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Solution - Adding, subtracting and finding the least common multiple

k=2
k=2
k=2
k=-2

Step by Step Solution

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     k/3-1/(3*k)-(1/k)=0 

Step by step solution :

Step  1  :

            1
 Simplify   —
            k

Equation at the end of step  1  :

   k     1     1
  (— -  ——) -  —  = 0 
   3    3k     k

Step  2  :

             1
 Simplify   ——
            3k

Equation at the end of step  2  :

   k     1     1
  (— -  ——) -  —  = 0 
   3    3k     k

Step  3  :

            k
 Simplify   —
            3

Equation at the end of step  3  :

   k     1     1
  (— -  ——) -  —  = 0 
   3    3k     k

Step  4  :

Calculating the Least Common Multiple :

 4.1    Find the Least Common Multiple

      The left denominator is :       3 

      The right denominator is :       3k 

        Number of times each prime factor
        appears in the factorization of:
 Prime 
 Factor 
 Left 
 Denominator 
 Right 
 Denominator 
 L.C.M = Max 
 {Left,Right} 
3111
 Product of all 
 Prime Factors 
333

                  Number of times each Algebraic Factor
            appears in the factorization of:
    Algebraic    
    Factor    
 Left 
 Denominator 
 Right 
 Denominator 
 L.C.M = Max 
 {Left,Right} 
 k 011


      Least Common Multiple:
      3k 

Calculating Multipliers :

 4.2    Calculate multipliers for the two fractions


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = k

   Right_M = L.C.M / R_Deno = 1

Making Equivalent Fractions :

 4.3      Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example :  1/2   and  2/4  are equivalent,  y/(y+1)2   and  (y2+y)/(y+1)3  are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

   L. Mult. • L. Num.      k • k
   ——————————————————  =   —————
         L.C.M              3k  

   R. Mult. • R. Num.       1
   ——————————————————  =   ——
         L.C.M             3k

Adding fractions that have a common denominator :

 4.4       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 k • k - (1)     k2 - 1
 ———————————  =  ——————
     3k            3k  

Equation at the end of step  4  :

  (k2 - 1)    1
  ———————— -  —  = 0 
     3k       k

Step  5  :

Trying to factor as a Difference of Squares :

 5.1      Factoring:  k2-1 

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =
         A2 - AB + BA - B2 =
         A2 - AB + AB - B2 =
         A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 1 is the square of 1
Check :  k2  is the square of  k1 

Factorization is :       (k + 1)  •  (k - 1) 

Calculating the Least Common Multiple :

 5.2    Find the Least Common Multiple

      The left denominator is :       3k 

      The right denominator is :       k 

        Number of times each prime factor
        appears in the factorization of:
 Prime 
 Factor 
 Left 
 Denominator 
 Right 
 Denominator 
 L.C.M = Max 
 {Left,Right} 
3101
 Product of all 
 Prime Factors 
313

                  Number of times each Algebraic Factor
            appears in the factorization of:
    Algebraic    
    Factor    
 Left 
 Denominator 
 Right 
 Denominator 
 L.C.M = Max 
 {Left,Right} 
 k 111


      Least Common Multiple:
      3k 

Calculating Multipliers :

 5.3    Calculate multipliers for the two fractions


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = 1

   Right_M = L.C.M / R_Deno = 3

Making Equivalent Fractions :

 5.4      Rewrite the two fractions into equivalent fractions

   L. Mult. • L. Num.      (k+1) • (k-1)
   ——————————————————  =   —————————————
         L.C.M                  3k      

   R. Mult. • R. Num.       3
   ——————————————————  =   ——
         L.C.M             3k

Adding fractions that have a common denominator :

 5.5       Adding up the two equivalent fractions

 (k+1) • (k-1) - (3)     k2 - 4
 ———————————————————  =  ——————
         3k                3k  

Trying to factor as a Difference of Squares :

 5.6      Factoring:  k2 - 4 

Check : 4 is the square of 2
Check :  k2  is the square of  k1 

Factorization is :       (k + 2)  •  (k - 2) 

Equation at the end of step  5  :

  (k + 2) • (k - 2)
  —————————————————  = 0 
         3k        

Step  6  :

When a fraction equals zero :

 6.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

  (k+2)•(k-2)
  ——————————— • 3k = 0 • 3k
      3k     

Now, on the left hand side, the  3k  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   (k+2)  •  (k-2)  = 0

Theory - Roots of a product :

 6.2    A product of several terms equals zero. 

 
When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 
We shall now solve each term = 0 separately 

 
In other words, we are going to solve as many equations as there are terms in the product 

 
Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 6.3      Solve  :    k+2 = 0 

 
Subtract  2  from both sides of the equation : 
 
                     k = -2

Solving a Single Variable Equation :

 6.4      Solve  :    k-2 = 0 

 
Add  2  to both sides of the equation : 
 
                     k = 2

Two solutions were found :

  1.  k = 2
  2.  k = -2

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