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Equations with fractions:   2nd Level

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Let us now compare equations with fractions with adding fractions.

To add these fractions --

2
a
 +   3
b
 +   4
c

-- their LCM is abc.

2
a
 +   3
b
 +   4
c
  =   2bc + 3ac + 4ab
        abc

We have multiplied each numerator -- 2, 3, 4 -- by those factors of abc that are missing from its denominator.

Now, when we clear this equation -- which has the same fractions:

2
a
 +   3
b
 =  4
c
 
2bc + 3ac  = 4ab

-- each term in the equation is the same as the numerator in the addition. Each numerator -- 2, 3, 4 -- is multiplied by those factors of the LCM that are missing from its denominator.

The terms of the cleared equation are the same as the numerators
of the added fractions.

Example 1.   Solve for x:

 1 
2x
 +     1   
x − 1
 =       1     
2(x − 1)
 
 Solution.   The LCM is  2x(x −1).
 
      Here is the cleared equation:
 
x − 1 + 2x  = x.
 
      Each numerator has been multiplied by those factors of the LCM that are missing from its denominator.   The solution follows:
 
2x  = 1
 
x  = 1
2

Again, when solving an equation with fractions, the very next statement you write -- the next line -- should have no fractions.

Problem 9.   Solve for x:

  _9_  
3x − 5
 +     _1_  
x + 2
 =     4  
x − 2
 
       The LCM is the product of the three denominators.  Here is the cleared equation and its solution:
 
9(x + 2)(x − 2) + (3x − 5)(x − 2)  =  4(3x − 5)(x + 2)
 
9(x² − 4) + 3x² − 11x + 10  =  4(3x² + x − 10)
 
9x² − 36 + 3x² − 11x + 10  =  12x² + 4x − 40
 
12x² + − 11x − 26  =  12x² + 4x − 40
 
−11x − 4x  =  −40 + 26
 
−15x  =  −14
 
x  =  14
15

The original equation is immediately transformed into an equation without fractions.  Each succesive statement -- each line -- follows from the previous line.  The transformations are a logical sequence of statements, as in Lesson 9.

Problem 10.   Solve for x:

1
x
  +     1   
x − 1
 =    1 
8x
 +     _1_   
8(x − 1)
 
       The LCM is  8x(x − 1).   Here is the cleared equation and the logical sequence that leads to the solution:
 
8(x − 1) + 8x  =  x − 1 + x
 
8x − 8 + 8x  =  2x − 1
 
16x − 2x  =  −1 + 8
 
14x  =  7
 
x  =  1
2

Problem 11.   Factor the denominators, clear of fractions, and solve for x:

   _1_   
x² − 2x
       _8_       
3x² − 5x − 2
=    _4_   
3x² + x
   _1_   
x(x − 2)
        _8_        
(3x + 1)(x − 2)
=     _4_    
x(3x + 1)

The LCM is  x(x − 2)(3x + 1).   Here is the cleared equation and its solution:

3x + 1 − 8x = 4(x − 2)
 
1 − 5x = 4x − 8
 
−5x − 4x = −8 − 1
 
−9x = −9
 
x = 1

Problem 12.   Factor the denominators, clear of fractions, and solve for x:

 x + 6 
x² − 9
  +        x − 9     
x² − 4x + 3
 =      _2x − 1_  
x² + 2 x − 3
  __x + 6__  
(x + 3)(x − 3)
  +         x − 9      
(x − 1)(x − 3)
 =       _2x − 1_   
(x + 3)(x − 1)

The LCM is  (x + 3)(x − 3)(x − 1).   Here is the cleared equation and its solution:

(x + 6)(x − 1) + (x − 9)(x + 3) = (2x − 1)(x − 3)
 
x² + 5x − 6 + x² − 6x − 27 = 2x² − 7x + 3
 
2x² − x − 33 = 2x² − 7x + 3
 
x + 7x = 3 + 33
 
6x = 36
x = 6
  Example 2. ax
 b
  =    c 
d

This is a simple fractional equation, which we saw in Lesson 9.  It is

  not necessary to clear of fractions.   a
b
 simply goes to the other side as its

reciprocal.

  x   =    bc 
ad

Whatever multiplies on one side will divide on the other.  And whatever divides on one side will multiply on the other.

  Example 3. 2s
3t
 =   pq
rx

We would like x to be in the numerator on the left.  Imagine placing it there.

Then  pq
r
stays on the right.   And  2s 
3t
 goes to join them as its

reciprocalexclamation

  x  =   3tpq
2sr

Problem 13.   Solve for x:

ab
cd
 =   mx 
npq
 
mx
npq
 =  ab
cd
   Exchange sides.
 
x  =  npqab
 mcd 

Problem 14.   Solve for x:

ab
c  
 =      _st_    
u(v + w)x
 
x  =    __cst__  
abu(v + w)

In each of the following, solve for x.

   Problem 15.  A  =  ½Bx
 
  2A  =  Bx
 
  x  =  2A
 B
   Problem 16.  s  =  ½(x + w)t
 
  (x + w)t  =  2s
 
  xt + wt  =  2s
 
  xt  =  2swt
 
  x  =  2swt
    t
   Problem 17.  s  =  sx
  at
 
  sat  =  sx
 
  x  =  ssat
   Problem 18.  A  =  B(   2x  
x − 2
 
  A(x − 2)  =  2Bx
 
  Ax − 2A  =  2Bx
 
  Ax − 2Bx  =  2A
 
  x(A − 2B)  =  2A
 
  x  =     2A   
A − 2B

Example 4.  Solving for the reciprocal.

  1
3
  +   1
x
 =   1
2

While we could solve this by clearing of fractions, there is the more

   elegant method of solving for  1
x
, and then taking reciprocals.

We have

  1
3
  +   1
x
  =   1
2
 
          This implies
 
1
x
  =   1
2
 −   1
3
 
    =  3 − 2
   6
 
1
x
  =   1
6
 
          Therefore, on taking reciprocals,
 
x   =  6.

For, if two numbers are equal, then their reciprocals are also equal.(Except, if the number is 0.)

In each of the following, solve for x by first solving for its reciprocal.

  Problem 19. 1
r
  +   1
p
 =   1
x
 
  p + r
pr   
 =  1
x
 
  x  =     pr  
p + r
  Problem 20. 1
a
 =  1
x
1
b
 
  1
a
 −  1
b
 =  1
x
 
  ba
ab   
 =  1
x
 
  x  =     ab  
ba
end

Next Lesson:  Word problems

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