3(w-3)≥9-3w if all real numbers can solve tell me why

 " w = (all real numbers) " —as explained below.

Step-by-step explanation:

Given the following "inequality":

 " -3(w – 3) ≥ 9 – 3w " ;   Solve for "w" ;

                                         and see if the answer ["value for "w"]; is:

                                           "all real numbers."

  →  " -3(w – 3) ≥ 9 – 3w " ;  

Method 1):

On the "right-hand side" of the "inequality";

  Factor out a "(-3)" ;

  " -3(w – 3) ≥  ( -3 * ? = 9?) – (-3 * ? = 3w? ) " ;

                        1)   " -3 * (what value?) = 9 "?

                          Let "x" be the 'unknown value' :

                             →  " -3x = 9 " ;

                                Divide each side of the equation by "(-3)"  ;

                                  to isolate "x"  on one of the equation;

                                 & to solve for "x" ;

                                  -3x / -3 = 9 / -3 ;

                              to get:  " x = -3 " ;  

                        2)   " -3 * (what value?) = 3w " ?

                        Let "x" be the 'unknown value' :                      

                           →   " -3x = 3w " ? ;

                         → Divide each side or the equation by ("-3").

                             → to isolate "x" on one side of the equation;

                                →  & to solve for "x" ;

                                    -3x / 3 = 3w / -3 ;

                                    to get:  x = -1w ;

So:    " -3(w – 3) ≥  -3* (? = 9?) – (-3 * ? = 3w?) " ;

Rewrite as:  

         " -3(w – 3) ≥  -3* [(-3 – (-1w)] " ;

Now, let us consider the Following Portion of the "right-hand side" of the inequality:;

                            " (-3 – (-1w)]  =  " (-3 + 1w) " ;  

                    → {since:  "subtracting a "negative value is the equivalent of adding that particular value's positive value"} ;

         and bring down the "(-3)" on the "right-hand side" of the inequality; and rewrite the inequality as:

           " -3(w – 3) ≥ -3(-3 + 1w) "  ;

        →  Now, divide EACH SIDE of the inequality by "(-3)" :

 {Note: Each time when one multiplies or divides an inequality by a "negative value"—the inequality sign flips to the other direction.}.

          →  [ -3(w – 3)]  / -3  ≥ [-3(-3 + 1w ] / -3 ;  

                 to get:

             →  " (w – 3) ≤ (-3 + 1w) " ;

Note: " (-3 + 1w) " ; ↔ " [(1w + (-3)]  =  " 1w – 3 " =  "w – 3 " ;

Rewrite the inequality:

             →  " (w – 3) ≤ (w – 3 ) " ;

             ↔ "  w – 3  ≤  w – 3  "  ;

        The same value is equal to each other: not less than or equal to each other:   For instance:

              →  " w – 3  ≤  w – 3 " ;

If we add "3" to each side of the equation:

              →  " w – 3 + 3  ≤  w – 3 + 3 " ;

  We get:    " w  ≤ w " .   "w = w".  "w is not "less than" itself.  So all real numbers apply!

Method 2)

Given the following "inequality":

 " -3(w – 3) ≥ 9 – 3w " ;   Solve for "w" ;

                                         and see if the answer ["value for "w"]; is:

                                           "all real numbers."

On the "left-hand side of the inequality;

we have:

 " -3(w – 3) " ;

Take note of the "distributive property of multiplication" :

   → " a(b + c) = ab + ac " ;

As such:

We can expand:

     " -3(w – 3) = (-3*w) + (-3*-3) " ;

                     =  -3w + (9) ;

Now, we can rewrite the original inequality:

     " -3w + 9 ≥ 9 – 3w " ;

Note: " -3w + 9 = 9 + (-3w) = 9 – 3w " ;  

                                      → {since: adding a "negative value" gets the same value as subtracting that value's "positive equivalent"};

       And we can rewrite our inequality as:

     " 9 – 3w  ≥ 9 – 3w " ;

Note: We have the same value on each side.

 "(9 – 3w)" is not greater than itself; it is "equal to itself".

 Any and all real numbers as values for "w" will result in the same value for any particular expression's Exact Same Expression!

 So:  "w = (all real numbers)" .

Hope this is helpful to you!

 Best wishes!

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8