algebraic fractions - unidad educativa stella...

$: Algebraic fractions - Unidad Educativa Stella Marissmaris.edu.ec/wp-content/uploads/2015/08/IB-MYP-5-Plus.295-308.pdf · ALGEBRAIC FRACTIONS (Chapter 12) 297 Simplify: a 2x2 4x b$
Algebraic fractions

12Chapter

Contents: A Simplifying algebraic fractions

B Multiplying and dividing algebraicfractions

C Adding and subtracting algebraicfractions

D More complicated fractions

IB10 plus 2nd edmagentacyan yellow black

0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\295IB_10P-2_12.CDR Monday, 4 February 2008 9:15:33 AM PETERDELL

296 ALGEBRAIC FRACTIONS (Chapter 12)

Fractions which involve unknowns are called algebraic fractions.

Algebraic fractions occur in many areas of mathematics. We have already seen them in

problems involving similar triangles.

CANCELLATION

We have observed previously that number fractions can be simplified by cancelling common

factors.

For example, 1228 = 4£3

4£7 = 37 where the common factor 4 was cancelled.

The same principle can be applied to algebraic fractions:

If the numerator and denominator of an algebraic fraction are both written in factored form

and common factors are found, we can simplify by cancelling the common factors.

For example,4ab

2a=

2 £ 2 £ a£ b

2 £ affully factorisedg

=2b

1fafter cancellationg

= 2b

For algebraic fractions, check both numerator

and denominator to see if they can be expressed

as the product of factors, then look for common

factors which can be cancelled.

ILLEGAL CANCELLATION

Take care with fractions such asa + 3

3:

The expression in the numerator, a + 3, cannot be written as the product of factors other

than 1 £ (a + 3): a and 3 are terms of the expression, not factors.

A typical error in illegal cancellation is:a + 3

3=

a + 1

1= a + 1.

You can check that this cancellation of terms is incorrect by substituting a value for a.

For example, if a = 3, LHS =a + 3

3=

3 + 3

3= 2,

whereas RHS = a + 1 = 4.

SIMPLIFYING ALGEBRAIC FRACTIONSA

When cancelling inalgebraic fractions,only factors can be

cancelled, not terms.

1 1

11

1

1

1

1


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\296IB_10P-2_12.CDR Tuesday, 5 February 2008 9:35:35 AM PETERDELL

ALGEBRAIC FRACTIONS (Chapter 12) 297

Simplify: a2x2

4xb

6xy

3x3c

x + y

x

a2x2

4x

=2 £ x£ x

4 £ x

=x

2

b6xy

3x3

=6 £ x£ y

3 £ x£ x£ x

=2y

x2

cx + y

xcannot be simplified

as x+ y is a sum,

not a product.

Simplify: a(x + 3)(x¡ 2)

4(x + 3)b

2(x + 3)2

x + 3

a(x + 3)(x¡ 2)

4(x + 3)

=x¡ 2

4

b2(x + 3)2

x + 3

=2(x + 3)(x + 3)

(x + 3)

= 2(x + 3)

EXERCISE 12A.1

1 Simplify if possible:

a6a

3b

10b

5c

3

6xd

8t

te

t + 2

t

f8a2

4ag

2b

4b2h

2x2

x2i

4a

12a3j

4x2

8x

kt2 + 8

tl

a2b

ab2m

a + b

a¡ cn

15x2y3

3xy4o

8abc2

4bc

p(2a)2

aq

(2a)2

4a2r

(3a2)2

3as

(3a2)2

9a2t

(3a2)2

18a3

2 Split the following expressions into two parts and simplify if possible.

For example,x + 9

x=

x

x+

9

x= 1 +

9

x:

ax + 3

3b

4a + 1

2c

a + b

cd

a + 2b

b

e2a + 4

2f

3a + 6b

3g

4m + 8n

4h

4m + 8n

2m

3 Which of the expressions in 2 produced a simplified answer and which did not? Explain

why this is so.

Example 2 Self Tutor


In these examplesis the

common factor.( + 3)x

1

2

1

1

2 1

11

1

11

1


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



4 Simplify:

a3(x + 2)

3b

4(x¡ 1)

2c

7(b + 2)

14

d2(n + 5)

12e

10

5(x + 2)f

15

5(3 ¡ a)

g6(x + 2)

(x + 2)h

x¡ 4

2(x¡ 4)i

2(x + 2)

x(x + 2)

jx(x¡ 5)2

3(x¡ 5)k

(x + 2)(x + 3)

2(x + 2)2l

(x + 2)(x + 5)

5(x + 5)

m(x + 2)(x¡ 1)

(x¡ 1)(x + 3)n

(x + 5)(2x¡ 1)

3(2x¡ 1)o

(x + 6)2

3(x + 6)

px2(x + 2)

x(x + 2)(x¡ 1)q

(x + 2)2(x + 1)

4(x + 2)r

(x + 2)2(x¡ 1)2

(x¡ 1)2x2

FACTORISATION AND SIMPLIFICATION

It is often necessary to factorise either the numerator or denominator before simplification

can be done.

Simplify: a4a + 8

4b

3

3a¡ 6b

a4a + 8

4

=4(a + 2)

4

=(a + 2)

1

= a + 2

b3

3a¡ 6b

=3

3(a¡ 2b)

=1

a¡ 2b

Simplify: aab¡ ac

b¡ cb

2x2 ¡ 4x

4x¡ 8

aab¡ ac

b¡ c

=a(b¡ c)

b¡ c

=a

1

= a

b2x2 ¡ 4x

4x¡ 8

=2x(x¡ 2)

4(x¡ 2)

=x

2



1

1

1

1

1

1

1

1

1

2


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Simplify: a3a¡ 3b

b¡ ab

ab2 ¡ ab

1 ¡ b

a3a¡ 3b

b¡ a

=3(a¡ b)

¡1(a¡ b)

= ¡3

bab2 ¡ ab

1 ¡ b

=ab(b¡ 1)

¡1(b¡ 1)

= ¡ab

Simplify:x2 ¡ x¡ 6

x2 ¡ 4x + 3

x2 ¡ x¡ 6

x2 ¡ 4x + 3

=(x + 2)(x¡ 3)

(x¡ 1)(x¡ 3)

=x + 2

x¡ 1

EXERCISE 12A.2

1 Simplify by cancelling common factors:

a6

2(x + 2)b

2x + 6

2c

3x + 12

3d

3x + 6

6

e5x + 20

10f

3a + 12

9g

xy + xz

xh

xy + xz

z + y

iab + bc

ab¡ bcj

3x¡ 12

6(x¡ 4)2k

(x + 3)2

6x + 18l

2(x¡ y)2

6(x¡ y)

2 Simplify:

a4x + 8

2x + 4b

mx + nx

2xc

mx + nx

m + nd

x + y

mx + my

e2x + 4

2f

x2 + 2x

xg

x2 + 2x

x + 2h

x

bx + cx

i3x2 + 6x

x + 2j

2x2 + 6x

2xk

2x2 + 6x

x + 3l

ax2 + bx

ax + b



1

Don’t forget to expandyour factorisations to

check them.

1 1

1

1

1

b¡ a = ¡1(a¡ b)is a useful rule for converting into .It can sometimes allow us to cancel common factors.

b a a b� � � �¡ ¡


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



3 Simplify, if possible:

a2a¡ 2b

b¡ ab

3a¡ 3b

6b¡ 6ac

a¡ b

b¡ ad

a + b

a¡ b

ex¡ 2y

4y ¡ 2xf

3m¡ 6n

2n¡mg

3x¡ 3

x¡ x2h

xy2 ¡ xy

3 ¡ 3y

i6x2 ¡ 3x

1 ¡ 2xj

4x + 6

4k

12x¡ 6

2x¡ x2l

x2 ¡ 4

x¡ 2

mx2 ¡ 4

x + 2n

x2 ¡ 4

2 ¡ xo

x + 3

x2 ¡ 9p

m2 ¡ n2

m + n

qm2 ¡ n2

n¡mr

3x + 6

4 ¡ x2s

16 ¡ x2

x2 ¡ 4xt

x2 ¡ 4

4 ¡ x2

u5x2 ¡ 5y2

10xy ¡ 10y2v

2d2 ¡ 2a2

a2 ¡ adw

4x2 ¡ 8x

x2 ¡ 4x

3x2 ¡ 6x

4 ¡ x2

4 Simplify:

ax2 ¡ x

x2 ¡ 1b

x2 + 2x + 1

x2 + 3x + 2c

x2 ¡ 4x + 4

2x2 ¡ 4x

dx2 + 4x + 3

x2 + 5x + 4e

x2 ¡ 4

x2 ¡ 3x¡ 10f

x2 + 7x + 12

2x2 + 6x

gx2 + 4x¡ 5

2x2 + 6x¡ 20h

x2 + 6x + 9

x2 + 3xi

2x2 ¡ 7x¡ 4

x2 ¡ 2x¡ 8

j3x2 + 5x¡ 2

3x2 ¡ 4x + 1k

2x2 ¡ 3x¡ 20

x2 ¡ x¡ 12l

8x2 + 14x + 3

2x2 ¡ x¡ 6

The rules for multiplying and dividing algebraic fractions are identical to those used with

numerical fractions. These are:

To multiply two or more fractions, we multiply

the numerators to form the new numerator, and

we multiply the denominators to form the new

denominator.

To divide by a fraction we multiply by its reciprocal.

a

b£ c

d=

a£ c

b £ d=

ac

bd

a

b¥ c

d=

a

b£ d

c=

ad

bc

MULTIPLICATION

Step 1: Multiply numerators and multiply denominators.

Step 2: Separate the factors.

Step 3: Cancel any common factors.

Step 4: Write in simplest form.

MULTIPLYING AND DIVIDINGALGEBRAIC FRACTIONS

B


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



For example,n2

3£ 6

n=

n2 £ 6

3 £ n

=n£ n£ 2 £ 3

3 £ n

=2n

1

= 2n

Simplify: a4

d£ d

8b

5

g£ g3

a4

d£ d

8=

4 £ d

d£ 8

= 12

b5

g£ g3 =

5

g£ g3

1

=5 £ g £ g £ g

g

=5g2

1

= 5g2

DIVISIONFor example,m

2¥ n

6=

m

2£ 6

n

=m£ 6

2 £ n

=m£ 6

2 £ n

=3m

n

Step 1: To divide by a fraction, multiply by

its reciprocal.

Step 2: Multiply numerators and multiply

denominators.

Step 3: Cancel any common factors.

Step 4: Write in simplest form.

Simplify: a6

x¥ 2

x2b

8

p¥ 2

a6

x¥ 2

x2=

6

x£ x2

2

=3 £ 2 £ x£ x

x£ 2

= 3x

b8

p¥ 2 =

8

p£ 1

2

=8 £ 1

p£ 2

=4

p



1

2

1

1

1

1

1 1

11

1

4

1 1

11

fStep 1g

fSteps 2 and 3g

fStep 4g

1

3

fStep 1g

fStep 2g

fStep 3g

fStep 4g


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



EXERCISE 12B

1 Simplify:

aa

2£ b

3b

x

4£ 2

xc

c

4£ 2

cd

a

2£ a

3

ea

b£ x

yf

x

y£ y

xg

x

3£ x h

x

4£ 8

y

in

2£ 1

n2j

6

p£ p

2k

m

x£ x

nl x£ 2

x

m5

t£ t2 n

µx

y

¶2

o

µ4

d

¶2

pa

b£ b

c£ c

a

2 Simplify:

ax

3¥ x

2b

3

y¥ 6

yc 3 ¥ 1

xd 6 ¥ 2

y

e3

p¥ 1

pf

c

n¥ n g d¥ 5

dh x¥ x

3

i 1 ¥ a

bj

3

d¥ 2 k

4

x¥ x2

2l

4

x¥ 8

x2

m a¥ a2

3n

x

y¥ x2

yo

5

a¥ a

2p

a2

5¥ a

3

The rules for addition and subtraction of algebraic fractions are identical to those used with

numerical fractions.

To add two or more fractions we obtain the

lowest common denominator and then add the

resulting numerators.

a

c+

b

c=

a + b

c

To subtract two or more fractions we obtain

the lowest common denominator and then

subtract the resulting numerators.

a

c¡ d

c=

a ¡ d

c

To find the lowest common denominator, we look for the lowest common multiple of the

denominators.

For example, when adding 34 + 2

3 , the lowest common denominator is 12,

when adding 23 + 1

6 , the lowest common denominator is 6.

The same method is used when there are variables in the denominator.

ADDING AND SUBTRACTINGALGEBRAIC FRACTIONS

C


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



For example, when adding4

x+

5

y, the lowest common denominator is xy,

when adding4

x+

3

2x, the lowest common denominator is 2x,

when adding1

3a+

2

5b, the lowest common denominator is 15ab.

To findx

2+

3x

5we find the LCD and then proceed in the same manner as for ordinary

fractions.

The LCM of 2 and 5 is 10, so the LCD is 10.

)x

2+

3x

5=

x£ 5

2 £ 5+

3x£ 2

5 £ 2

=5x

10+

6x

10

=11x

10

Simplify: ax

3+

5x

6b

3b

4¡ 2b

3

ax

+5x

6

=7x

6

b3b ¡ 2b

=12

Simplify: a2

a+

3

cb

7

x¡ 5

2x

a2

+3

=2c

ac+

3a

ac

=2c + 3a

ac

b7 ¡ 5

2

=14

2x¡ 5

2x

=9

2x



3 6

=x£ 2

3 £ 2+

5x

6fLCD = 6g

=2x

6+

5x

6

=2x + 5x

4 3

=3b£ 3

4 £ 3¡ 2b£ 4

3 £ 4fLCD = 12g

=9b

12¡ 8b

12

b

a c

=2 £ c

a£ c+

3 £ a

c£ afLCD = acg

x x

=7 £ 2

x£ 2¡ 5

2xfLCD = 2xg


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Simplify: ab

3+ 1 b

a

4¡ a

=3

=4¡

4

=¡3a

4or ¡ 3a

4

EXERCISE 12C

1 Simplify by writing as a single fraction:

ax

2+

x

5b

x

3¡ x

6c

x

4+

3x

5d

x

2¡ x

5

e2t

3¡ 7t

12f

11n

21¡ n

7g

a

2+

a

3h

a

2+

b

4

in

3+

2n

15j

5g

6¡ g

3k

4s

5¡ 2s

3l a¡ 3a

5

mx

3+

x

2+

x

6n

y

2+

y

4¡ y

3o

z

4+

z

6¡ z

3p 2q ¡ q

3+

2q

7

2 Simplify:

a3

a+

2

bb

4

a+

3

dc

5

a¡ 3

bd

3a

m¡ 2a

m

ea

y+

b

3yf

4

a¡ 5

2ag

3

a¡ 2

abh

c

a+

b

d

i4

b+

a

bj

2

a¡ c

dk

5

x+

x

3l

p

6¡ 2

d

mm

3+

n

mn

2m

p¡ m

no

3b

5+

b

4p

5b

3¡ 3b

5

3 Simplify:

ax

3+ 2 b

m

2¡ 1 c

a

3+ a d

b

5¡ 2

ex

6¡ 3 f 3 +

x

4g 5 ¡ x

6h 2 +

3

x

i 6 ¡ 3

xj b +

3

bk

5

x+ x l

y

6¡ 2y

4 Simplify:

ax

3+

3x

5b

3x

5¡ 2x

7c

5

a+

1

2ad

6

y¡ 3

4y

e3

b+

4

cf

5

4a¡ 6

bg

x

10+ 3 h 4 ¡ x

3


ab

3+ 1 =

b

3+

3

3b + 3

ba

4¡ a =

a

4¡ a£ 4

1 £ 4a 4a


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Addition and subtraction of more complicated algebraic fractions can be made relatively

For example:x + 2

3+

5 ¡ 2x

2=

2

2

µx + 2

3

¶+

3

3

µ5 ¡ 2x

2

¶fachieves LCD of 6g

=2(x + 2)

6+

3(5 ¡ 2x)

6fsimplify each fractiong

We can then write the expression as a single fraction and simplify the numerator.

Write as a single fraction: ax

12+

x¡ 1

4b

x¡ 1

3¡ x + 2

7

ax

12+

x¡ 1

4

=x

12+

3

3

µx¡ 1

4

¶=

x + 3(x¡ 1)

12

=x + 3x¡ 3

12

=4x¡ 3

12

bx¡ 1

3¡ x + 2

7

=7

7

µx¡ 1

3

¶¡ 3

3

µx + 2

7

¶=

7(x¡ 1)

21¡ 3(x + 2)

21

=7(x¡ 1) ¡ 3(x + 2)

21

=7x¡ 7 ¡ 3x¡ 6

21

=4x¡ 13

21

Write as a single fraction: a2

x+

1

x + 2b

5

x + 2¡ 1

x¡ 1

a2

x+

1

x + 2

=2

x

µx + 2

x + 2

¶+

µ1

x + 2

¶x

x

fLCD = x(x + 2)g

=2(x + 2) + x

x(x + 2)

=2x + 4 + x

x(x + 2)

=3x + 4

x(x + 2)

b5

x + 2¡ 1

x¡ 1

=

µ5

x + 2

¶µx¡ 1

x¡ 1

¶¡µ

1

x¡ 1

¶µx + 2

x + 2

¶fLCD = (x + 2)(x¡ 1)g

=5(x¡ 1) ¡ 1(x + 2)

(x + 2)(x¡ 1)

=5x¡ 5 ¡ x¡ 2

(x + 2)(x¡ 1)

=4x¡ 7

(x + 2)(x¡ 1)

MORE COMPLICATED FRACTIONSD



straightforward if we adopt a consistent approach.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



EXERCISE 12D.1

1 Write as a single fraction:

ax

4+

x¡ 1

5b

2x + 5

3+

x

6c

x

7+

2x¡ 1

6

da + b

2+

b¡ a

3e

x¡ 1

4+

2x¡ 1

5f

x + 1

2+

2 ¡ x

7

gx

5¡ x¡ 3

6h

x¡ 1

6¡ x

7i

x

10¡ 2x¡ 1

5

jx

6¡ 1 ¡ x

12k

x¡ 1

3¡ x¡ 2

5l

2x + 1

3¡ 1 ¡ 3x

8


a2

x + 1+

3

x¡ 2b

5

x + 1+

7

x + 2c

5

x¡ 1¡ 4

x + 2

d2

x + 2¡ 4

2x + 1e

3

x¡ 1+

4

x + 4f

7

1 ¡ x¡ 8

x + 2

g1

x + 1+

3

xh

5

x¡ 2

x + 3i

x

x + 2+

3

x¡ 4

j 2 +4

x¡ 3k

3x

x + 2¡ 1 l

x

x + 3+

x¡ 1

x + 2

m2

x(x + 1)+

1

x + 1n

1

x¡ 1¡ 1

x+

1

x + 1

o2

x + 1¡ 1

x¡ 1+

3

x + 2p

x

x¡ 1¡ 1

x+

x

x + 1

PROPERTIES OF ALGEBRAIC FRACTIONS

Writing expressions as a single fraction can help us to find when the expression is zero.

However, we need to be careful when we cancel common factors, as we can sometimes lose

values when an expression is undefined.

Write as a single fraction: a3

(x + 2)(x¡ 1)+

x

x¡ 1b

¡3

(x + 2)(x¡ 1)+

x

x¡ 1

a3

(x + 2)(x¡ 1)+

x

x¡ 1

=3

(x + 2)(x¡ 1)+

µx

x¡ 1

¶µx + 2

x + 2

¶fLCD = (x + 2)(x¡ 1)g

=3 + x(x + 2)

(x + 2)(x¡ 1)

=x2 + 2x + 3

(x + 2)(x¡ 1)which we cannot simplify further.



0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


REVIEW SET 12A


b¡3

(x + 2)(x¡ 1)+

x

x¡ 1

=¡3

(x + 2)(x¡ 1)+

µx

x¡ 1

¶µx + 2

x + 2

¶fLCD = (x + 2)(x¡ 1)g

=¡3 + x(x + 2)

(x + 2)(x¡ 1)

=x2 + 2x¡ 3

(x + 2)(x¡ 1)

=(x + 3)(x¡ 1)

(x + 2)(x¡ 1)

=x + 3

x + 2

EXERCISE 12D.2


a2

x(x + 1)+

1

x + 1b

2

x(x + 1)+

x

x + 1c

2x

x¡ 3+

4

(x + 2)(x¡ 3)

d2x

x¡ 3¡ 30

(x + 2)(x¡ 3)e

3

(x¡ 2)(x + 3)+

x

x + 3f

x

x + 3¡ 15

(x¡ 2)(x + 3)

g2x

x + 4¡ 40

(x¡ 1)(x + 4)h

x + 5

x¡ 2¡ 63

(x¡ 2)(x + 7)

2 a Write2

(x + 2)(x¡ 3)+

2x

x¡ 3as a single fraction.

b Hence, find x when2

(x + 2)(x¡ 3)+

2x

x¡ 3is i undefined ii zero.

3 Simplify: a

xx¡2 ¡ 3

x¡ 3b

3xx+4 ¡ 1

x¡ 2c

x2

x+2 ¡ 1

x + 1

d

x2

2¡x+ 9

x¡ 3e

1x2 ¡ 1

4

x¡ 2f

x¡3x2 ¡ 1

16

x¡ 4

1 Simplify:

a6x2

2xb 6 £ n

2c

x

2¥ 3 d

8x

(2x)2


a8

4(c + 3)b

3x + 8

4c

4x + 8

4d

x(x + 1)

3(x + 1)(x + 2)


a2x

3+

3x

5b

2x

3£ 3x

5c

2x

3¥ 3x

5d

2x

3¡ 3x

5

The expression is zerowhen . The

expression is undefinedwhen and alsowhen We can

see this from theoriginal expression.

x

x

x :

� �

� ��

= 3

= 2=1

¡

¡

1

1


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


REVIEW SET 12B


4 Simplify by factorisation:

a4x + 8

x + 2b

5 ¡ 10x

2x¡ 1c

4x2 + 6x

2x + 3


ax + 3

4+

2x¡ 2

3b

x¡ 1

7¡ 1 ¡ 2x

2c

2

x + 2+

1

x


a8 ¡ 2x

x2 ¡ 16b

x2 + 7x + 12

x2 + 4xc

2x2 ¡ 3x¡ 2

3x2 ¡ 4x¡ 4

7 a Write3x

x¡ 4¡ 60

(x + 1)(x¡ 4)as a single fraction.

b Hence, find x when3x

x¡ 4¡ 60

(x + 1)(x¡ 4)is i undefined ii zero.

1 Simplify:

a4a

6ab

x

3£ 6 c 3 ¥ 1

nd

12x2

6x


a3x + 15

5b

3x + 15

3c

2(a + 4)

(a + 4)2d

abc

2ac(b¡ a)


a3x

4+ 2x b

3x

4¡ 2x c

3x

4£ 2x d

3x

4¥ 2x


a3 ¡ x

x¡ 3b

5x + 10

2x + 4c

3x2 ¡ 9x

ax¡ 3a


ax

5+

2x¡ 1

3b

x

6¡ 1 + 2x

2c

3

2x¡ 1

x + 2


a2x2 ¡ 8

x + 2b

x2 ¡ 5x¡ 14

x2 ¡ 4c

3x2 ¡ 5x¡ 2

4x2 ¡ 7x¡ 2

7 a Write2x

x + 5¡ 70

(x + 5)(x¡ 2)as a single fraction.

b Hence, find x when2x

x + 5¡ 70

(x + 5)(x¡ 2)is i zero ii undefined.

8 Simplify: a

3x+7x¡1 ¡ 13

x¡ 2b

x2

3¡x¡ 1

2

x¡ 1.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


algebraic fractions - unidad educativa stella...

Documents