Balik Aral (Algebra)

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1st Balik-Aral: Academic Review Program 1

CHAPTER I: INTEGER EXPONENTS

For any real number a, with an integer exponent n, the nth power of a is an.

•

If n is a positive integer, then an

= a⋅a⋅a… a (n times).• -an and (-a)n are not always eual.

-!" = -!#!#!#! = -1$

(-!)" = (-!)(-!)(-!)(-!) = 1$

• If a % &, then a& = 1.

• If n ' & and a % &, then a-n =1

n.

*+ F /00+

• an # am = an+m xample2 22⋅2

4=22+4=2

6

x 3 # x -4 = x -" =1

x

4

• (an) m = anm xample2 (53)! = 51&

• (ab)n = anbn xample2 (-!⋅5)" = (-!)"⋅(5)"

x 6 # 6 = ( x )6

• If a % &, thena

n

am

= an-m xample278

74

= 67-" = 6"

y3

y−10

= y3−(−10)= y

13

• If b % &, then ( a

b )n

=a

n

bn

xample2 ( x

y )2

= x

2

y2

5

3

153=(

5

15 )3

= 1

27

8/+

•

36 x6 y

3 z

7

32 x

5 y

9 z

7 =

22⋅3

2

32

x6−5

y3−9

z7−7

¿4 x1 y−6

z0

¿ 4 x

y6

(−5 x3 y

−4)4( x5 y

2)−5=(−5)4 x3(4 )

y−4 (4 )

x5(−5)

y2(−5)

¿625 x12

y−16

x−25

y−10

¿ 625

x13

y26

•

x−1+ y

−1

( x+ y )−1=

1

x+ 1

y

1

x+ y






o multiply polynomials, apply the *#st"#!+t#, law o- '+lt#&l#$at#on o,"

a**#t#on and the laws o- %&onnts.

8/+

• y2−3 (4−2 y

2 )= y2−12+6 y

2

¿7 y2−12

• (2 x−3 y ) (5 x2− xy+4 y

2)=5 x2− xy+4 y

2 (1)

2 x−3 y (!)

10 x3−2 x

2 y+8 x y

2

−15 x2 y+3 xy

2−12 y3

10 x3−17 x

2 y+11 x y

2−12 y3

I. +/:I /9>:+

• /rodu;t of a sum and diAeren;e

( x+ y ) ( x− y )= x

2

− y

2

xample2

2a

(¿¿ 2)2−(b)2=4a4−b

2

(2a2+b )( 2a

2−b )=¿• +uare of a binomial

x

(¿¿2±2 xy+ y2)

( x ± y )2=¿

xample2

4 a

(¿¿ 2)[¿¿2+2(4 a2)(b)+b2]=16a

4+8a2

b+b2

¿(4 a

2+b )2=¿• /rodu;t of binomials I

( x+m) ( x+n )= x2+(m+n) x+mn

xample2

(a)[¿¿2+ (1+3 )a+ (1+3 )]=a

2+4a+3

( a+1 ) ( a+3 )=¿• /rodu;t of binomials II

(mx+n ) (ox+ p )=mox2+(mp+on ) x+np

xample2

(3⋅2)a

[¿¿2+ (3 ⋅1+5 ⋅2 ) a+(5 ⋅1 )]=6 a2+13a+5(3a+5 ) (2a+1 )=¿

• :ube of a binomial

x

(¿¿3±3 x2 y+3 xy

2± y

3)( x ± y )3=¿

8ultiply (1) by ea;h term in

dd the liCe terms



1st Balik-Aral: Academic Review Program "

xample2¿=a

3−6 a2+12a−8

(a−2 )3=¿

DO NOT DISTRIBTE EXPONENTS O/ER A SM: ( x+ y)n≠ x

n+ yn

IDI+I0 F /@08I+

• 8onomial ivisor E use x ± y

z =

x

z ±

y

z and laws o- %&onnts.

xample26 x

3 y

2+12 x2 y

3

3 x2

y2

=6 x

3 y

2

3 x2 y

2+12 x

2 y

3

3 x2 y

2 =2 x+4 y

• ong ivision

xample2 (4 x3−12 x+20) ÷ (2 x+3)=2 x

2−3 x−1+ 23

2 x+3

F:9II00a$to"#ng is the pro;ess of <nding the fa;tors of a given polynomial.

• :ommon monomial fa;tor

(ax+ay )=a ( x+ y )

xamples2 6 x3

y2

−3 x2

y+9 xy=3 xy (2 x2

y− x+3)

+1 x ¿¿

x ¿ ¿ ( x+1 )( x2−2−4 )

• iAeren;e of two suares

( x2− y2 )=( x+ y)( x− y)

xamples2 (4 a2−b

2 )=(2a)2−(b)2

¿(2a+b)(2a−b) (16m

4−n12 )=(4m

2)2−(n6)2

¿ (4 m2+n6 )( 4m2−n6 ) ¿(4m

2+n6)(2m+n

3)(2m−n3)

• /erfe;t suare trinomial

x

(¿¿2±2 xy+ y2)¿ ( x ± y )2

¿ xample2 (4 a

2+4 ab+b2 )=(2a)2+2 (2a ) ( b )+(b)2




¿(2a+b)2

• Fa;toring by grouping E it may be possible to group terms in su;h a way that ea;h

GgroupH has a ;ommon fa;tor.

xamples2 ax+bx−ay−by=(a+b ) x−(a+b ) y

¿ (a+b )( x+ y) x y

3+2 y2− xy−2= y

2 ( xy+2 )−( xy+2 )

y

(¿¿ 2−1)( xy+2)¿¿

¿( y+1)( y−1)( xy+2)

9:I++ II

. +implify the A. polynomials by

performing the indi;ated operation(s).

1. 4 x

2

−5 x+6 x

2

−2 x!.

1

2a3

b2(2a

2+5ab−b2)

5. 3 x2n( xn+1−4 x

n+5)". (w+6)(w−6)

3.

t

t (¿¿ 2+9)(¿¿ 2−5)¿

¿

$. 35u

2v3−20u

3v2

−5u2

v

6. 16 t

4n−64 t 6n

2 t 2n

8. a6−b6

a−b?. Fa;tor the A. polynomials ;ompletely.

1. a4b

3−a3b

4+a2

b6

!. 4 s2−25r

2

5. x4n− y

6n

". 16 x2−8 x+1

3. 10a3+25a−4 a

2−10

$. (2 x−3 y)2−16

1. abx+acx−bcy−aby+bcx−acy

CHAPTER III: LINEAR AND 2ADRATIC E2ATIONS

n alg!"a#$ 3+at#on is a statement that two algebrai; expressions are eual.

xamples are2

2 x+1= x−72 x

e

5=

4

x+1 sol+t#on or "oot of the euation is a single value of the variable that maCes the

euation true. If there are many solutions to a given euation, then they are ;alled

the sol+t#on st.

I09 >I0+L#na" 3+at#ons are euations whi;h only involve polynomials of degree 1.

8/+

• 5 x−5=2 x+7

• 5 x−2 x=7+5• 3 x=12

• x=4

•



1st Balik-Aral: Academic Review Program $

•

•

3

x+4=

2

3 x−2

•

( x+4 ) (3 x−2 ) 3

x+4=

2

3 x−2( x+4 ) (3 x−2 )

• 3 (3 x−2 )=2 ( x+4 )• 9 x−6=2 x+8• x=2

•

• >9I: >I0+

• 2+a*"at#$ 3+at#ons are euations whi;h only involve polynomials of degree

! and whi;h ;an be written in the form2 a x2+bx+c=0 . o get its roots, several

methods ;an be used.

• Fa;toring

a. +olve for x in x2+5 x=−6 .

• x2+5 x+6=0 :he;Cing2

• ( x+3 ) ( x+2 )=0 If

x=−3: (−3 )2+5 (−3 )=−6

• x+3=0 ; x+2=0 If x=−2: (−2 )2+5 (−2 )=−6

• x=−3∨ x=−2 ∴ +olution +et2

{−3,−2 }b. +olve for x in 6 x

2+19 x−7=0 .

• 6 x2+19 x−7=0 :he;Cing2

• (2 x+7 ) (3 x−1 )=0 If

x=−72

:6

(−72 )

2

+19(−72 )−

7=0

• 2 x+7=0 ;3 x−1=0 If

x=1

3:6( 13 )

2

+19 (13 )−7=0

• x=−7

2∨ x=

1

3 ∴ +olution +et2

{−7

2,1

3}

•

•

uadrati; Formula E used when the polynomial is not -a$to"a!l. For apolynomial in the form ax

2+bx+c where a , b , c

• x=−b ±√ b2−4ac

2a

•

a. +olve for x in x2+5 x=−6 .




• x2+5 x+6=0

• a=1,b=5,c=6

•

x=

−5±√ 52−4 (1)(6)

2(1)

• x=−3∨ x=−2 (+ame as what we got from fa;toriJation)

b. +olve for x in x2+6 x−8=0 .

• a=1,b=6,c=−8 :he;Cing2

• x=−6±√ 6

2−4(1)(−8)2(1)

If x=−3±√ 17:

• x=−3± √ 17

(−3±√ 17)2+6 (−3±√ 17)−8=0

• ∴ +olution +et2

{−3±√ 17}•

• +@+8 F >I0+

• syst' o- 3+at#ons is ;omposed of two or more euations in several

variables. Kiven variables x∧ y , it has the form

• a1 x+b

1 y=c

1

• a2 x+b

2 y=c

2

•

• wherein a1

, a2

, b1, b

2, c

1∧c

2 are ;onstant real numbers. In <nding the solution

set of the system, two methods ;an be used.• +ubstitution method E transform one of the euations in terms of one variable and

substitute this to the other euation.

a. etermine the solution set of the euations 2 x+ y=3∧5 x+3 y=10.

•

• +olve the <rst euation in terms of y.

• y=3−2 x

• +ubstitute this value to the se;ond euation.

• 5 x+3 (3−2 x)=10

• − x+9=10

• x=−1

• +ubstitute this value of x in the <rst euation to get y .

• y=3−2(−1)• y=5

• ∴ +olution +et2 {−1,5 }

• limination method E multiply both euations by a nonJero real number in su;h a

way that when these euations are added, one of the variables will be

eliminated.

a. etermine the solution set of the euations 2 x+ y=3∧5 x+3 y=10.

L




•

• o eliminate one of the variables, say y , we must multiply the <rst

euation by -5 and the se;ond euation by 1.

• −3 ⋅ (2 x+ y=3 ) →−6 x−3 y=−9

• 1⋅ (5 x+3 y=10 ) →5 x+3 y=10

• dd the resulting two euations together.

• −6 x−3 y=−9

• 5 x+3 y=10

• − x=1

• x=−1

• +ubstitute this value of x in any of the euations, say the se;ond

one, to get y .

• 5 (−1 )+3 y=10

• y=5

• ∴ +olution +et2 {−1,5 }

b. etermine the solution set of the euations3 x−2 y=13∧4 x+7 y=−2.

•

• o eliminate one of the variables, say x , we must multiply the <rst

euation by " and the se;ond euation by -5.

• 4 ⋅ (3 x−2 y=13 )→12 x−8 y=52

• −3 ⋅ (4 x+7 y=−2 )→−12 x−21 y=6

• dd the resulting two euations together.

• 12 x−8 y=52

• −12 x−21 y=6

• −29 y=58

• y=−2

• +ubstitute this value of y in any of the euations, say the se;ond

one, to get x .

• 4 x+7 (−2 )=−2

• x=3

• ∴ +olution +et2 {3,−2}

•

• *9 /9?8+

• Keneral problems

a. If a re;tangle has a length that is 5 ;m less than four times its width and

its perimeter is 14 ;m, what are its dimensionsM

•

• et w 2 width of the re;tangle

• l=4w−3 2 length of the re;tangle

• +in;e the perimeter of a re;tangle is given as / = !(width) N !(length)

and substituting the variables to the euation,

• 19=2 (w )+2(4 w−3)• 19=10w−6




• w=2.5 ;m

• l=7 ;m

•

b. dmission ti;Cets to a motion pi;ture theater were pri;ed at O" for adults

and O5 for students. If 71& ti;Cets were sold and the total re;eipts wereO!,735, how many of ea;h type of ti;Cets were soldM

•

• et x 2 number of ti;Cets sold to adults

• y : number of ti;Cets sold to students

• +in;e there are two variables, we would need two euations. If the

total number of ti;Cets sold is 71&, writing it mathemati;ally gives

• x+ y=810

• lso, sin;e the total revenue for the ti;Cets sold was O!,735, the

se;ond euation is given as

• 4 x+3 y=2,853

• >sing elimination method to remove x , we multiply the <rst

euation by -" and the se;ond one by 1.

• −4 x−4 y=−3,240

• 4 x+3 y=2,853

• y=387

• +ubstituting this value of y to any euation, say the <rst one, x is eual

to

• x+387=810

• x=423

• ∴ 576 student ti;Cets and "!5 adult ti;Cets were sold.

•

• 8ixture problems E problems whi;h involve ;ombining solutions of diAerent

;on;entrations to obtain solution of a parti;ular ;on;entration.a. etermine how many liters of a 6P and 1!P a;id solutions should be

mixed to obtain $ liters of a 1&P a;id solution.

•

• *e are looCing for the number of liters of ea;h solution should be used

given that the <nal solution is $ liters.

• et x 2 number of liters of the 6P a;id solution

• 6− x 2 number of liters of the 1!P a;id solution

• +in;e we Cnow that a;id ;on;. x number of liters of solution = number

of liters of a;id, we ;onstru;t a table for the given solutions and the <nalmixture.

• • ;id ;on;. (P) • iters of

solution

• iters of a;id

• 6P a;id

solQn

• 6 • x • 0.07 x

• 1!P

a;id solQn

• 1! • 6− x • 0.12(6− x)



1st Balik-Aral: Academic Review Program 1&

• Final

mixture

• 1& • $ • &.1&($)

•

• dding the values found in the last ;olumn gives us the euation

• 0.07 x+0.12 (6− x )=0.10 (6) • −0.05 x=−0.12

• x=2.4 and 6−(2.4) = 5.$

• ∴ !." of 6P a;id solution and 5.$ of the 1!P a;id solution must

be used.

•

• >niformEmotion problems - problems whi;h involve using the formula

• r ⋅t =d• where r , t ∧d are the uniform rate, time of travel and distan;e travelled,

respe;tively. In applying the formula, be $ons#stnt with the units of

measurement used.

a. ne runner tooC 5 min. "3 se;. to ;omplete a ra;e while another runnerreuired " min. to run the same ra;e. he rate of the faster runner is &."

mRse;. than the rate of the slower one. Find their rates.

•

• et r 2 rate of the slower runner (in mRse;)

• r+0.4: rate of the faster runner (in mRse;)

• +in;e the runners ran the same ra;e, they travelled eual distan;es.

pplying the formula above

• • 9ate (in

mRse;)

• ime (in se;) • istan;e (in

m)

• Faster

runner

• r+0.4 • 225 •

225(r+0.4 )• +lower

runner

• r • 240 • 240r

•

• uating the values found in the last ;olumn, we have

• 225 (r+0.4 )=240 r

• 15 r=90

• r=6 and r+0.4=6.4

• ∴ he rate of the slower runner is $ mRse; while the faster one is

$." mRse;.

•

• 9:I++ III

• . Find the solution set of the

A. euations.

1. 2 (t −5 )=3−(4+t )!. 3 (4 y+9 )=7 (2−5 y )−2 y

5. x2=8 x−15

". 8w2+10w−3=0

3. 49 x2+84 x+36=0

4. 5 y2−4 y−2=0

•

• ?. Find the solution set of the

A. systems of euations.

1. 5 x+3 y=3



(bs;issa)

(rdinate)

uadrant I(N, N)

uadrant II(-, N)

uadrant ID(N, -)

adrant III(-, -)


• x+9 y=2

!. 3 x+4 y−4=0

• 6 x−2 y−3=0

5. 6

x +

3

y=−2

4

x+7

y=−2

• :. Find the solution(s) to the A. problems.1. he smaller of two numbers is 4 less than the larger, and their sum is 56. Find the

numbers.!. he pro<ts of a business are shared among three sto;Cholders. he <rst one

re;eives twi;e as mu;h as the se;ond and the se;ond re;eives three times as

mu;h as the third. If the pro<ts for last year were O!$,"&&, how mu;h did ea;h

sto;Cholder re;eiveM5. group of women de;ided to ;ontribute eual amounts toward obtaining a

speaCer for a booC review. If there were 1& more women, ea;h would have paid

O! less. Sowever, if there were 3 less women, ea;h would have paid O! more.

Sow many women were in the group and how mu;h was the speaCer paidM". ea worth O".1& per pound is to be mixed with tea worth O".4& per pound. Sow

many pounds of ea;h should be used to obtain !3 lb of a blend worth O"."& per

poundM6. A "a*#ato" $onta#ns 76 3+a"ts o- a wat" an* ant#-"9 sol+t#on o-

w;#$; 4<= >!y ,ol+'? #s ant#-"9. How '+$; o- t;#s sol+t#on s;o+l*

! *"a#n* an* "&la$* w#t; wat" -o" t; nw sol+t#on to ! @<=

ant#-"9$. wo airplanes, travelling in opposite dire;tions, leave an airport at the same time.

If one plane averages "7& miRhr and the other averages 3!& miRhr, how long will

it taCe before they are !&&& mi apartM1. On ;o+" a-t" a t"+$) ;as l-t on an o,"n#g;t ;a+l a 'ssng" on a

'oto"$y$l la,s -"o' t; sa' sta"t#ng &o#nt to o,"ta) t; t"+$). I-

t; 'ssng" t"a,ls at an a,"ag "at o- 46 '#;" an* o,"ta)s t;

t"+$) #n @ ;". w;at #s t; a,"ag "at o- t; t"+$)

•

•

•

•

CHAPTER I/: 7DIMENSIONAL COORDINATE SYSTEM: LINES•

• n o"*"* &a#" ( x , y ) (or $oo"*#nats) is any

;ombination of two real numbers where order is signi<;ant.

Kiven an ordered pair P in a :artesian ;oordinate plane, the

<rst number x represents the %$oo"*#nat or the a!s$#ssawhile the se;ond number y represents the y$oo"*#nat or

the o"*#nat.

•

•

•

• he x and y axes are ;alled the $oo"*#nat a%s

wherein their interse;tion is ;alled the o"#g#n O whi;h





(&, "&&)

(1&&, $&&)

(!&&, 7&&)

(5&&, 1&&&)

("&&, 1!&&)


they are solved when y is set to & while y−¿ inter;epts have the form (0, y)sin;e they are solved when x is set to &.

• y=3 x−2

•

x

•

•

y • •

•

• iCe stated above, this euation has an

unlimited number of solutions. ?ut for linear

euations liCe this, only two diAerent ;oordinates

(but getting the x−¿ y−¿ inter;eptRs would

be easiest) to ;onstru;t its graph.

•

• y= x2−3

• x •

•

•

y

•

•

•

• For uadrati; euations, a minimum of

three diAerent ;oordinates is needed to

;onstru;t its graph.

•

• I0+

•

• Kiven an euation y=2 x+400 , it ;an be seen that for

ea;h 1&&-unit in;rease in x results to !&&-in;rease unit in

y or in lowest terms, for ea;h 1-unit in;rease in x, y

in;reases by ! units. his ;onstant ratio between the rate

of ;hange y of with respe;t to x is ;alled the slo& of the

line. etting m euals the slope, its formula is given as

• m= y

2− y

1

x2− x1

, where x2

≠ x1

.

•

• aCe note that verti;al lines have +n*n* slo& sin;e

all the ;oordinates in the line have the same x-

;oordinates while horiJontal lines have < slo& sin;e all the ;oordinates in the

line have the same y-;oordinates.

•

• Find the slope of the line through (!, 1) and ?(",6).






•

• From the line l , we ;an dedu;e that m=2 .

+in;e they are parallel, the line we are looCing for

has also the same slope. >sing the point-slope

form,• y−(1)=2( x−1)• y=2 x−1

•

• wo distin;t non-verti;al lines, l1∧l

2 , are said to be perpendi;ular to one

another if ml1ml2=−1.

• Find the euation of the line that passes through (1, 1)

and is perpendi;ular to l : y=2 x+3 .

•

• From the line l , we ;an dedu;e that m=2 .

Kiven that they are perpendi;ular, the line we are

looCing for has the slope m=−1

2. >sing the point-

slope form,

• y−(1)=−1

2( x−1)

• y=−1

2 x+

3

2

•

• 9:I++ ID

• . raw sCet;h of the graph of the A. euations2

1. 2 x+5 y+10=0 !. y=4− x

2

5.

". y2−9 x=0 ?. Find an euation of the line satisfying the given ;onditions.

1. he slope is " and through the points (!, -5).

!. hrough the point (-!, 5) and parallel to the line 2 x− y−2=0 .

5. hrough the point (!, ") and perpendi;ular to the line whose euation is

x−5 y+10=0 .

@. T;"o+g; t; o"#g#n an* !#s$t#ng t; angl !twn t; a%s #n t; "st

an* t;#"* 3+a*"ants.

3. :. +olve the A. problems.

1. produ;erQs total ;osts ;onsist of a manufa;turing ;ost of O!& per unit and a

<xed daily overhead. Kiven that the total ;ost of produ;ing !&& units in one day is

O",3&&, determine the <xed daily overhead.7. 0#n* t; ,al+ o- ) s+$; t;at t; l#ns w;os 3+at#ons a"

3 x+6 !y=7∧9!x+8 y=15are∥.

4.

1.

8. A. RE0ERENCES

• eithold, . (!&&!). !ollege Algebra and "rigonometr. +ingapore2 /earson

du;ation sia /te td.

=2 x−1

=2 x−1=2 x−1l : =2 x+3

l : =2 x+3






!. {( 23 , 1

2 )}5.

{(

−15

4,−

15

2

)}!6.

!7. :.

1. 23,14

!. # 15840,#7920,#2640

5. 20 ; #120

". 155

8lbo$ #4.10 tea

3. 81

3%uarts

$. 2hr .

6. 52 mihr

!4.

5&. xer;ises ID

51. .

1.

5!.

55.

5".

53.

5$.

!.

51.

57.

54."&.

"1.5.

"!.

"5.

"".

"3.

"$. ?.

1. 4 x− y−11=0

!. 2 x− y+7=0

5. 5 x+ y−14=0

". x− y=0

"6.

"7. :.

1. # 500

!.

5. ± 2

3

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Balik Aral (Algebra)