Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "0.5" was replaced by "(5/10)". 3 more similar replacement(s)

Step by step solution :

Step  1  :

            1
 Simplify   —
            2

Equation at the end of step  1  :

     6         1     1
  ((——•(x2))-(——•x))-—  = 0 
    10        10     2

Step  2  :

             1
 Simplify   ——
            10

Equation at the end of step  2  :

     6         1     1
  ((——•(x2))-(——•x))-—  = 0 
    10        10     2
 Step  3  :
            3
 Simplify   —
            5

Equation at the end of step  3  :

    3           x     1
  ((— • x2) -  ——) -  —  = 0 
    5          10     2

Step  4  :

Equation at the end of step  4  :

   3x2     x     1
  (——— -  ——) -  —  = 0 
    5     10     2

Step  5  :

Calculating the Least Common Multiple :

 5.1    Find the Least Common Multiple

      The left denominator is :       5 

      The right denominator is :       10 

      Least Common Multiple:
      10 

Calculating Multipliers :

 5.2    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 = 2

   Right_M = L.C.M / R_Deno = 1

Making Equivalent Fractions :

 5.3      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.      3x2 • 2
   ——————————————————  =   ———————
         L.C.M               10   

   R. Mult. • R. Num.       x
   ——————————————————  =   ——
         L.C.M             10

Adding fractions that have a common denominator :

 5.4       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:

 3x2 • 2 - (x)     6x2 - x
 —————————————  =  ———————
      10             10   

Equation at the end of step  5  :

  (6x2 - x)    1
  ————————— -  —  = 0 
     10        2

Step  6  :

Step  7  :

Pulling out like terms :

 7.1     Pull out like factors :

   6x² - x  =   x • (6x - 1) 

Calculating the Least Common Multiple :

 7.2    Find the Least Common Multiple

      The left denominator is :       10 

      The right denominator is :       2 

      Least Common Multiple:
      10 

Calculating Multipliers :

 7.3    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 = 5

Making Equivalent Fractions :

 7.4      Rewrite the two fractions into equivalent fractions

   L. Mult. • L. Num.      x • (6x-1)
   ——————————————————  =   ——————————
         L.C.M                 10    

   R. Mult. • R. Num.       5
   ——————————————————  =   ——
         L.C.M             10

Adding fractions that have a common denominator :

 7.5       Adding up the two equivalent fractions

 x • (6x-1) - (5)     6x2 - x - 5
 ————————————————  =  ———————————
        10                10     

Trying to factor by splitting the middle term

 7.6     Factoring  6x² - x - 5 

The first term is,  6x²  its coefficient is  6 .
The middle term is,  -x  its coefficient is  -1 .
The last term, "the constant", is  -5 

Step-1 : Multiply the coefficient of the first term by the constant   6 • -5 = -30 

Step-2 : Find two factors of  -30  whose sum equals the coefficient of the middle term, which is   -1 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -6  and  5 
                     6x² - 6x + 5x - 5

Step-4 : Add up the first 2 terms, pulling out like factors :
                    6x • (x-1)
              Add up the last 2 terms, pulling out common factors :
                    5 • (x-1)
Step-5 : Add up the four terms of step 4 :
                    (6x+5)  •  (x-1)
             Which is the desired factorization

Equation at the end of step  7  :

  (x - 1) • (6x + 5)
  ——————————————————  = 0 
          10        

Step  8  :

When a fraction equals zero :

 8.1    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:

  (x-1)•(6x+5)
  ———————————— • 10 = 0 • 10
       10     

Now, on the left hand side, the  10  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   (x-1)  •  (6x+5)  = 0

Theory - Roots of a product :

 8.2    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 :

 8.3      Solve  :    x-1 = 0 

 Add  1  to both sides of the equation : 
                      x = 1

Solving a Single Variable Equation :

 8.4      Solve  :    6x+5 = 0 

 Subtract  5  from both sides of the equation : 
                      6x = -5
Divide both sides of the equation by 6:
                     x = -5/6 = -0.833

Supplement : Solving Quadratic Equation Directly

Solving    6x2-x-5  = 0   directly 

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

 9.1      Find the Vertex of   y = 6x²-x-5

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 6 , is positive (greater than zero). 

 Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. 

 Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. 

 For any parabola,Ax²+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is   0.0833  

 Plugging into the parabola formula   0.0833  for  x  we can calculate the  y -coordinate : 
  y = 6.0 * 0.08 * 0.08 - 1.0 * 0.08 - 5.0
or   y = -5.042

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 6x²-x-5
Axis of Symmetry (dashed)  {x}={ 0.08} 
Vertex at  {x,y} = { 0.08,-5.04} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-0.83, 0.00} 
Root 2 at  {x,y} = { 1.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 9.2     Solving   6x²-x-5 = 0 by Completing The Square .

 Divide both sides of the equation by  6  to have 1 as the coefficient of the first term :
   x²-(1/6)x-(5/6) = 0

Add  5/6  to both side of the equation :
   x²-(1/6)x = 5/6

Now the clever bit: Take the coefficient of  x , which is  1/6 , divide by two, giving  1/12 , and finally square it giving  1/144 

Add  1/144  to both sides of the equation :
  On the right hand side we have :
   5/6  +  1/144   The common denominator of the two fractions is  144   Adding  (120/144)+(1/144)  gives  121/144 
  So adding to both sides we finally get :
   x²-(1/6)x+(1/144) = 121/144

Adding  1/144  has completed the left hand side into a perfect square :
   x²-(1/6)x+(1/144)  =
   (x-(1/12)) • (x-(1/12))  =
  (x-(1/12))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(1/6)x+(1/144) = 121/144 and
   x²-(1/6)x+(1/144) = (x-(1/12))²
then, according to the law of transitivity,
   (x-(1/12))² = 121/144

We'll refer to this Equation as  Eq. #9.2.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (x-(1/12))²   is
   (x-(1/12))^2/2 =
  (x-(1/12))¹ =
   x-(1/12)

Now, applying the Square Root Principle to  Eq. #9.2.1  we get:
   x-(1/12) = √ 121/144

Add  1/12  to both sides to obtain:
   x = 1/12 + √ 121/144

Since a square root has two values, one positive and the other negative
   x² - (1/6)x - (5/6) = 0
   has two solutions:
  x = 1/12 + √ 121/144
   or
  x = 1/12 - √ 121/144

Note that  √ 121/144 can be written as
  √ 121  / √ 144   which is 11 / 12

Solve Quadratic Equation using the Quadratic Formula

 9.3     Solving    6x²-x-5 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  x  , the solution for   Ax²+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  x =   ————————
                      2A

  In our case,  A   =     6
                      B   =    -1
                      C   =   -5

Accordingly,  B²  -  4AC   =
                     1 - (-120) =
                     121

Applying the quadratic formula :

               1 ± √ 121
   x  =    —————
                    12

Can  √ 121 be simplified ?

Yes!   The prime factorization of  121   is
   11•11 
To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 121   =  √ 11•11   =
                ±  11 • √ 1   =
                ±  11

So now we are looking at:
           x  =  ( 1 ± 11) / 12

Two real solutions:

x =(1+√121)/12=(1+11)/12= 1.000

or:

x =(1-√121)/12=(1-11)/12= -0.833

Two solutions were found :

1.  x = -5/6 = -0.833
   
2.  x = 1