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04 - 22- 23

ALGEBRA: Division of Algebraic Expressions
(Sample Problems & Solutions) - Part 007
1
...


� − 1 �2 + 1 ÷

Solution:

Rewrite the expressions as follows,
� − 1 �2 + 1
�−1 � +1 ÷ �−1 �+1 =
�−1 �+1
2

Simplifying the expressions, cancel out � − 1 , since they’re
both present in the numerator and denominator, so we have,
� − � �2 + 1
�� + �
=
�−� �+1
�+�

Therefore, the quotient of the given problem is

2
...


��+�
�+�


...

�� −��

3
...

�+1 2

2

�2 − 1
� �2 − 1
÷
� �+1
�+1 2

Take the reciprocal of the second term, then proceed
to multiplication, so we have,
2

�2 − 1
� �2 − 1
÷
� �+1
�+1 2

2

�2 − 1
�+1 2
×
� �+1
� �2 − 1

Rewrite the expressions for further simplification as
follows,
�2 − 1 ∙ �2 − 1
�+1 ∙ �+1
×
� �+1
� �2 − 1

Then cancel out some expressions, such as,
�2 − 1 ∙ �� − �
�+1 ∙ �+�
×
� �+�
� �� − �
�2 − 1
�+1
×
�
�
�� − � � + �
��

Therefore, the quotient of the given problem is

�4
4
...

�+1 2

�4
�3
÷
�+1
�+1

2

�� −� �+�
��


...


5
...

�2 −1

� + 1 4 �4
�2 − 1

Take the reciprocal of the second term, then proceed
to multiplication, so we have,

2

�3 �2 − 1
�+1

�3 �2 − 1
�+1

2

�3 �2 − 1
�+1

2

÷

� + � � ��
�� − �

�� − �
×
� + � � ��

Rewrite the expressions for further simplification as
follows,
�2 − 1
×
� + 1 4 �4
2
�� �2 − 1
�2 − 1
×
�+1
� + 1 4 �� ∙ �

Then cancel out some expressions, such as,
�� �2 − 1
�+1

2

�2 − 1
�+1

�2 − 1
×
� + 1 4 �� ∙ �

2

�2 − 1
×
� + 1 4�

�� − �

� �+�

�

�

Therefore, the quotient of the given problem is

�� −�

�

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