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Grade 8 – Mathematics
Quarter I
ILLUSTRATING RATIONAL
ALGEBRAIC EXPRESSIONS
1. illustrate rational algebraic
expressions;
2. evaluate rational algebraic
expressions, and
3. find every value of the variables
that makes a rational expression
undefined.
RATIONAL ALGEBRAIC EXPRESSIONS
- A rational expression in one variable is
an expression that can be written in the
form
𝑷
𝑸
, where P and Q are polynomials.
πŸ’
𝒙 βˆ’πŸ
π’™πŸ
+ πŸπ’™ βˆ’ πŸ‘πŸ“
πŸ‘π’™ + πŸ’
πŸ’π’™
π’™πŸ + πŸ—
not UNDEFINED
Remember, Q β‰  0,
EVALUATING ALGEBRAIC EXPRESSIONS
Evaluate the expression
π’™πŸβˆ’ πŸ’
𝒙 βˆ’πŸ
when…
=
π’™πŸβˆ’ πŸ’
𝒙 βˆ’πŸ
x = 0
=
( )πŸβˆ’ πŸ’
( ) βˆ’πŸ
0
0
=
βˆ’ πŸ’
βˆ’πŸ
= 2
=
π’™πŸβˆ’ πŸ’
𝒙 βˆ’πŸ
x = 1
=
( )πŸβˆ’ πŸ’
( ) βˆ’πŸ
=
βˆ’ πŸ‘
βˆ’πŸ
= 3
1
1
=
πŸβˆ’ πŸ’
πŸβˆ’πŸ
EVALUATING ALGEBRAIC EXPRESSIONS
Evaluate the expression
π’™πŸβˆ’πŸ
𝒙 βˆ’πŸ’
when…
=
π’™πŸβˆ’πŸ
𝒙 βˆ’πŸ’
x = 1
=
( )πŸβˆ’πŸ
βˆ’ πŸ’
1
1
=
πŸβˆ’πŸ
πŸβˆ’πŸ’
=
𝟏
πŸ‘
=
π’™πŸβˆ’ πŸ’
𝒙 βˆ’πŸ
x = 2
=
( )πŸβˆ’πŸ
βˆ’πŸ’
=
𝟐
βˆ’πŸ
= -1
2
2
=
πŸ’ βˆ’πŸ
𝟐 βˆ’ πŸ’
=
βˆ’πŸ
βˆ’πŸ‘
EVALUATING ALGEBRAIC EXPRESSIONS
Evaluate the expression
π’‚πŸβˆ’πŸπ’ƒ
𝒂 βˆ’π’ƒ
when…
=
π’‚πŸβˆ’πŸπ’ƒ
𝒂 βˆ’π’ƒ
a = 1 b = 2
=
( )πŸβˆ’πŸ( )
βˆ’( )
1
1
=
πŸβˆ’πŸ’
πŸβˆ’πŸ
= 3
=
βˆ’πŸ‘
βˆ’πŸ
2
2
=
π’‚πŸβˆ’πŸπ’ƒ
𝒂 βˆ’π’ƒ
a = 2 b = 3
=
( )πŸβˆ’πŸ( )
βˆ’( )
2
2
=
πŸ’βˆ’πŸ”
πŸβˆ’πŸ‘
= 2
=
βˆ’πŸ
βˆ’πŸ
3
3
UNDEFINED
To make a rational expression
π’™πŸβˆ’πŸ
𝒙 βˆ’πŸ’
UNDEFINED,
x – 4 = 0
equate the denominator to zero
solve for the variable x = 0 + 4
x = 4
The expression
π’™πŸβˆ’πŸ
𝒙 βˆ’πŸ’
is UNDEFINED if x is replaced by
4, because the denominator would be 0.
UNDEFINED
To make a rational expression
π’™πŸβˆ’πŸ’
𝒙 βˆ’πŸ
UNDEFINED,
x – 2 = 0
equate the denominator to zero
solve for the variable x = 0 + 2
x = 2
The expression
π’™πŸβˆ’πŸ’
𝒙 βˆ’πŸ
is UNDEFINED if x is replaced by
2, because the denominator would be 0.
UNDEFINED
To make a rational expression
π’™πŸβˆ’π’š
𝒙 βˆ’π’š
UNDEFINED,
x – y = 0
equate the denominator to zero
solve for the variable x = 0 + y
x = y
The expression
π’™πŸβˆ’π’š
𝒙 βˆ’π’š
is UNDEFINED if the variables
x and y are replaced by equal values.
Test Yourself.
1. Explain how to evaluate an algebraic
expression.
2. What causes a rational expression to
be undefined?
3. How do you find the values that may
cause a rational expression to be
undefined?
Test Yourself.
I. Evaluate each rational expression if a = 1
and b = 2
1.
π‘Ž
𝑏2 2.
2π‘Žβˆ’π‘
3π‘Žπ‘
II. Find the value of the variable that makes
the expression undefined.
1.
3π‘Ž
4𝑏
2.
2π‘Žβˆ’π‘
π‘βˆ’3

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illustrating RAE.pdf

  • 1. Grade 8 – Mathematics Quarter I ILLUSTRATING RATIONAL ALGEBRAIC EXPRESSIONS
  • 2. 1. illustrate rational algebraic expressions; 2. evaluate rational algebraic expressions, and 3. find every value of the variables that makes a rational expression undefined.
  • 3. RATIONAL ALGEBRAIC EXPRESSIONS - A rational expression in one variable is an expression that can be written in the form 𝑷 𝑸 , where P and Q are polynomials. πŸ’ 𝒙 βˆ’πŸ π’™πŸ + πŸπ’™ βˆ’ πŸ‘πŸ“ πŸ‘π’™ + πŸ’ πŸ’π’™ π’™πŸ + πŸ— not UNDEFINED Remember, Q β‰  0,
  • 4. EVALUATING ALGEBRAIC EXPRESSIONS Evaluate the expression π’™πŸβˆ’ πŸ’ 𝒙 βˆ’πŸ when… = π’™πŸβˆ’ πŸ’ 𝒙 βˆ’πŸ x = 0 = ( )πŸβˆ’ πŸ’ ( ) βˆ’πŸ 0 0 = βˆ’ πŸ’ βˆ’πŸ = 2 = π’™πŸβˆ’ πŸ’ 𝒙 βˆ’πŸ x = 1 = ( )πŸβˆ’ πŸ’ ( ) βˆ’πŸ = βˆ’ πŸ‘ βˆ’πŸ = 3 1 1 = πŸβˆ’ πŸ’ πŸβˆ’πŸ
  • 5. EVALUATING ALGEBRAIC EXPRESSIONS Evaluate the expression π’™πŸβˆ’πŸ 𝒙 βˆ’πŸ’ when… = π’™πŸβˆ’πŸ 𝒙 βˆ’πŸ’ x = 1 = ( )πŸβˆ’πŸ βˆ’ πŸ’ 1 1 = πŸβˆ’πŸ πŸβˆ’πŸ’ = 𝟏 πŸ‘ = π’™πŸβˆ’ πŸ’ 𝒙 βˆ’πŸ x = 2 = ( )πŸβˆ’πŸ βˆ’πŸ’ = 𝟐 βˆ’πŸ = -1 2 2 = πŸ’ βˆ’πŸ 𝟐 βˆ’ πŸ’ = βˆ’πŸ βˆ’πŸ‘
  • 6. EVALUATING ALGEBRAIC EXPRESSIONS Evaluate the expression π’‚πŸβˆ’πŸπ’ƒ 𝒂 βˆ’π’ƒ when… = π’‚πŸβˆ’πŸπ’ƒ 𝒂 βˆ’π’ƒ a = 1 b = 2 = ( )πŸβˆ’πŸ( ) βˆ’( ) 1 1 = πŸβˆ’πŸ’ πŸβˆ’πŸ = 3 = βˆ’πŸ‘ βˆ’πŸ 2 2 = π’‚πŸβˆ’πŸπ’ƒ 𝒂 βˆ’π’ƒ a = 2 b = 3 = ( )πŸβˆ’πŸ( ) βˆ’( ) 2 2 = πŸ’βˆ’πŸ” πŸβˆ’πŸ‘ = 2 = βˆ’πŸ βˆ’πŸ 3 3
  • 7. UNDEFINED To make a rational expression π’™πŸβˆ’πŸ 𝒙 βˆ’πŸ’ UNDEFINED, x – 4 = 0 equate the denominator to zero solve for the variable x = 0 + 4 x = 4 The expression π’™πŸβˆ’πŸ 𝒙 βˆ’πŸ’ is UNDEFINED if x is replaced by 4, because the denominator would be 0.
  • 8. UNDEFINED To make a rational expression π’™πŸβˆ’πŸ’ 𝒙 βˆ’πŸ UNDEFINED, x – 2 = 0 equate the denominator to zero solve for the variable x = 0 + 2 x = 2 The expression π’™πŸβˆ’πŸ’ 𝒙 βˆ’πŸ is UNDEFINED if x is replaced by 2, because the denominator would be 0.
  • 9. UNDEFINED To make a rational expression π’™πŸβˆ’π’š 𝒙 βˆ’π’š UNDEFINED, x – y = 0 equate the denominator to zero solve for the variable x = 0 + y x = y The expression π’™πŸβˆ’π’š 𝒙 βˆ’π’š is UNDEFINED if the variables x and y are replaced by equal values.
  • 10. Test Yourself. 1. Explain how to evaluate an algebraic expression. 2. What causes a rational expression to be undefined? 3. How do you find the values that may cause a rational expression to be undefined?
  • 11. Test Yourself. I. Evaluate each rational expression if a = 1 and b = 2 1. π‘Ž 𝑏2 2. 2π‘Žβˆ’π‘ 3π‘Žπ‘ II. Find the value of the variable that makes the expression undefined. 1. 3π‘Ž 4𝑏 2. 2π‘Žβˆ’π‘ π‘βˆ’3