Olympic College Topic 13 Rational Expressions
Topic 13 Rational Expressions
Introduction:
Definition:
A rational expression is one of the form
the top term x2 – 9 is called
the numerator and bottom term x2 – 5x + 6 is called the denominator.
Rational expressions can occur when you divide two polynomials or when you use a
formula that contains fractions. Topic 13 deals with how to deal with such expressions
and in particular we will look at the following processes.







Finding Undefined Values of Rational Expressions.
How to Simplify a Rational Expression.
How to Multiply a Rational Expression.
How to Divide a Rational Expression.
How to Add and Subtract a Rational Expression.
How to Solve a Rational Equations.
Applications involving Rational Expressions.
1. Finding Undefined Values of Rational Expressions.
We can usually evaluate a rational expression for any given value of the variable but
there are sometimes specific values of the variable which generate undefined values for
the expression if you use them. This situation will only occur when you use a value of the
variable and it generates a denominator of zero.
Example 1: List all numbers for which
Solution:
is undefined.
The expression will be undefined when the denominator (bottom) is equal
to zero.
x–6
=
0
Solution is x = 6
So when x = 6 the rational expression
will have an undefined value.
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Olympic College Topic 13 Rational Expressions
Example 2: List all numbers for which
Solution:
is undefined.
The expression will be undefined when the denominator (bottom) is equal
to zero.
x2 – 8x – 20 =
(x + 2)(x – 10) =
0
0
Solution is x = – 2 and x = 10
So when x = – 2 or x = 10 the rational expression
undefined value.
Example 3: List all numbers for which
Solution:
will have an
is undefined.
The expression will be undefined when the denominator (bottom) is equal
to zero.
y2 + 8y + 16 =
(y + 4)(y + 4) =
0
0
Solution is y = – 4
So when y = – 4 the rational expression
value.
Example 4: List all numbers for which
Solution:
will have an undefined
is undefined.
The expression will be undefined when the denominator (bottom) is equal
to zero.
y2 – 9 =
(y + 3)(y – 3)=
0
0
Solution is y = – 3 and y = 3
So when y = – 3 and y = 3 the rational expression
will have an
undefined value.
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Olympic College Topic 13 Rational Expressions
Example 5: List all numbers for which
Solution:
is undefined.
The expression will be undefined when the denominator (bottom) is equal
to zero.
y2 + 9 =
0
Since y2 and + 9 are both positive we can never get zero as an answer, so
we can conclude that there are no values of y which will generate an
undefined value for the rational expression
Example 6: List all numbers for which
Solution:
.
is undefined.
The expression will be undefined when the denominator (bottom) is equal
to zero.
2x5 – 50x3 =
2x3 (x2 – 25) =
2x3 (x + 5)(x – 5) =
0
0
0
Solution is x = 0, x = – 5 and x = 5
So when x = 0 , x = – 5 or x = 5 the rational expression
an undefined value.
will have
Exercise 1: List all the values of the variable for which the following are undefined.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
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Olympic College Topic 13 Rational Expressions
2. Simplifying a Rational Expression.
Definition:
A simplified rational expression is one in which the numerator and
denominator have no shared common factors.
In order to simplify a rational expression we do the following steps.
Step 1:
Step 2:
Step 3:
Factor the numerator and the denominator.
Cancel all shared factors.
Write down the terms which are left.
Example 1: Simplify the expression
Solution:
=
Factor the numerator and denominator
=
Cancel the shared factor (x+4)
=
Write down the terms which are left.
Example 2: Simplify the expression
Solution:
=
Factor the numerator and denominator
=
Cancel the shared factor of (e + 3)
=
Write down the terms which are left.
Example 3: Simplify the expression
Solution:
=
Factor the numerator and denominator
=
Cancel the shared factor x
=
Write down the terms which are left.
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Olympic College Topic 13 Rational Expressions
Example 4: Simplify the expression
Solution:
=
Factor the numerator and denominator
=
Cancel the shared factor (b – 4)
=
Write down the terms which are left.
Example 5: Simplify the expression
Solution:
=
Factor the numerator and denominator
=
Cancel the shared factor of (d + 10)
=
Write down the terms which are left.
Example 6: Simplify the expression
Solution:
=
Take out a common factor of e
=
Factor the numerator and denominator
=
Cancel the shared factor of (e + 3)
=
Write down the terms which are left.
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Olympic College Topic 13 Rational Expressions
Example 7: Simplify the expression
Solution:
=
=
Take out a common factor of 2
=
Factor the trinomial x2 + 5x + 6
=
Cancel the shared factor (x+3)
=
Write down the terms which are left.
=
2(x + 2)
Example 8: Simplify the expression
Solution:
=
Factor the numerator and denominator
=
Cancel the shared factor (x – 3) and (3 – x)
=
Write down the terms which are left
=
– 2(x+10)
Notice that (x – 3) and (3 – x) can cancel each other out but they leave behind a negative
value since 3 – x = –1(– 3 + x) = – (x – 3)
Exercise 2:
Simplify the following expression.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
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Olympic College Topic 13 Rational Expressions
3. Multiplication of Rational Expressions.
There are two forms the multiplication of rational expressions takes and each requires a
different technique.
The first type of question involves multiplication of exponential terms.
Example 1: Multiply the expressions
Solution (Method 1):
=
Separate the terms
=
Use the division rule
=
=
Using the property that
and a0 = 1
Combining the terms
=
Solution (Method 2):
=
=
Divide by 5
=
=
Cancel the d2
=
=
Cancel the x3
=
=
Cancel the y4
=
Combining the terms
Notice that in method 2 you do not need to write each step of the process, we typically
do all of it in one step, but since this would be a little messy we have shown each part
separately.
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Olympic College Topic 13 Rational Expressions
In order to multiply rational expression we use a process which is very similar to the one
used to simplify rational expressions. The process involve the following three steps.
Step 1:
Step 2:
Step 3:
Factor the numerator and the denominator of each fraction.
Cancel all shared factors.
Combine the term which are left.
Example 2: Multiply the expressions
Solution:
=
Factor the numerators and denominators
=
Cancel the shared factors of (d + 3) and (d – 2)
=
Combine the terms which are left
Example 3: Multiply the expressions
Solution:
=
Factor the numerators and denominators
=
Cancel the shared factors of d and (d + 10)
=
Combine the terms which are left
Example 4: Multiply the expressions
Solution:
=
Factor the numerators and denominators
=
Cancel the shared factors of (3y +1) and (y – 2)
=
Combine the terms which are left
This is the final solution you cannot simplify the result by cancelling out the y’s as these
are not factors so it would be wrong to take
and do this
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Olympic College Topic 13 Rational Expressions
Example 5: Multiply the expressions
Solution:
=
Factor the numerators and denominators
=
Cancel the shared factors of x2 and (x – m)
=
Combine the terms which are left
Example 6: Multiply the expressions
Solution:
=
Factor the numerators and denominators
=
Simplify the
to
=
=
Cancel the shared factors of x2
=
=
Cancel the shared factors of (x + 8)
=
=
Combine the terms which are left
Notice that you do not need to write each step of the process, we typically do all of it in
one step, but since this would be a little messy we have shown each part separately.
It is possible to check your answer by knowing that in all cases there can only be one
instance of each term, so for example, if you have an (2x + 1) in the numerator then you
cannot have a (2x + 1) also in the denominator.
For example the expression
is not simplified as the denominator and
numerator both have the factor (2x +1) it can however be simplified by cancelling the
(2x + 1) term to give
.
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Olympic College Topic 13 Rational Expressions
Exercise 3: Multiply the following Rational Expressions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
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Olympic College Topic 13 Rational Expressions
4. Division of Rational Expressions.
The division of rational expressions uses the following fact.
The division
is equivalent to the multiplication
This means that a division question is essentially transformed into an equivalent multiplication.
When we divide rational expressions that involve exponentials terms we convert it into its
equivalent multiplication and then solve in the usual manner.
Example 1: Simplify the expressions
Solution (Method 1):
=
Convert the division into an equivalent multiplication
=
Separate the terms
=
=
Use the division rule
=
Using the property that
=
Combining the terms
=
Convert the division into a equivalent multiplication
Solution (Method 2):
=
=
Divide by 2
=
=
Cancel the x2
=
=
Cancel the y3
=
=
Cancel the d3
=
Combining the terms
Notice that in method 2 you do not need to write each step of the process, we typically do all of it
in one step, but since this would be a little messy we have shown each part separately.
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Olympic College Topic 13 Rational Expressions
In order to divide a rational expression we first convert it into an equivalent multiplication we
then use a process which is very similar to the one used to simplify rational expressions. The
process involves the following four steps.
Step 1:
Step 2:
Step 3:
Step 4:
Convert the question into an equivalent multiplication.
Factor the numerator and the denominator of each fraction.
Cancel all shared factors.
Combine the term which are left.
Example 2: Divide the expressions
Solution:
=
Convert the division into an equivalent multiplication
=
Factor the numerators and denominators
=
Cancel the shared factors of (d + 8) and (d – 3)
=
=
Combine the terms which are left
Example 3: Multiply the expressions
Solution:
=
Convert the division into an equivalent multiplication
=
Factor the numerators and denominators
=
Cancel the shared factors of x,(x + 4) and (x+10)
=
Combine the terms which are left
Notice that you do not need to write each step of the process, we typically do all of it in one step,
but since this would be a little messy we have shown each part separately.
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Olympic College Topic 13 Rational Expressions
Exercise 4: Divide the following Rational Expressions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
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Olympic College Topic 13 Rational Expressions
5. How to Add and Subtract a Rational Expression.
Definition:
The Lowest Common Multiple LCM of two numbers a and b is the smallest
number that is a multiple of both.
Example 1: Find the Lowest Common Multiple of the following numbers.
(a) 7 and 3
(b) 6 and 8
(c) 5 and 30
(d) 60,80 and 10
Solution(a): Multiples of 7 are 7,14,21,28,35,…..
Multiples of 3 are 3,6,9,12,15,18,21…..
So the LCM of 7 and 3 is 21
Solution(b): Multiples of 6 are 6,12,18,24,30,…..
Multiples of 8 are 8,16,24…..
So the LCM of 6 and 8 is 24
Solution(c): Multiples of 5 are 5,10,15,20,25,30,…..
Multiples of 30 are 30,….
So the LCM of 5 and 30 is 30
Solution(d): Multiples of 60 are 60,120,180,240,300,…..
Multiples of 80 are 80,160,240,320,400,
Multiples of 10 are 10,20,30,……,240,……..
So the LCM of 60,80 and 10 is 240
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Olympic College Topic 13 Rational Expressions
Note the previous method works well with smaller numbers but can be laborious for larger
values. In these situations it may help to find the Prime Factorization of each number and then
get the LCM after this. The following example shows in detail how this can be done.
Example 2: Find the LCM of 120 and 144
Solution:
We get the prime factorization of 120, in order to do this we use the prime numbers
2,3,5,7,11,13,17, etc….
120 =
Using the prime number 2 we factor 120 into 2 times 60
=
Using the prime number 2 we factor 60 into 2 times 30
=
Using the prime number 2 we factor 30 into 2 times 15
=
Using the prime number 3 we factor 15 into 3 times 5
=
Since all the factors are prime numbers we are finished
=
Simplifying the result into powers of each factor.
144 =
Using the prime number 2 we factor 144 into 2 times 72
=
Using the prime number 2 we factor 72 into 2 times 36
=
Using the prime number 2 we factor 36 into 2 times 18
=
Using the prime number 2 we factor 18 into 2 times 9
=
Using the prime number 3 we factor 9 into 3 times 3
=
Since all the factors are prime numbers we are finished
=
Simplifying the result into powers of each factor.
The final step is to use these prime factorizations to get the LCM, in order to do this we take
each prime factor and use the larger power.
120 =
144 =
The largest factors are 24, 32 and 5 the LCM is formed by multiplying each of these terms
together So the LCM of 120 and 144 is
= 720
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Olympic College Topic 13 Rational Expressions
To calculate the LCM of two or more algebraic expressions involves a more complex process.
Find the LCM of two numbers in the expressions.
Find the LCM of the exponential terms.
Combine the two parts together.
Example 3: Find the LCM of 15x2y5z and 9x3y2z10.
Solution:
First we find the LCM of 15 and 9
Multiples of 15 are 15,30,45,60,75,90,…..
Multiples of 9 are 9,18,27,36,45,54,63,72,81,90,…..
LCM of 15 and 9 is 45
Or using the prime factorization of 15 and 9
and 9 =
So the LCM of 15 and 9 is
= 45
Next we find the LCM of the exponential terms.
LCM of x2 and x3 = x3
LCM of y5 and y2 = y5
LCM of z and z10 = z10
The LCM of x2y5z and x3y2z10 is x3y5z10
Lastly we combine the 2 parts to get the LCM of 15x2y5z and 9x3y2z10 = 45x3y5z10
Example 4: Find the LCM of 10a3bc5 and 3ab12.
Solution:
First we find the LCM of 10 and 3
Multiples of 10 are 10,20,30,…..
Multiples of 9 are 3,6,9,12,15,18,21,24,27,30,…..
LCM of 10 and 3 is 30
Next we find the LCM of the exponential terms.
LCM of a3 and a = a3
LCM of b and b12 = b12
LCM of c5 and 1= c5
The LCM of 10a3bc5 and 2ab12 = 30a3b12c5
Lastly we combine the 2 parts to get the LCM of 10a3bc5 and 3ab12 = 30a3b12c5
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Olympic College Topic 13 Rational Expressions
Example 5: Find the LCM of 4x3(x+1)2 and 6x(x + 1)3.
Solution:
First we find the LCM of 4 and 6
Multiples of 4 are 4,8,12,…..
Multiples of 6 are 6,12,…..
LCM of 4 and 6 is 12
Next we find the LCM of the exponential terms.
LCM of x3 and x = x3
LCM of (x + 1)2 and (x + 1)3 = (x + 1)3
The LCM of 4x3(x+1)2 and 6x(x + 1)3 = x3(x + 1)3
Lastly we combine the 2 parts to get the LCM of 4x3(x+1)2 and 6x(x + 1)3 = 12x3(x + 1)3
Notice that (x +1)2 and (x+1)3 are both powers of (x + 1) and so we get a an LCM of (x + 1)3
Example 6: Find the LCM of x(2x – 1)4 and x2(1 – 2x)3
Solution:
Since there are no constant terms we can omit the first step.
Next we find the LCM of the exponential terms. In this situation we have the
unusual situation that the two factors (2x – 1)4 and (1 – 2x)3 at first sight seem to be
different but (2x – 1) and (1 – 2x) are both equivalent with the exception of sign as
(1 – 2x) = – (2x – 1) this allows us to do the following.
LCM of x and x2 = x2
LCM of (2x – 1)4 and (1 – 2x)3 = (2x – 1)4
Hence the LCM of x(2x – 1)4 and x2(1 – 2x)3 = x2(2x – 1)4
Exercise 5A.
Find the LCM of the following terms.
1. 5 and 24
2.
16 and 50
3. 30 and 25
4.
5, 12 and 32
5.
6.
8a5b6c2 and 10a5b12
5x3y4z2 and 4x2y4z7
7. 2y3(y – 1)5 and 6y2(y – 1)3
9.
x6(2x – 1)7 and x5(1 – 2x)8
8. 14x2(x+1)2 and 21x4(x + 1)
10. y2(y – 5)4 and y2(5 – y)3
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Olympic College Topic 13 Rational Expressions
Definition:
An Equivalent Fraction is defined as two fractions which are equal in value but use
different numerators and denominators.
For example, , , and are all equivalent fractions.
One very special equivalent fraction is generated by the number 1, here are some
equivalent fractions to
.
A useful property of fraction multiplication is that if you multiply any fraction by one of the
versions of 1 then you will have an equivalent fraction of it.
For example, if we take the fraction
the fraction
and multiply by it by the equivalent fraction of 1=
is equivalent to
.
If on the other hand we multiply it by the equivalent fraction of 1 =
the equivalent fraction
Definition:
then
we will change
into
.
The Lowest Common Denominator (LCD) of two fractions is the Lowest Common
Multiple of their denominators.
Example 7: Find the LCD of the fractions and
Solution:
First we find the Lowest Common Multiple of the denominators 5 and 8.
Multiples of 5 = 5,10,15,20,25,30,35,40,…….
Multiples of 8 = 8,16,24,32,40,…..
So the LCD of and is 40.
Example 8: Find the LCD of the fractions
Solution:
and
First we find the Lowest Common Multiple of the denominators 5x and 8x3 = 40x3.
Multiples of 5 = 5,10,15,20,25,30,35,40,…….
Multiples of 8 = 8,16,24,32,40,…..
LCM of x and x3 = x3
So the LCD of
and
is 40x3.
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Olympic College Topic 13 Rational Expressions
In order to add or subtract two fractions we need both fractions to have the same denominator.
The method used involves finding the common denominator term called the Lowest Common
Denominator (LCD) that both fractions can be converted into and then using the method of
multiplying each fraction by its own unique equivalent version of 1 in order to achieve this.
Example 9: Find the value of
Solution:
.
The LCD of these fractions = LCM of 6 and 8 = 24
We then need to convert both these fractions into
To convert into
we need to multiply it by .
To convert into
we need to multiply it by .
th
=
=
=
Example 10: Find the value of
Solution:
.
The LCD of these fractions = LCM of 12 and 9 = 36
We then need to convert both these fractions into th
To convert
into
To convert into
we need to multiply it by .
we need to multiply it by .
=
=
=
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Olympic College Topic 13 Rational Expressions
Example 11: Find the value of
Solution:
.
The LCD of these fractions = LCM of y and 5x = 5xy
th
We then need to convert both these fractions into
To convert into
To convert
we need to multiply it by
into
.
we need to multiply it by .
=
=
=
Example 12: Find the value of
Solution:
.
We start by factoring the denominators to get
The LCD of these fractions = LCM of x(x+2) and 3(x+1) = 3x(x+2)(x+1)
th
We then need to convert both these fractions into
To convert
into
we need to multiply it by
.
To convert
into
we need to multiply it by
.
=
=
=
=
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Olympic College Topic 13 Rational Expressions
Example 13: Find the value of
Solution:
.
The LCD of these fractions = LCM of 2x3(x +1) and 5x2(x+1)2 = 10 x3(x+1)2
th
We then need to convert both these fractions into
To convert
To convert
into
into
we need to multiply it by
.
we need to multiply it by
.
=
=
=
=
Example 14: Find the value of
Solution:
.
This question involves an extra initial step where we convert the term (3 – 2x)
into –(2x – 3) so the question is now changed into the following equivalent fraction
=
the above addition is now converted into an
equivalent subtraction.
=
The LCD of these fractions = LCM of
and
We then need to convert both these fractions into
= 10x2(2x – 3)2
th
To convert
into
we need to multiply it by
To convert
into
we need to multiply it by
.
.
=
=
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Olympic College Topic 13 Rational Expressions
Example 15:
Solution:
Find the value of
The LCD of these fractions = LCM of 3(4x2 – 3) and
We then need to convert both these fractions into
To convert
into
To convert
into
= 6x(4x2 – 3)
th
we need to multiply it by
we need to multiply it by
.
.
=
=
=
=
=
=
Example 16:
Solution:
Simplify by dividing by 2
Find the value of
We start by expressing 3 as the fraction so the question becomes
=
The LCD of these fractions = LCM of 1 and
We then need to convert both these fractions into
To convert into
To convert
we need to multiply it by
into
th
.
we need to multiply it by .
=
=
=
=
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Olympic College Topic 13 Rational Expressions
Example 17:
Solution:
Find the value of
We start by factoring the denominator
=
= x(x + 2)(x – 4)
The LCD of these fractions = LCM of x(x + 2) and
We then need to convert both these fractions into
To convert
into
To convert
th
we need to multiply it by
into
.
we need to multiply it by .
=
=
=
=
Exercise 5B.
1.
Find the LCD of the following fractions
(a)
2.
and
(b)
and
(c)
and
(d)
and
.
Evaluate the following simplifying your results if necessary.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)
.
.
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Olympic College Topic 13 Rational Expressions
6. How to Solve Rational Equations.
Definition:
A Rational Equation is an equation that contains a rational expression, the
following are all examples of rational equations.
=
=1
=0
=
When you are asked to solve a rational equation you are trying to find value(s) of x
that satisfy the given initial equation.
We have found in other situations that some equations have one solution, for
example we can show that the rational equation
=
has one solution
x = 6.
There are other rational equations which have more than one solution, for example
= 1 has two solutions x = – 7 and x = 3.
There are other rational equations rational equation has no solutions, for example
= 0 has no solutions.
The last situation is that the rational equation has a “Phantom Solution” this
situation occurs when one or more of the solutions cannot be used as it will
generate an undefined value in the original equation. For example, the rational
equation
=
when solved algebraically seems to have two solutions x = 4
and x = – 2 yet when you check these solutions you will find that for x = – 2 the
solution works but that for x = 4 we get an undefined value in the original equation
and so x = 4 is not a solution it’s a “Phantom Solution” – its not really there.
Testing possible solution x = – 2 we get
=
=
=
=
2
Testing possible solution x = 4 we get
=
2
It Works!!
=
=
=
Undefined value!!
It is therefore very important to check all possible solutions to be sure that they are
viable solutions and not Phantom Solutions.
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Olympic College Topic 13 Rational Expressions
There are two main techniques used to solve rational equations
Method 1:
Method 2:
Use the “Cross Multiplication”.
Multiply by the LCD.
We use Method 1 in all those situations where the rational equation consists of two fractions
equal to each other such as
=
or
=
or
=
The cross multiplication method uses the fact that
is equivalent to
ad = bc
provided b and d are not zero.
The steps involved in using the cross multiplication method are as follows.
Step 1: Convert any terms to a fraction if necessary.
Step 2: Cross multiply the two fractions.
Step 3: Solve the resulting equation.
Step 4: Check your solutions.
Example 1: Solve the rational equation
Solution:
=
Step 1 is not necessary since both are already fractions.
7(2x – 8)
14x – 56
14x
10x
x
=
=
=
=
=
=
=
4(x +1)
4x + 4
4x+ 60
60
6
Step 2 Cross Multiply.
Use the Distributive Law.
Add 56 to both sides.
Subtract 4x from both sides.
Divide both sides by 10
Step 3 Solve the equation
Step 4 Check the solution.
Substitute x = 6 into the original equation
=
=
=
1
=
1
Since both sides of the equation agree we can conclude that x = 6 is a valid solution
Page 25
Olympic College Topic 13 Rational Expressions
Example 2: Solve the rational equation
Solution:
= x
Step 1 Convert the x into the fraction to transform the equation into
=
2
5x =
x2 =
x
x
4x2 + 16
16
=
=
Step 2 Cross Multiply.
Subtract 4x2 from both sides.
take the square root of both sides
Step 3 Solve the equation
Step 4 Check the solutions.
Substitute x = 4 into the original equation
= x
= 4
= 4
= 4
= 4
1
=
1
Since both sides of the equation agree we can conclude that x = 4 is a valid solution
Substitute x = – 4 into the original equation
= x
= –4
= –4
= –4
= –4
1
=
1
Since both sides of the equation agree we can conclude that x = – 4 is also a valid
Solution to the rational equation.
Note:
Step 1 was necessary in this equation as we had x on its own, we would also need to do
it in situations where the fraction equaled a number of an algebraic expression so for
example.
= 7
would become
=
= 2x + 5
would become
=
Page 26
Olympic College Topic 13 Rational Expressions
Example 3: Solve the rational equation
Solution:
=
Step 1 is not necessary since both are already fractions.
=
(2x + 4)(5x – 1) =
4(x2 + 5x)
Step 2 Cross Multiply.
10x2 – 2x + 20x – 4
=
4x2 + 20x
Use F.O.I.L.
10x2 + 18x – 4
=
4x2 + 20x
Collect like terms and simplify
6x2 + 18x – 4
=
20x
Subtract 4x2 from both sides.
0
Subtract 20x from both sides.
=
0
Take out a Common Factor of 2
2(3x + 2)(x – 1) =
0
Factor the trinomial
6x2 – 2x – 4 =
2(3x2 – x – 2)
x =
and x = 1
Step 3 Solve the equation
Step 4 Check the solutions.
Substitute x =
into the original equation
=
=
=
=
1
=
1
Since both sides of the equation agree we can conclude that x = 6 is a valid solution
Page 27
Olympic College Topic 13 Rational Expressions
Method 2 – is typically used if there are more than two fractions in the equation.
This method involves 3 steps.
Step 1: Find The LCD of all the fractions.
Step 2: Multiply all the terms in the equation by the LCD to remove all fractions.
Step 3: Solve the new equivalent equation.
Example 4: Solve the rational equation
Solution:
=
We will use Method 2 in this situation.
Step 1: To find the LCD we will write all the terms as fractions so
=
will become the equivalent equation
=
The LCD of x and 1 is x so all fractions in the equation will be multiplied by x.
Step 2: We now multiply all the terms in the equation
=
by the LCD x
=
2x2
x2
x
x
=
=
=
=
=
40
32
16
4 or x = – 4
Subtract 8 from both sides.
Divide both sides by 2.
Take the square root of both sides.
The last thing we do is check our solutions.
Substitute x =
into the original equation
=
=
=
10 =
10
10
Since both sides of the equation agree we can conclude that x = 4 is a valid solution
Substitute x =
into the original equation
=
=
=
10 =
10
10
Since both sides of the equation agree we can conclude that x = 4 is a valid solution
Page 28
Olympic College Topic 13 Rational Expressions
Example 5: Solve the rational equation
Solution:
=
Step 1: The LCD of x and (x + 1) is x(x+1) so all fractions in the equation will be
multiplied by x(x+1)
Step 2: Multiply all the terms in the
=
by the LCD x(x+1)
=
=
8x + 8 + 6x =
14x + 8
=
2x + 8
=
2x
=
x
=
12(x +1)
12x + 12
12x + 12
12
4
2
Multiply out the brackets.
Subtract 12x from both sides.
Subtract 8 from both sides.
Divide both sides by 2.
The last thing we do is check our solution.
Substitute x =
into the original equation
=
=
6
=
=
6
6
Since both sides of the equation agree we can conclude that x = 2 is a valid
solution.
Exercise 6: Solve the following Rational Equations.
1.
=
2.
4.
= x
5.
7.
=
8.
9.
=
10.
12.
=
13.
=
3.
=
y
=
6.
= 4
9.
=
11.
=
=
=
14.
=
= 4
Page 29
Olympic College Topic 13 Rational Expressions
7. Application Problems.
The application problems that involve rational expressions frequently contain the concept of a
“Rate” this is a simple formula that shows the relationship between quantities.
Examples of rates could be
Speed
=
Unit Cost
=
Work Rate
=
If the question involves Speed, Distance and Time we can solve the problem by using the
following steps
Step 1: Read over the question and identify the three quantities that make up the Rate.
Step 2: Identify the variable x, this will usually be the thing you are being asked to find.
Step 3: Create a table which will contain all the information linking the Rate.
Step 4: Set up a rational equation that contains x.
Step 5: Solve the equation.
If the question involves Work Rate we can solve the problem by using the following steps
Step 1: Read over the question and identify the individual work rates of each person using
x = the total amount of work done.
Step 2: Combine the work rates to get the total work rate.
Step 3: Find the time taken to complete the work by using the formula
Time to complete work
=
Page 30
Olympic College Topic 13 Rational Expressions
Example 1:
Vecca can paddle her kayak 4 mph in still water. It takes her as long to paddle 8
miles upstream as it takes her to paddle 12 miles downstream. Find the speed of the
water current.
Solution:
Step 1: The three quantities that make up the rate are Speed, Distance and Time.
Step 2: We are being asked to find the speed of the water current so we will let
x = speed of current. This means that the upstream speed will be 4 – x
Step 3: The table below shoes the interconnection of all the information.
Upstream
Distance
Speed
8
4–x
Downstream 12
Time
4+x
Step 4: Since the times are equal we have the equation that
12
4 x
8
4 x
=
12
4 x
Step 5: Solving the equation
=
8(4 + x) =
12(4 – x)
Cross Multiplication.
32 + 8x =
48 – 12x
Using the Distributive Law.
8x =
12 – 12x
Subtracting 48 from both sides.
20x =
12
Add 12x to both sides.
x
=
x
=
So the speed of the water current is
Divide both sides by 20.
.
Page 31
Olympic College Topic 13 Rational Expressions
Example 2:
John is in training for a marathon as part of this training every day he jogs for 8
miles and then he runs for 10 miles. He can run at 6 mph faster than he jog and the
total time of his training is 3 hours. What is the speed John can run and jog?
Solution:
Step 1: The three quantities that make up the rate are Speed, Distance and Time.
Step 2: We are being asked to find the speed John can run and jog?
x = speed John can run
this means that the speed that John can Jog = x – 6
Step 3: The table below shoes the interconnection of all the information’
Distance
Speed
Run
10
x
Jog
8
x–6
Time
8
x 6
Step 4: Since the total time was 3 hours we have the equation that
=
3
Step 5: Solving the equation
=
10(x – 6) + 8x
10x – 60 + 8x
18x – 60
– 60
0
0
0
=
=
=
=
=
=
=
=
3
3x(x – 6)
3x2 – 18x
3x2 – 18x
3x2 – 18x
3x2 – 36x
3x2 – 36x + 60
3(x2 – 12x + 10)
3(x – 2)(x –10)
Multiplying by the LCD.
Multiply all the terms.
Using the Distributive Law.
Subtracting 18x from both sides.
Add 60 to both sides.
Take out a common factor of 3.
Factor the trinomial
We have two possible solutions x = 10 and x = 2
We can use the solution x = 10 since this will give us usable results since we get.
Speed John runs =
and Speed John jogs =
x = 10 mph
x – 6 = 10 – 6 = 4 mph.
We cannot use the solution x = 2 since this would give us unusable results.
Speed John runs =
and Speed John Jogs =
x = 2 mph
x – 6 = 2 – 6 = – 4 mph which makes no sense.
Page 32
Olympic College Topic 13 Rational Expressions
Example 3:
Peter can paint a wall in 20 hours, Anne can paint the same wall in 10 hours.
How long will it take to paint the wall if they both work together?
Solution:
Step 1: We find the individual work rates for both Peter and Anne.
If we assume that the total amount of work done in painting the wall = x
then:-
Work Rate for John
=
=
Work Rate for Anne =
=
Step 2: We now combine the two work rates together to get the total work rate.
Total work rate
=
=
=
Step 3: We find the time would take to paint the wall using the total work rate.
Time to complete work
=
=
=
=
=
hours
Note we can make a short cut in finding the time to complete the work by noticing
that in these type of questions the time taken it is the same value as the reciprocal of
the total work rate without the x term.
So for example if the Total work rate is
Total work rate is
Total work rate is
then the time taken will be
hours.
then the time taken will be hours.
then the time taken will be
hours.
Page 33
Olympic College Topic 13 Rational Expressions
Example 4:
A large pump takes 80 minutes to empty a swimming pool.
A medium pump will take 120 minutes to empty a swimming pool.
A smaller pump will take 240 minutes to empty a swimming pool.
How long will it take to empty the swimming pool if all three pumps are used.
Solution:
Step 1: We find the individual work rates for both the large, medium and small
pump.
If we assume that the total amount of work done in emptying the pool = x
then:Work Rate Large pump
=
=
Work Rate Medium pump
=
=
Work Rate for Small Pump
=
=
Step 2: We now combine the two work rates together to get the total work rate.
Total work rate
=
=
=
=
Step 3: We find the time would take to empty the pool
Time to complete work
=
=
=
=
=
hours
Note we can make a short cut in finding the timer to complete the work by noticing
that it is the same value as the reciprocal of the total work rate = 40 hours.
Page 34
Olympic College Topic 13 Rational Expressions
Exercise 7
1.
Mary can paddle her canoe 5 mph in still water. It takes her as long to paddle 12 miles
upstream as it takes her to paddle 6 miles downstream. Find the speed of the water current.
2.
The speed of the water current is 4 mph. Peter takes as long to paddle 12 miles upstream as
it takes to paddle 6 miles downstream. Find the speed that Peter can paddle if there were no
current.
3.
John is in training for a marathon as part of this training every day he jogs for 8 miles and
then he runs for 10 miles. He can run at 6 mph faster than he jog and the total time of his
training is 3 hours. What is the speed John can run and jog?
4.
A worker is travelling from Abbyville to Bellshill a distance of 120 miles and then on to
Cardiff a further 50 miles. The road from Abbyville to Bellshill was quiet and he made good
progress while the road from Bellshill to Cardiff was busy and his speed was 10 mph slower
The total time of the journey was 3 hours, what are the average speeds for the two parts of
the journey.
5.
Robert can paint a house in 12 hours, Janice can paint the same house in 4 hours.
How long will it take to paint the wall if they both work together?
6.
A worker can empty a full truck in 3 hours , his coworker can empty the same truck in 4
hours .How long would it take to empty the truck if they worked together.
Page 35
Olympic College Topic 13 Rational Expressions
Solutions:
Exercise 1: (a) x = 5
(d) x = 2 and x = 7
(g) x = 3 and x =
(j) y = – 2 , y = 0 , y = 2
(b) x = – 5
(e) y = – 1 and y = 0
(h) z = – 5 and z = 0
(k) y = 0 and y = 9
(c) x = – 3 and x = 4
(f) y = – 4 and y = 4
(i) t = – 1 and t = 1
(l) there are none
Exercise 2: 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
2. 2x5z
3.
4.
5.
6.
7
8.
9. 2(d + 5)
10.
11.
12.
13.
14.
Exercise 3: 1. 3d2x2y6
Exercise 4: 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
13.
14.
11.
1
12.
Page 36
Olympic College Topic 13 Rational Expressions
Exercise 5A.
Exercise 5B.
1.120
2.
400
3. 150
4.
480
5.20x3y4
6.
40a5b12c2
7. 6y3(y – 1)5
8.
42x4(x+1)2
9. x6(2x – 1)8
10. y2(y – 5)4
1.(a)
(b)
(c)
(d) 14x4
2.(a)
(b)
(c)
2.(d)
(e)
(f)
2.(g)
(h)
(i)
2.(j)
(k)
(l)
2.(m)
(n)
(o)
2.
y = – 22
3.
x=4
x = – 1 or x = 1
5.
y = – 5 or y = 5
6.
x= –
7.
d = 0 or d = 1
8.
x=
9.
x = – 2 or x = 2
10.
y = – 1 or y = 1
11. x =
13.
x= – 2
14. x = 4
Exercise 6: 1. x =
4.
Exercise 7: 1.
or x =
12. y = 3
Speed of current = mph
2.
Speed can paddle = 12 mph
3.
Speed Jog = 4 mph Speed Run = 10 mph
4.
Speeds 60 mph and 50 mph
5.
Time = 3 hours
6.
Time =
hours
Page 37