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Equivalent fractions:  2nd Level

Back to Level 1

Example 1.   Write the missing numerator.

3
4
 =  ___?__
8x − 12

Answer.   What times 4 produced 8x − 12?  In other words, we must expect that 4, the denominator on the left, will be a factor on the right.

3
4
 =  ___?__
8x − 12
 =  ___?___
4(2x − 3)

The denominator 4 has been multiplied by 2x − 3; therefore, the numerator 3 also must be multiplied by 2x − 3.

3
4
 =  3(2x − 3)
4(2x − 3)
 =   6x − 9 
8x − 12

Example 2.   Write the missing numerator.

__5_
x − 3
 =  ____?____
x2 − 4x + 3

Answer.   What times  x − 3  produced  x2 − 4x + 3?   For, we must expect that x − 3 will be a factor of x2 − 4x + 3.

__5_
x − 3
 =  ____?____
x2 − 4x + 3
 =  _____?_____
(x − 3)(x − 1)

x − 3 has been multiplied by x − 1; therefore, 5 also must be multiplied by x − 1:

__5_
x − 3
= __5(x − 1)__
(x − 3)(x − 1)
= __5x − 5__
x2 − 4x + 3

Problem 1.   Factor the new denominator, and write the missing numerator.

 a)   2
5
= ___?___
15x + 20
 =  ___?___
5(3x + 4)
 =  2(3x + 4)
5(3x + 4)
 =  _6x + 8_
15x + 20
 b)   2
x
= __?__
x2x
= __  ? __
x(x − 1)
= 2(x − 1)
x(x − 1)
= 2x − 2
x2x
 c)   _5_
2x2
= ____?___
4x3 + 6x2
 =  __ __?_ __
2x2(2x + 3)
 =   5(2x + 3)
2x2(2x + 3)
= 10x + 15
4x3 + 6x2
 d)   4
x
= ___?___
8x2 − 3x
= _ __?___
x(8x − 3)
= 4(8x − 3)
x(8x − 3)
= 32x − 12
8x2 − 3x
 e)   _2_
3x4
= ____?___
6x6 + 3x4
 =  ___ _?_ __
3x4(2x2 + 1)
 =  _2(2x2 + 1)_
3x4(2x2 + 1)
 
  = _4x2 + 2_
6x6 + 3x4

The following problems assumes skill with Quadratic Trinomials,  Perfect Square Trinomials, and the Difference of Two Squares.

Problem 2.   Factor the new denominator, and write the missing numerator.

 a)   __2__
x + 4
= ____?____
x2 + 6x + 8
= __ 2(x + 2)__
(x + 4)(x + 2)
= __2x + 4__
x2 + 6x + 8
 b)   x + 3
x − 2
= ____?____
x2 + x − 6
= __ (x + 3)2__
(x − 2)(x + 3)
= x2 + 6x + 9
x2 + x − 6
 c)   __1__
x − 2
= __?__
x2 − 4
= ___ x + 2___
(x − 2)(x + 2)
=  x + 2
x2 − 4
 d)   __x__
x + 2
= ____?____
x2 + 5x + 6
= __x(x + 3)__
(x + 2)(x + 3)
= __x2 + 3x_
x2 + 5x + 6
 e)   x + 3
x − 1
= ____?____
x2 − 4x + 3
= (x + 3)(x − 3)
(x − 1)(x − 3)
= __x2 − 9__
x2 − 4x + 3
 f)   x − 5
x + 5
= ___?__
x2 − 25
= __ (x − 5)2__
(x + 5)(x − 5)
= x2 − 10x + 25
    x2 − 25
 g)   2x − 1
x − 5
= ____?____
2x2 − 9x − 5
= (2x − 1)(2x + 1)
 (x − 5)(2x + 1)
= __4x2 − 1__
2x2 − 9x − 5

Problem 3.   Reduce to lowest terms. See Lesson 19, Problems 10, 12, 14.

 a)   x5 − 1
 x − 1
= x4 + x3 + x2 + x + 1

x − 1 is a factor of x5 − 1.

 b)   x4 − 1
 x − 1
= x3 + x2 + x + 1
 c)   x3 − 1
 x − 1
= x2 + x + 1
 d)   x7 − 1
 x − 1
= x6 + x5 + x4 + x3 + x2 + x + 1

The following problem depends on knowing that  ba  is the negative of ab.  (Lesson 7.)

Problem 4.   Simplify.

 a)   2 − x
x − 2
= −1   b)   x(x − 3)
  3 − x
 =  x(−1) = −x
  c)   __1 − x_
x2(x − 1)
  =    1 
x2
(−1)   =    1 
x2
  d)   3x − 6
 2 − x
  =   3(x − 2)
  2 − x
  =   3(−1)   =   −3
  e)   x2 − 7x + 12
    4 −x
  =   (x − 4)(x − 3)
     4 −x
  =  −1(x − 3)   =  3 − x
  f)   __1 − x__ 
x2 − 2x + 1
  =   ___1 − x__  
(x − 1)(x − 1)
  =    −1  
x − 1
  =        1     
−(x − 1)
  =      1   
1 − x

end

Next Lesson:  Negative exponents

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