Solution - Adding, subtracting and finding the least common multiple

Step by step solution :

Step  1  :

              1  
 Simplify   —————
            x + 1

Equation at the end of step  1  :

   3      1  
  —— +  —————  = 0 
  2x    x + 1

Step  2  :

             3
 Simplify   ——
            2x

Equation at the end of step  2  :

   3      1  
  —— +  —————  = 0 
  2x    x + 1

Step  3  :

Calculating the Least Common Multiple :

 3.1    Find the Least Common Multiple

      The left denominator is :       2x 

      The right denominator is :       x+1 

      Least Common Multiple:
      2x • (x+1) 

Calculating Multipliers :

 3.2    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 = x+1

   Right_M = L.C.M / R_Deno = 2x

Making Equivalent Fractions :

 3.3      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.       3 • (x+1)
   ——————————————————  =   ——————————
         L.C.M             2x • (x+1)

   R. Mult. • R. Num.          2x    
   ——————————————————  =   ——————————
         L.C.M             2x • (x+1)

Adding fractions that have a common denominator :

 3.4       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:

 3 • (x+1) + 2x        5x + 3   
 ——————————————  =  ————————————
   2x • (x+1)       2x • (x + 1)

Equation at the end of step  3  :

     5x + 3   
  ————————————  = 0 
  2x • (x + 1)

Step  4  :

When a fraction equals zero :

 4.1    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:

    5x+3  
  ———————— • 2x•(x+1) = 0 • 2x•(x+1)
  2x•(x+1)

Now, on the left hand side, the  2x • x+1  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   5x+3  = 0

Solving a Single Variable Equation :

 4.2      Solve  :    5x+3 = 0 

 Subtract  3  from both sides of the equation : 
                      5x = -3
Divide both sides of the equation by 5:
                     x = -3/5 = -0.600

One solution was found :