look like your graph?

Answer 3

first of all, realize that the 2 absolute values are going to make this problem a bit longer than most would enjoy, because you have to accommodate for x being positive and negative....
first of all, start by solving when x is positive....
you get 6x-1 + 3x+2 > 7-3x (the absolute values go away)
solving this, you get 9x +1 > 7 - 3x......and then 12x > 6, so therefore x >2 is one peice of the interval of solutions.
now, if x is negative you will need to switch the greater than sign, and make the right hand quantity negative, just as if you were solving standard absolute value problems...
so: (6x-1) + (3x+2) < - (7-3x)
so, 9x +1 < 3x -7
6x < -8
x< -8/6
so now you have an intrval for x: whevenver x > 2 or whenver x < -8/6 the equation will be true, you can pick a value for x to verify.
x=5 : (29) + (17) > 7 - 15 (TRUE)
x= -2: (13) + (4) > 7 + 6 (TRUE)
x= 0 (not in your interval).... (1) + (2) > 7 - 0 (NOT TRUE)
so there, we checked all three parts of the interval and they work out perfect.
Hope this helped you!

Answer 4

I can give you one way to do it but it may not be the best way.
If 6x - 1 and 3x + 2 are both positive then just remove the modulus signs. This will give you three conditions which must all be simultaneously true. These are x>1/6 and x>-2/3 and x>1/2 which means that the last is the controlling condition.
Then if 6x - 1 is negative but not 3x + 2 replace 6x - 1 by 1 - 6x and again remove modulus signs. Again you will have three conditions to satisfy. I'll leave this to you.
Do the same thing for 6x - 1 positive but 3x + 2 negative replacing 3x + 2 by -(3x + 2).
Finally if both are negative make both of the above replacements.

This method is long-winded so I hope that someone comes up with a better answer.

Answer 5

I'm not clear as to whether the "|" you have put in , are indeed "|" or meant to be brackets. If they are "|" then that indicates that the result between the "|" is to be considered positive, regardless of the actual result ... eg. | 2-3| is 1, not -1.


6x - 1 + 3x + 2 > 7 - 3x

= 6x + 3x - 1 + 2 > 7 - 3x
= 9x +1 > 7 - 3x
then move the x's to one side of the equation, and the numbers to the other, remember when you swap sides the sign changes so...
= 9x + 3x > 7 -1
= 12 x > 6
then reduce to 1 x by doing the same thing to both sides of the equation .. that is divide by 12
= x > 6/12
= x > 1/2 OR x > 0.5


M : )

Answer 6

| 6x - 1 | + | 3x + 2 | > 7 - 3x

Rewrite as:
| 6x - 1 | + | 3x + 2 | - 7 + 3x > 0

Now we need to look at this piece wise in three pieces.

(i) x > 1/6 Both absolute values are positive.
6x - 1 + 3x + 2 - 7 + 3x > 0
12x - 6 > 0
12x > 6
x > 1/2

(ii) -2/3 < x ≤ 1/6 One absolute value is positive and one negative.
- 6x + 1 + 3x + 2 - 7 + 3x > 0
-4 > 0
So this is false on the entire interval.

(iii) x ≤ -2/3 Both absolute values are negative.
- 6x + 1 - 3x - 2 - 7 + 3x > 0
- 6x - 8 > 0
6x < -8
x < -8/6 = -4/3

The equation is true for
x < - 4/3 or x > 1/2

Answer 7

(6x+5)/3 = (3x/7)-2 multiply every time period by ability of the least undemanding diverse of the denominators, which occurs to be 21 (6x+5)/3 * 21 = 42x +35 3x/7 * 21 = 9x -2 * 21 = 40 2 so that then you definately have 42x + 35 = 9x - 40 2 subtract 9x from both area 33x +35 = -40 2 subtract 35 from both area 33x = -seventy seven divide both area by ability of 33 x = -2 and a million/3 or x = -2.33333333333333... dixtirix is inaccurate

Answer 8

│6x - 1│ + │3x + 2│ > 7 - 3x

6x - 1 + 3x + 2 > 7 - 3x

9x + 1 > 7 - 3x

9x + 1 + 3x > 7 - 3x + 3x

12x + 1 > 7

12x + 1 - 1 > 7 - `1

12x > 6

12x / 12 > 6 / 12

x > 6 / 12

x > 1/2

x > 0.5

- - - - - - - - -s-

Answer 9

6x + 3x +3x>7-2+1
12X>6
x>6/12
x>0.5
to check if it is right replace X for "0.5" and calculate if both sides has the same value then its ok

Answer 10

simple.

you can solve the problem by getting a book and read about inequalities