The number is:  " 12 ".Â
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 Let "x" represent "the unknown number" (for which we wish to solve.
The expression:
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x  − 6  =  2  ;  Solve for "x" ; Â
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Method 1)Â
  Add "6" to EACH SIDE of the equation;
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    → Â
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x  − 6  + 6 =  2 + 6 ;
to get:
   → Â
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x = 8 ;
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Multiply each side of the equation by "
![\frac{3}{2}](/tpl/images/0092/0491/410fe.png)
" ; to isolate "x" on one side of the equation ; and to solve for "x" ;
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   →
![\frac{3}{2}](/tpl/images/0092/0491/410fe.png)
 *
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x = 8 *
![\frac{3}{2}](/tpl/images/0092/0491/410fe.png)
;
    →  x = 8 *
![\frac{3}{2}](/tpl/images/0092/0491/410fe.png)
;
        =Â
![\frac{8}{1}](/tpl/images/0092/0491/39134.png)
*
![\frac{3}{2}](/tpl/images/0092/0491/410fe.png)
;
        =Â
![\frac{8*3}{1*2}](/tpl/images/0092/0491/3320c.png)
;
   Â
        =Â
![\frac{24}{2}](/tpl/images/0092/0491/dd41b.png)
;
Â
        = 12 .
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 x =  12 .
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Method 2)
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![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x  − 6  =  2  ;  Solve for "x" ;Â
  Add "6" to EACH SIDE of the equation;
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    → Â
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x  − 6  + 6 =  2 + 6 ;
to get:
   → Â
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x = 8 ;
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Multiply each side of the equation by "3" ; to get rid of the "fraction" ;
         → 3 *
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x = 8 * 3 Â ;
        →Â
![\frac{3}{1}](/tpl/images/0092/0491/89cb8.png)
*
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x = 8 * 3 ;
        →Â
![\frac{3*2}{1*3}](/tpl/images/0092/0491/070b6.png)
 x = 8 * 3Â
        →Â
![\frac{6}{3}](/tpl/images/0092/0491/a6483.png)
x = 24 ;Â
        → 2x = 24 ;
 →  Divide each side of the equation by "2" ; to isolate "x" on one side of the equation; & to solve for "x" :Â
Â
          2x / 2 = 24 / 2  ;
            x = 12 .
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Method 3).
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![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x  − 6  =  2  ;  Solve for "x" ; Â
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Add "6" to EACH SIDE of the equation;
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    → Â
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x  − 6  + 6 =  2 + 6 ;
to get:
   → Â
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x = 8 ;
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Now, divide each side of the equation by "
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
" ;
 to isolate "x" on one side of the equation; & to solve for "x" ;
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{
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
x } Â / Â {
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
} Â = Â 8 / {
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
} ;
to get: Â x =Â Â 8 / {
![\frac{2}{3}](/tpl/images/0092/0491/5f13e.png)
} ;
        =  8 * (
![\frac{3}{2}](/tpl/images/0092/0491/410fe.png)
;
        = Â
![\frac{8}{1}](/tpl/images/0092/0491/8adb5.png)
 * Â
![\frac{3}{2}](/tpl/images/0092/0491/5dd0e.png)
;
        = Â
![\frac{8*3}{1*2}](/tpl/images/0092/0491/3320c.png)
;
        = Â
![\frac{24}{2}](/tpl/images/0092/0491/dd41b.png)
;
        = 12 ;Â
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             x = 12 .
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NOTE: Â Variant: Â (in "Methods 2 & 3") :
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At the point where:
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 =  8 * (
![\frac{3}{2}](/tpl/images/0092/0491/410fe.png)
) ;
 = Â
![\frac{8}{1}](/tpl/images/0092/0491/8adb5.png)
 * Â
![\frac{3}{2}](/tpl/images/0092/0491/5dd0e.png)
;
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 We can cancel out the "2" to a "1" ; and we can cancel out the "8" to a "4" ;
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 {since: "8÷2 = 4" ; and since:  "2÷2 =1" } ;
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and we can rewrite the expression:
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Â
![\frac{8}{1}](/tpl/images/0092/0491/8adb5.png)
 * Â
![\frac{3}{2}](/tpl/images/0092/0491/5dd0e.png)
;
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as: Â Â
![\frac{4}{1}](/tpl/images/0092/0491/2a287.png)
 * Â
![\frac{3}{1}](/tpl/images/0092/0491/ddba7.png)
;Â
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which equals:
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→ Â
![\frac{4*3}{1*1}](/tpl/images/0092/0491/ed0f1.png)
;Â
  =  Â
![\frac{12}{1}](/tpl/images/0092/0491/67c96.png)
;
      =  12 .
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     x = 12 .Â
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 The number is:  " 12 ".Â
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