Test 13: Algebraic Fractions II

Question 1: If you failed question 1 in the test, then you must complete these 3 extra questions:

(a):  Simplify:		(i):  a  ×  bc			(ii):  4x  ÷  2x 
			      2b    3a			       3      3
  
Clue: Part (i): First, cancel ANY 'top' with ANY 'bottom':
 
                Cancelling the 'b's:
 
			      a  ×  bc 
			      2b    3a
 
                And cancelling the 'a's:
			     ┌———————————————————————— 	Since there is nothing left at the top of this
			     ▼				fraction, we leave a 1 there...
			     1a  ×  bc 
			      2b    3a 
 
                Finally, multiplying the tops and multiplying the bottoms:
 
			      1  ×  c  =   c  .
			      2     3     ...
 
 
     Part (ii): We can change the 'divide' to a 'multiply' if we turn the second fraction upside down:
 
			     4x  ×  3 . 
			      3     2x
  
                Next, cancel ANY 'top' with ANY 'bottom':
 
                Cancelling the 'x's:
 
			      4x ×  3  
			       3    2x
 
                And cancelling the '3's:
			           ┌—————————————————— 	Since there is nothing left at the top of this
			           ▼			fraction, we leave a 1 there... 
			      4x × 13 .
			       31   2x 
 
                And cancelling the '2's...
 
                Finally, multiplying the tops and multiplying the bottoms...
 
 
 
(b): Simplify:		(i):  3b ×  8c			(ii):  5x  ÷  3 
			      4c    6			       4      4
  
Clue: Part (i): First, cancel ANY 'top' with ANY 'bottom':
 
                Cancelling the 'c's:
 
			      3b ×  8c 
			      4c    6
 
                And cancelling down the '4' & '8':
 
			      3b × 28c 
			     14c    6
 
                And cancelling the '3's...
 
                Finally, multiplying the tops and multiplying the bottoms...
 
 
     Part (ii): We can change the 'divide' to a 'multiply' if we turn the second fraction upside down:
 
			     5x  ×  4. 
			      4     3
  
                Next, cancel ANY 'top' with ANY 'bottom':
  
                Cancelling the '4's:
 
			      5x × 14.
			       41   3 
  
                Finally, multiplying the tops and multiplying the bottoms...
 
 
 
(c):  Simplify:		(i):  2c  ×  d			(ii):  1   ÷   3 . 
			      5d     4			       6x     2xy
 
Clue: Sorry, no more help with this part....
 
 

Question 2: If you failed question 2 in the test, then you must complete these 3 extra questions:

(a):  Simplify:		(i):  a  ×  b			(ii):  x  ÷  y 
			      3     2			       4     3
  
Clue: Part (i): We can't cancel anything, so multiplying the tops and multiplying the bottoms:
 
			      a  ×  b  =   ab .
			      3     2     ...
 
 
     Part (ii): We can change the 'divide' to a 'multiply' if we turn the second fraction upside down:
 
			      x  ×  3 . 
			      4     y
  
                We can't cancel anything, so multiplying the tops and multiplying the bottoms...
 
 
  
(b):  Simplify:		(i):  5  ×  2			(ii):  7  ÷  1 
			      bc    c			       xy    x
 
Clue: Sorry, there's no help for this part...
 
 
 
(c):  Simplify:		(i):  1  ×  1			(ii):  3  ÷  1 
			      cd    d			       a     a
 
Clue: Sorry, you have to do this bit un-aided...
 
 

Question 3: If you failed question 3 in the test, then you must complete these 3 extra questions:

(a):  Simplify:		(i):  2  ×   3 . 		(ii):  2x  ÷   6 . 
			      3a    a-2			        5     x-2
  
Clue: Part (i): To start with, it is important that we put in brackets (we can't cancel down a
                fraction if there are and +/- sings sticking out of brackets...)
 
			      2  ×   3
			      3a   (a-2) 
 
                Cancelling the '3's:
 
			      2  ×   31
			     13a   (a-2)
  
                Finally, multiplying the tops and multiplying the bottoms:
 
			      2  ×  1     =   ...  .
			      a    (a-2)     a(a-2)
 
 
     Part (ii): We can change the 'divide' to a 'multiply' if we turn the second fraction upside down:
 
			     2x  ×  (x-2). 
			      5       6
  
                Next, cancel ANY 'top' with ANY 'bottom':
 
                Cancelling down the '2' & '6':
 
			     12x ×  (x-2) 
			       5      63
 
                Finally, multiplying the tops and multiplying the bottoms...
 
 
 
(b):  Simplify:		(i):  b(b+1) ×  b+2		(ii): 2(x+3) ÷ 3(x+3) 
			       b+2       b		        5       10x
 
Clue: Part (i): Firstly, make sure all of the brackets are in place:
 
			      b(b+1) × (b+2) 
			       (b+2)     b
 
                Then cancel down:
 
			      b(b+1) × (b+2) 
			       (b+2)     b1
 
                And again:
 
			       (b+1) × (b+2)1 
			      1(b+2)     1
 
                Finally, multiplying the tops and multiplying the bottoms...
 
 
     Part (ii): We can change the 'divide' to a 'multiply' if we turn the second fraction upside down:
 
			     2(x+3)  ×  10x. 
			       5       3(x+3) 
  
                Next, cancel ANY 'top' with ANY 'bottom':
 
                Cancelling the '(x+3)'s:
 
			     2(x+3)  ×  10x. 
			       5       3(x+3) 
 
                Cancelling down the '10' & '5':
 
                Finally, multiplying the tops and multiplying the bottoms...
 
 
 
(c):  Simplify:		(i):    c    × 6(c+2)		(ii):  x-1 ÷ 2(x-1)
			      3(c+2)     2		        x      5
 
Clue: Sorry - you'll have to do this on your own...
 
 

Question 4: If you failed question 4 in the test, then you must complete these 3 extra questions:

(a):  Simplify:		(i):   a + 4  ×  3a + 6		(ii):    3x    ÷    4
			      2a + 4     a + 4		       3x - 9     2x - 6 
 
Clue: Part (i): I know you are tempted to cancel it down like this:
 
			       a + 4  ×  3a + 6
			      2a + 4     a + 4
 
                But DON'T - IT IS WRONG!
 
                The first step is always to ensure that there are BRACKETS around any sums in the
                fractions - a fraction CANNOT be cancelled down if there are any +/- signs sticking out
                of the brackets:
 
			       a + 4  ×  3a + 6
			      2a + 4     a + 4
 
                So, we have to put in brackets...
                But, in this case, while we are putting in BRACKETS, we might as well factorise:
 
			      (a+4)  ×  3(a+2)
			      2(a+2)     (a+4)
 
                Now we can cancel down (ANY 'top' with ANY 'bottom')
 
			     1(a+4)  ×  3(a+2)
			      2(a+2)    1(a+4)
 
                And again:
 
			     1(a+4)  ×  3(a+2)
			      2(a+2)    1(a+4)
 
                Finally, multiplying the tops and multiplying the bottoms...
 
 
     Part (ii): We can change the 'divide' to a 'multiply' if we turn the second fraction upside down:
 
			        3x    ×   2x - 6. 
			      3x - 9        4 
  
                We can't cancel down until we put in the brackets and anyway, we need to FACTORISE:
 
			        3x    ×   2(x-3). 
			      3(x-3)        4 
 
                Now cancel (ANY top with ANY bottom):
 
			        3x    ×   2(x-3). 
			      3(x-3)        4 
 
                And again:
 
			        3x    ×   2(x-3). 
			      3(x-3)        4 
 
                And again:
 
			        3x    ×  12(x-3). 
			      3(x-3)       24 
 
                Finally, multiplying the tops and multiplying the bottoms...
 
 
 
(b):  Simplify:		(i):    b    ×  2b + 10		(ii):  x - 5  ÷  2x - 10  
			      b + 5        b²		         4x        6x²
 
Clue: Part (i): The first step is always to ensure that there are BRACKETS around any sums in the
                fractions (a fraction CANNOT be cancelled down if there are any +/- signs sticking out
                of the brackets)...
 
                But, in this case, while we are putting in BRACKETS, we might as well factorise:
 
			        b   ×  2(b+5)
			      (b+5)       b²
 
                Now we can cancel down (ANY 'top' with ANY 'bottom')
 
			       1b   ×  2(b+5)
			      (b+5)       b²
 
                And again:
  
                Finally, multiplying the tops and multiplying the bottoms...
 
 
     Part (ii): We can change the 'divide' to a 'multiply' if we turn the second fraction upside down:
 
			      x - 5   ×     6x². 
			        4x        2x - 10  
  
                We can't cancel down until we put in the brackets and anyway, we need to FACTORISE:
 
			       (x-5)  ×     6x². 
			        4x        2(x-5) 
 
                Now cancel (ANY top with ANY bottom)...
  
                And again...
  
                And again...
  
                Finally, multiplying the tops and multiplying the bottoms...
 
 
 
(c):  Simplify:		(i):  5c + 10  ×  3c²		(ii):  3x + 3  ÷  x + 1 
			         2c      c + 2			  6         5
  
Clue: Sorry, no more help here - follow the method from part (a)
 
 

Question 5: If you failed question 5 in the test, then you must complete these 3 extra questions:

(a):  Simplify		(i):  a²+3a  ×   1		(ii):   1   ÷     1
			        2       a+3 		       x+2     x²+3x+2 
 
Clue: Part (i): The first step is always to ensure that there are BRACKETS around any sums in the
                fractions (a fraction CANNOT be cancelled down if there are any +/- signs sticking out
                of the brackets)...
 
                But, in this case, while we are putting in BRACKETS, we might as well factorise:
 
			      a(a+3)  ×    1
			         2       (a+3)
 
                Now we can cancel down (ANY 'top' with ANY 'bottom')
 
			      a(a+3)  ×    1
			         2      1(a+3)
  
                Finally, multiplying the tops and multiplying the bottoms...
 
 
     Part (ii): We can change the 'divide' to a 'multiply' if we turn the second fraction upside down:
 
			       1    ×  x²+3x+2.  
			      x+2         1
  
                We can't cancel down until we put in the brackets and anyway, we need to FACTORISE:
 
			        1   ×  (x+2)(x+1).  
			      (x+2)        1
 
                Now cancel (ANY top with ANY bottom):
 
			        1   ×  (x+2)(x+1).  
			     1(x+2)        1
  
                Finally, multiplying the tops and multiplying the bottoms...
 
 
 
(b):  Simplify:		(i): 2b²+2b  ×  9		(ii):    2    ÷    4
			        3      2b+2		       x²-25     x²-5x
 
Clue: Part (i): The first step is always to ensure that there are BRACKETS around any sums in the
                fractions (a fraction CANNOT be cancelled down if there are any +/- signs sticking out
                of the brackets)...
 
                But, in this case, while we are putting in BRACKETS, we might as well factorise:
 
			      2b(b+1)  ×    9
			         3        2(b+1)
 
                Now we can cancel down (ANY 'top' with ANY 'bottom')
 
			      2b(b+1)  ×    9
			         3        2(b+1)
  
                And again...
 
 			      2b(b+1)  ×    9
			         3       12(b+1)

                And again (the '3' and the '9')...
 
                Finally, multiplying the tops and multiplying the bottoms...
 
 
     Part (ii): We can change the 'divide' to a 'multiply' if we turn the second fraction upside down:
 
			        1    ×  x²-5x.  
			      x²-25       4
  
                We can't cancel down until we put in the brackets and anyway, we need to FACTORISE:
 
			      1      ×  x(x-5).  
			 (x+5)(x-5)       4
 
                Now cancel (ANY top with ANY bottom)...
   
                Finally, multiplying the tops and multiplying the bottoms...
 
 
 
(c):  Simplify:		(i):    4    ×   c-2		(ii):    3   ÷   3
			      c²-2c       8		       x²-4     2x-4 
 
Clue: Sorry, I can't hold your hand for this question...
 
 

 

Note: No answers are given for these extra revision questions. If you are stuck, then make sure you have a serious attempt the question and then stay behind at the end of the lesson and ask me for help...