Find the common fraction equivalent to 0.12 the 12 is repeating. show all steps i have answer just can't figure out steps! !

The answer is:  "  "  .
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Given:  0.1212121212..... repeating ;  write that value as a fraction;
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In other words; we are given:  "0.1212121212..... repeating infinitely" ;  

→ that is to say; "0.12 ...  ;  {the "12" decimal portion repeats infinitely} ; 
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→ We write this value, as a fraction, as:  "12/99" ;
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→Explanation:
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Note:  "0.99999999...... repeating infinitely;  =  "1" .
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→Since:

Let us say that we have: 

"x = 0.999999 ; repeating infinitely;  

In order words, let us say we have: "x = 0.9.... ;  the "9" decimal repeats infinitely; 
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   Then "10x" ;  (that is: "10" multlipled by "x";  or "10*x" or "10x" );  is equal to:

"10" * (0.999999.....)  = 9.99999999...... (the "9" decimal repeats infinitely);

in other words:  10x = 9.99999999....

Divide each side by "10" ;

to get "x = 0.9999999....." ; the decimal "9" repeats indefinitely...." ;

But if you have:  "10x = 10" ;  divide each side of the equation by "10" ; 
   you get: "x = 1" . 
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Also,  if "x = 0.9999...(repeating infinitely); 

then:  10x = 9.99999.
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           10x  =  9.999999999999999......
       −     x  =  0.999999999999999.......
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            9x  =  9.00000000000000000000.....

 →  9x = 9 ; 

Divide each side of the equation; to get; 

 9x /9 = 9/ 9 ;  to get:  x = 1 ; and we have: x = 0.9999.... ;  so
  x= 0.99999.... = 1 ; 
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So, if the numbers "12" is repeating, we divie "12" by "99" ; 
  that is; we divide "12" by "two 9's" ;  since "12" is a "TWO-digit number"; a "two-digit number" is being repeated infinitely.
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            So;  0.12121212.....(the "12" is the decimal that repeats infinitely);            
                   
=  12/99 ;  which can be simplified;

Divide each side (both the numerator AND the denominator); by "3" ;
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  " 12/99 "  =  "(12÷3) / (99÷3) = 4/33 " .
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The answer is:    .
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The linear equation is: y=ax+b