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Mat he m at i c s 1 1 6

SupplementaryMaterials

Pre-Algebra:S t r e n g t h e n i n g

Ari thmet ic

Unders tanding

The Meaning of the Equal Sign,Order of Operations and

Decompos i t ion of Numbers

Teachers Guide

Pr epa r ed by :

1 ) P et er B a lyt a . .. .. .. .. .. .. .. .. .. .. Ma s s e y Va n i er H igh S c h oo l

2 ) Geor ge Ca lde r .. .. .. .. .. .. .. .. ..Howar d S . B illings High Schoo l

3 ) An d r e D el Ca s t ilh o .. .. .. .. .. . Ce n t en n ia l Re gio n a l High S c h ool

4 ) To n y Ro s cia n o . .. .. .. .. .. .. .. .. .P en fie ld Ac a d e m y

5 ) Colin DSouza .. .. .. .. .. .. .. .. .. .Macdona l d Car t ie r High Schoo l

Resource Per sonnel :

Fran oise Bou lan ger . . .. . .. . MEQ

Louise Gauthier

Carolyn Gould. . . . . . . . . . . . . . . . . .South Shore School Board

Math 116 materials prepared by teachers of the South Shore, Chateauguay Valley (Protestant), District of Bedford,

LEau-Vive and Brossard School Boards with funding provided by the Direction rgionale de la Montrgie (MEQ).

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I n t r oduc t i on

The MEQ Cur r icu lum makes the fo l lowing s t a tement wi th r espect toGenera l Object ive 1 : To h e lp s tu den t s acqu ire sk i lls prerequ is i t e to thes tudy of algebra.

M a t h e m a t i c s 5 6 8 - 1 1 6 t a k e s a t w o - s t e p a p p r o a c h t o p r e p a r i n gs tu den t s for the s tu dy of a lgebra : th e fir s t s t ep in volves fac i lit a t ing th et rans i t ion to a lgebra by ensur ing tha t the necessary ar i thmet ic sk i l l s haveb e e n m a s t e r e d .

I n s t e p o n e t h e p r i m a r y e m p h a s i s i s o n e x p a n d i n g t h e s t u d e n t s u nd ers tan ding of th e equ al s ign . To th is end , cer t a in a spect s of Termina lObjec t ives 2 .1 and 2 .2 not only g ive s tudent s the oppor tuni ty to apply therules for wr i t ing the order of operat ions , but also help them real ize that theequal s ign does not mean do this , or do that , but rather the express ionon the r ight has the same value as the express ion on the lef t and vice versa.Equal i t i es can be crea ted through concre te appl i ca t ion of the commutat ive ,

a s s oc i a t i ve and d i s t r i bu t i ve p r ope r t i e s , wh i ch a r e us ed ex t ens i ve l y i na l gebr a .

At the Secondary 1 level , the ar i thmet ic component i s included not todo ca l cu l a t i ons t o ob t a i n a s i ng l e number ans wer bu t r a t he r t o p r epa r es tu den t s for workin g with a lgebra ic express ion s . For examp le , an an sweri n a l gebr a may be an expr es s i on o f s eve r a l t e r ms r a t he r t han a s i ng l enu mb er . Thu s s t uden t s s hou ld wor k on expr es s ing nu mb er s in d iffe r en t

forms u s ing var ious operat ions . Eg. 30 = 2 2 5 + (12 2)

Cons iderable d i scuss ion may be r equi red to agree on how to wr i t e

such ar rangements so tha t they do ac tua l ly make sense by g iv ing cor rec tar ra n gemen ts . Th is sor t of act ivity n ot only gives the s tu den ts exper ien cein ma n ipu la t i on bu t r evea l s t he s t u den t s g r as p o f t h e o r de r in wh i choper a t i ons a r e done and t he need f o r b r acke t s bo t h us e f u l s k i l l s f o ra l g e b r a !

The ac t iv i t i es presented in these documents a re meant to increaset h e s t u den t s kn owledge of a r i t h m et ic s o t ha t t h is kn owledge can bet rans formed gradual ly so tha t they can bui ld for themselves the not ion of an a lgebra equat ion or express ion .

By s tar t ing f rom ar i thmet ic , the s tudent can cons t ruc t expl i c i t ly a

meaning for the a lgebra ic equat ion , thus a l lowing the s tudent to grasp theconcep t in tu i t ive ly before it becom es form al ized sym bol ica l ly . Su ch ac o n s t r u c t i o n a n c h o r s t h e c o n c e p t o f e q u a t i o n i n a r i t h m e t i c , t h e r e b ymaking the not ion meaningful1

1 Mathematics Teacher Novmber 1980, Constructing Meaning for the Concept of Equation Herscovics,N. and C. Kieran.



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Types of Act iv i t ies

1 . E xp a n d in g t h e Mea n in g of t h e E qu a l S ign

(1 ) Decom p os it ion of Nu m b er s

In doing the ac t iv i t i es on pages 1 16 , s tudent s move beyondt he no t i on t ha t t he equa l s i gn means f i nd t he ans wer t om a n i p u l a t i n g n u m b e r s a n d o p e r a t i o n s t o d e r i v e v a r i o u sequivalent express ions .

(2 ) Id en tit ies

T h e n o t i o n o f e q u a l i t y i s e x p a n d e d t o i n c l u d e m u l t i p l eoperat ions on both s ides of the equal s ign.

(3 ) Is ola tin g a Term

It i s di f f icul t for s tudents to be able to i solate a var iable in anequat ion when they have had no pr ior exper ience in doing th i sw i t h n u m b e r s .



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1

Decompos i t ion of NumbersToday is the 1 0 t h day o f t he m ont h .

1 . (a ) Exp res s th is nu m ber in 5 d ifferen t wa ys .

E xa m ple : 1 0 = 8 + 2

i. 1 0 = 2 5 ii. 10 = 7 + 3 iii. 1 0 = 2 0 2

iv. 10 = 1 1 0 v. 1 0 = 1 2 2

(b ) E xp r e s s th is n u m b e r (in 5 d iffe r en t w a ys ) u s in g 1 a d d it io n a n d1 s ub t r ac t i on .

E xa m p le : 1 0 = 7 + 5 2

i. 1 0 = 12 + 2 4 ii. 1 0 = 9 + 6 5 iii. 10 = 7 + 4 1

iv. 1 0 = 6 + 7 3 v. 10 = 5 + 11 6

(c ) E xp r es s t h is n u m b e r (5 d iffe re n t wa ys ) u s in g 1 a d d it ion a n d 1mul t ip l i ca t ion .

E xa m ple : 10 = 6 + 2 2

i. 1 0 = 4 + 3 2 ii. 10 = 0 + 5 2 iii. 1 0 = 2 + 2 4iv. etc. v. e tc.

(d ) E xp r e s s th i s n u m b e r (5 d iffe r en t w a ys ) u s in g 1 d ivis io n a n d 1s u b t r a c t i o n .

E xa m ple : 1 0 = 3 0 2 5

i. 1 0 = 22 2 1 ii. 1 0 = 3 6 3 2 iii. 10 = 24 2 2

iv. etc. v. e tc.

(e ) E xp r es s t h is n u m b e r (5 d iffe re n t wa ys ) u s i n g a n y 3 op e ra t io n s .

E xa m ple : 1 2 2 + 7 3 = 10

i. 1 0 = 4 2 + 5 3 ii. 10 = 3 0 6 + 7 2 iii. et c.

iv. etc. v. e tc.

(f) E xp r es s t h is n u m b e r (5 d iffe re n t wa ys ) u s i n g a ll 4 o p er a t io n s .

E xa m ple : (3 5 1) 2 + 3 = 10

i. 4 7 14 1 + 9 = 10 ii. e t c . iii. etc.

iv. etc. v. etc.

. .. / 2



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2

Decompos i t ion of NumbersToday is the 1 0 t h day o f t he m ont h .

(g ) E xp r es s t h is n u m b e r (in 5 d iffe re n t wa ys ) u s in g 2 o p er a t ion sand 1 se t of parenthes i s .

E xa m p le : 1 0 = (2 + 3 ) 2

i. 1 0 = (1 0 5 ) 2 ii. 1 0 = (4 + 6 ) 1 iii. et c.

iv. etc. v. e tc.

(h ) E xp r es s t h is n u m b e r (in 5 d i ffe re n t wa ys ) u s i n g a t le a s t on ee x p o n e n t .

E xa m p le : 1 0 = 3 2 + 1

i. 1 0 = 2 3 + 2 ii. 1 0 = 4 2 6 iii. et c.

iv. etc. v. e tc.

(i) E xp r es s t h is n u m b e r (5 d iffe re n t wa ys ) u s i n g a t le a s t 2exponen t s and o t he r ope r a t i ons .

E xa m p le : 3 2 + 1 2 = 104 2 2 2 = 1 0

i. 2 3 + 2 1 = 10 ii. 3 2 + 3 0 = 10 iii. et c.

iv. etc. v. e tc.

(j) E xp r es s t h is n u m b e r (5 d iffe re n t wa ys ) u s i n g a t le a s t 1 d ec im a l.

E xa m ple : 1 0 = 4 .5 0.9 + (3 + 2)

i. 1 0 = 5 .1 + 4 .9 ii. 1 0 = 6 .6 1.1 + 4

iii. et c. iv. etc.

v. etc.

(k ) E xp r es s t h is n u m b e r (5 d iffe re n t wa ys ) u s i n g:

th e 4 opera t ion s (+, , , ) a t lea s t 1 expon en t a t lea s t 1 d ecim a l a t lea s t 1 set of p a ren th es is

E xa m p le : 1 0 = (3 + 2 1 ) 23 + (0.6 10) + 3

i. a n swers will va ry ii. _________________________

iii. _________________________ iv. _________________________

v. _________________________

. .. / 3



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3

(Group work.)

2 . Given th e n u m b er s 3 , 4 an d 5 . Us in g a n y com b in a t ion of t h e

opera t ion s + , , , and any exponen t i a t i on , f o r m anexpress ion whose answer i s :

(answers wil l vary)

(a ) 60 = 5 4 3

(b ) 8 = 4 2 3 5

(c ) 14 4 = 5

3

+ 4

2

+3

(d ) 48 = 4 2 3 5 0

3 . Given th e n u m b er s 6 , 7 an d 8 . Us in g a n y com b in a t ion of t h e

opera t ion s + , , , and exponent ia t ion , form an express ionwhose answer i s :

(Parentheses are permi t t ed as wel l . )

(a ) 2 = (8 + 6 ) 7

(b ) 10 4 = (7 + 6 ) 8

(c ) 44 1 = (8 + 7 + 6 )2

(d ) 5 = 8 - 7 - 6

(e ) 2 .5 = (8 + 7 ) 6

. .. / 4



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4

Decomposit ion of Negat ive Numbers

4 . E xp r es s 8 in 5 d i ffe re n t wa ys u s in g on ly 1 a d d it ion a n d 1s u b t r a c t i o n .

i. 8 = 3 + 4 9 ii. 8 = 4 + 5 9 iii. e t c .

iv. e t c . v. et c.

5 . E xp r es s 1 0 in 5 d iffe re n t wa ys u s in g on ly 1 a d d it ion a n d 1mul t ip l i ca t ion .

i. 1 0 = 2 + 4 2 ii. 1 0 = (3 )2 + 19 iii. 1 0 = 3 3 19

iv. e t c . v. e t c .

6 . E xp r es s 1 0 in 5 d i ffe re n t wa ys u s i n g o n ly 1 d ivis ion a n d 1s u b t r a c t i o n .

i. 10 2 5 = 10 ii. 2 7 3 1 = 1 0 iii. e t c .


7 . E xp r es s 8 in 5 d iffe re n t wa ys u s in g on ly 2 op e ra t ion s a n d 1 s e t of p a r e n t h e s i s .

i. (3 + 5 ) 4 = 8 ii. (2 1 4 ) 2 = 8 iii. e t c .


8 . E xp r es s 1 3 in 5 d iffe re n t wa ys u s in g a ll 4 o pe ra t ion s .

i. 5 2 + 10 5 1 = 13 ii. (6 3 + 10 5) 1 = 13iii. e t c . iv. et c .

v. e t c .

. .. / 5Math 116 materials prepared by teachers of the South Shore, Chateauguay Valley (Protestant), District of Bedford,


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1 4 . Ma k e ch a n ge fo r $ 1 in 5 d iffe re n t wa ys .

(a ) $ 1 = 4 0.25 (b ) $ 1 = 2 .10 + 0 .50 + 0 .25 + 0 .05

(c ) $ 1 = 2 0 .2 5 + 0 .5 0 (d ) $ 1 = 4 0.10 + 7 0.05 + 0 .25

(e ) $ 1 = 7 0.10 + 0 .25 +0.05

1 5 . Ma k e c h a n ge fo r $ 0 . 9 0 u s in g on l y n i ck e ls a n d d im e s in 5 d i ffe r en tways .

(a ) $ 0 .90 = 6 0.05 + 6 .10 (b ) $ 0 .9 0 = 5 0.10 + 8 0 . 0 5

(c ) $ 0.9 0 = 7 0.10 + 4 0 .0 5 (d ) $ 0.90 = 8 0.10 + 2 0 . 0 5

(e ) $ 0.9 0 = 1 0.10 + 16 0 . 0 5

1 6 . J o h n h a s 1 6 co in s in q u a r t er s , n ic k els a n d p en n ie s . If t h e t ot a l va lu e

is $2.28 , h ow ma n y coin s of each does he h ave? 8 Q 5 N 3 P

1 7 . F in d t h e le a s t n u m b e r o f c oin s n e e d ed to t ot a l $ 1 . 8 7 . (Us e o n ly

qu ar ters , dimes , n ickels an d pen n ies . ) 7 Q 1D 2 P

1 8 . G ive n t h e p e r im e t e r s fin d t h e m is s in g s id e s fo r e a c h figu r e . (Th e

len gths ar e al l whole n u m bers .) (An swers wil l var y.)

(a ) P = 1 7 cm a = 3 , b = 4 (b ) P = 2 6 cm b = 5 , a = 3 or a = 5 , b = 4

a

b

a

b

(c ) P = 20 cm a = 2 , b = 5 , c = 3 (d ) P = 27 cm a = 6 , b = 3 , c = 2

a

b

c order can vary

5b

a

d

order can vary. .. / 8



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19 . Ther e a r e fou r pos s ib le s cor ing p lays in an NFL foot ba ll gam e .

Touchdown.. . . . . . . . . . . . . . . . . . . . .6 points

Point After Touchdown...... 1 pointField Goal..... ............. ......3 points

Safety. . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 points

(a ) Lis t fou r w a ys t h a t a te a m cou l d s c or e 1 1 p oin t s .I I I I I I IV1 TD (6) 1 TD (6) 3 FG (9 ) 4 S (8)1 Pa TD (1 ) 1 FG (3 ) 1 S (2) 1 FG (3)

2 S (4) 1 S (2 )

(b ) It is no t pos s ib le t o ob t a in a fina l s co re o f 1 poin t . Ar e t he r ean y other fin al scores which are im poss ible? 1

(c ) If t h e 2 - p oin t s a fe ty d id n o t e xis t , lis t 5 s c or e s t h a t w ou l d n o tbe poss ible. 1 , 2 , 4 , 5 , 8

(d ) If t he 3 - po in t fie ld goa l d id n o t ex is t , lis t 4 s co r es t h a t wou ldn ot be poss ible. 1 , 3 , 5

(e ) Lis t 1 2 s cor es t h a t c a n n o t b e m a d e if a t e a m d oe s n t s cor e asa fety or a field goal . 1, 2, 3, 4, 5, 8, 9, 10 , 11, 15 , 16, 17

20 . The fo llowing a r e ways of ob t a in i ng poin t s i n a ba s ke t ba ll gam e .

Free Throw... ................ 1 point

Regular Shot................ 2 points3-Point Shot.... ............. 3 points

(a ) Lis t 5 wa ys a t ea m c ou ld s cor e 2 1 p oin t s .

7 3, 10 2 + 1, 21 1 , 9 2 + 3, (1 11) + (5 2) an swers willva r y

(b ) If t he fr ee t h r ow d id no t exis t , wou ld an y s co res be impos s i b leto obta in?

(c ) A t eams fina l s cor e was 25 po in t s . They ha d fou r 3 -po in t s ho t s .Given tha t both f r ee throws and regular shot s were t aken , whatare the poss ib le combinat ions needed in order to obta in the 25po i n t s ?

(5 1) + (4 2), (5 2) + (3 1), (11 1) + (1 2 )

(1 1) + (6 2)), (3 2) + (7 1), (9 1) + (2 2 ). .. / 9



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2 1 . Mike and Daves Day at the Golf Course

1st Hole

Par 5

2nd Hole

Par 3

3rd Hole

Par 4

B i r dy . .. . . 1

Bo g ey . . . . +1

Eag l e . . . . . 2

Double

Bo g ey . . . . +2

P a r . . . . . . . . . 0

means 1 stroke less than par

means 1 stroke more than par

means 2 strokes less than par

means 2 strokes over par

Legend

Note : The player with the lowest score wins .

(a ) If o n t h e fir s t h o le yo u s c or e d a d ou b le b oge y, h o w m a n yst rokes did it tak e you to s ink the ba ll? 7

(b ) At wha t ho le wou ld you be a t if it t ook you fou r s t r okes t o s ink the ba ll an d you ended u p with a bogey? 2 n d

(c ) O n th e fir s t h o le D a ve h a d a n e a gle . O n th e s a m e h o le Mik e

h ad a bi rdy. Wh o h ad th e bet ter score? Dave

(d ) D a ve s cor ed a n e a gle on t h e fir s t h ole , a p a r o n t h e s ec on d a n da bi rdy on th e thi rd hole. Wh at was h is fin al score? 3 + 3 + 3 = 9

(e ) Aft er a r ou n d of t h r e e h o le s , wh o h a d t h e b e tt er s c or e c a r d ?

Dave

1st Hole.....Birdy 4

2nd Hole.... Birdy 2

3rd Hole.....Birdy 3 9

Mike

1st Hole.....P ar 5

2nd Hole.... Double Bogey 5

3rd Hole.....B ir dy 3 13Dave

(f) At t he end o f t he gam e Dave ob t a ined t he fo llowing r es u lt s :

1st Hole.....Pa r 52nd Hole.... Birdy 2

3rd Hole.....Bir dy 3 10

Wh at poss ib le combina t ions m ight a ppea r on Mikes score cardif h e won th e gam e? Eagle on all 3 holes. Birdy, Birdy, Eagle.Birdy on al l 3 h oles. Bird y, Ea gle, Ea gle.

. . ./ 1 0



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2 2 . P la c e t h e n u m b e rs 1 , 4 , 6 , 7 , 8 , 9 a n d 1 0 in t h e cir cle s to m a k e e a chs ide of the pentagon add to 19 .

5

2

3

8

1

1 0

7

9

4

6

. . ./ 1 1



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Create an ident i ty where both s ides of the = s ign are equal to the givenn u m b e r .

Exam ple : The nu mb er is 10 . Use 1 addi t ion on 1 s ide an d 1 su bt ra c t ion

on t he o t he r .

8 + 2 = 12 2

14 4 = 6 + 4

17 7 = 1 + 9

2 3 . E xp r es s 2 0 w it h a n id e n t it y u s in g 1 m u lt ip lic a t ion o n o n e s id e a n d 1divis ion on the other s ide of the equal s ign.

a ) __________________________________________________________________

b ) __________________________________________________________________

c ) __________________________________________________________________

2 4 . E xp r es s 1 5 wit h a n id e n t it y u s i n g 1 a d d it io n a n d 1 s u b t r a ct ion o nboth s ides of the equal s ign.

a ) __________________________________________________________________

b ) __________________________________________________________________

c ) __________________________________________________________________

2 5 . E xp r e s s 2 4 w it h a n id e n t it y u s i n g 1 m u lt ip lic a t io n a n d 1 d ivis io n o nboth s ides of the equal s ign.

a ) __________________________________________________________________

b ) __________________________________________________________________

c ) __________________________________________________________________

.. ./ 1 2



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2 6 ) E xp r es s 3 6 wit h a n id e n t it y u s i n g 1 a d d it io n a n d 1 m u lt ip lic a t ion onboth s ides of th e equ al s ign. (Remem ber th e order of operat ions .)

a ) 4 + 8 4 = 6 5 + 6

b ) __________________________________________________________________

c ) __________________________________________________________________

2 7 ) E xp r es s 5 w it h a n id e n t it y u s i n g 1 d ivis io n a n d 1 s u b t r a ct io n o n b o ths ides of th e equ al s ign. (Remem ber th e order of operat ions .)

a )

2 0

2

5 8 12 4 =

b ) __________________________________________________________________

c ) __________________________________________________________________

2 8 ) E xp r es s 8 w it h a n id e n t it y u s i n g 3 d iffe r e n t o p er a t io n s o n b o t h s id e sof th e equ al s ign . (Remem ber the order of operat ions .)

a ) 3 4 20 4 = 35 7 + 3 1

b ) __________________________________________________________________

c ) __________________________________________________________________

2 9 ) E xp r es s 6 w it h a n id e n t it y u s i n g 3 d iffe r e n t o p er a t io n s o n b o t h s id e sof th e equ al s ign . (Remem ber the order of operat ions .)

a )

2 0

4

1 8 7 9 3 6 2

+ = + +

b ) __________________________________________________________________

c ) __________________________________________________________________

.. ./ 1 3



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3 0 . Re a rr a n ge th e give n n u m b er s to m a k e a tr u e s ta t em e n t .

(a ) 12 = 3 + 2 + 7

eg. i. 7 = 1 2 3 2 ii. 2 = 12 3 7

iii. 3 = 1 2 2 7

(b ) 15 = 5 + 3 + 7

i. 5 = 1 5 3 7 ii. 3 = 15 5 7

iii. 7 = 1 5 5 3

(c ) 12 3 = 5 0 + 25 + 3 0 + 1 8

i. 50 = 12 3 25 3 0 18 ii. 3 0 = 1 23 5 0 25 18

iii. 25 = 12 3 50 3 0 18 iv. 1 8 = 1 23 5 0 25 3 0

(d ) 12 0 = 5 + 15 +11 0 10

i. 5 = 12 0 1 5 11 0 + 1 0 ii. 15 = 120 5 1 10 + 1 0

iii. 11 0 = 1 20 5 1 5 + 1 0 iv. 1 0 = 5 + 1 5 + 1 10 1 20

(e ) 93 = 9 0 + 13 1 0

i. 90 = 9 3 + 10 1 3 ii. 1 0 = 90 + 1 3 93

iii. 1 3 = 9 3 + 1 0 9 0

(f) 20 = 14 + 1 2 6

i. 14 = 20 1 2 + 6 ii. 1 2 = 2 0 14 + 6

iii. 6 = 14 + 12 2 0 iv. 6 = 20 14 1 2

.. ./ 1 4



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(g) 17 = 7 + 11 1

i. 7 = 1 7 11 + 1 ii. 11 = 17 7 + 1

iii. 1 = 7 + 1 1 17 iv. 1 = 17 7 11

(h ) 18 = 6 3

i. 6 = 1 8 3 ii. 3 = 18 6

(i) 18 = 2 9

i. 2 = 1 8 9 ii. 9 = 18 2

(j) 20 = 4 0 1

2

i. 40 = 2 01

2 ii .

1

2= 20 40

(k ) 10 = 30 1

3

i. 30 = 1 01

3 ii .

1

3= 10 30

(l) 49 = 2 24 + 1

i. 1 = 49 (2 2 4 ) ii. 2 = (4 9 1 ) 2 4

iii. 2 4 = (4 9 1 ) 2

.. ./ 1 5



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(m ) 25 = 3 6 + 7

i. 7 = 2 5 (3 .6 ) ii. 3 = (2 5 7 ) 6

iii. 6 = (2 5 7 ) 3

(n ) 3 0 = (2 ) (5 ) + 2 0

i. 2 = (3 0 - 2 0 ) 5 ii. 5 = (3 0 - 20 ) 2

iii. 2 0 = 3 0 (2 5)

(o) 26 = 2 12 + 5 3

i. 5 = 2 6 + 3 (2 1 2 ) ii. 3 = 2 1 2 + 5 2 6

iii. 2 = (2 6 5 + 3 ) 12 iv. 1 2 = (2 6 5 + 3 ) 2

(p ) 2 2 = (2 ) (1 3) 4

i. 13 = (2 2 + 4 ) 2 ii. 2 = (2 2 + 4 ) 13

iii. 4 = 2 1 3 2 2

(q) 88 = 3 33 11

i. 11 = 3 3 3 8 8 ii. 3 3 = (8 8 + 1 1 ) 3

iii. 3 = (8 8 + 1 1 ) 3 3

(r ) 3 (5 + 2 ) = 2 1

i. 3 = 2 1 (5 + 2 ) ii. (5 + 2 ) = 2 1 3

iii. 5 = 2 1 3 2 iv. 2 = 21 3 5

. . ./ 1 6



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(s ) 6 (4 + 3 ) = 42

i. 6 = 4 2 (4 + 3 ) ii. (4 + 3 ) = 4 2 6

iii. 4 = 4 2 6 3 iv. 3 = (42 6) 4

(t ) 3 (8 2 ) = 18

i. 3 = 18 (8 2 ) ii. (8 2 ) = 18 3

iii. 8 = 1 8 3 + 2 iv. 2 = 8 (1 8 3)

v. 2 = 18 3 8

(u ) 7 (9 3 ) = 42

i. 7 = 42 (9 3 ) ii. (9 3 ) = 42 7

iii. 9 = 4 2 7 + 3 iv. 3 = 9 (4 2 7)

v. 3 = 42 7 9

(v)

5

6=

1

3+

1

2

i.

1

3=

5

6

1

2 ii .

1

2=

5

6

1

3

(w )

6

8=

1

2+

1

4

i.

1

4=

6

8

1

2 ii .

1

2=

6

8

1

4



116pre-algtg

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