1. Multiplying and Dividing Rational Expressions
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Chapter 4
1.
Multiplying and Dividing Rational Expressions
| Student Learning Objectives: |
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Why
Understanding rational expressions helps extend fraction operations into algebra. This lesson explores how polynomial ratios can be simplified, multiplied, divided, and used in more complex expressions, while also emphasizing the importance of excluded values. By learning how to factor expressions, cancel common factors, and keep track of restrictions, students can work more confidently with rational expressions in both abstract problems and real-world situations involving fuel efficiency, area, packaging, and ratios.
Lesson Settings & Tools
| | 12 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Multiplying and Dividing Rational Expressions
Slide of 12
As with rational numbers, it is possible to write two polynomial expressions as a ratio, so long as the divisor is not zero. This lesson will explore various aspects of this type of algebraic expression, including how mathematical operations such as multiplication and division can be applied to it.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Combined Fuel Efficiency
The distance that a vehicle can travel per one gallon of fuel is measured as its mile per gallon (mpg) fuel economy. Each car has two fuel economy numbers, one measuring its efficiency for city driving and the other for highway driving. The combined fuel economy C for x mpg in the city and y mpg on the highway is computed by the following formula. C = 1/12( 1x + 1y )
a If a car travels 28 miles per gallon in the city and 36 mpg on the highway, what is the combined fuel economy for the car?
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b Let x be the city fuel economy for a new car model. If the highway fuel economy in terms of x is 10xx-20, write a simplified expression for the combined fuel economy for such a car.
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Discussion
Rational Expression
A rational expression is a fraction where both the numerator and the denominator are polynomials.
p(x)/q(x)
Here, p(x) and q(x) are polynomials and q(x)≠ 0. The expression below is an example of a rational expression. x^2-7/x^3+5x A rational expression is said to be written in its simplest form if the numerator and denominator have no common factors.
| Rational Expressions | |
|---|---|
| Not in Simplest Form | In Simplest Form |
| xy/x(x-3)(y+2) | x-1/x+1 |
| x^4+x^2/x^2+1 | x^3+7/x^2-x-6 |
Notice that for some of the expressions shown in the table, there are some x-values that make the denominator 0. For example, the denominator of x-1x+1 is 0 when x=- 1. Any value of a variable for which a rational expression is undefined is called an excluded value.
| Expression | Restriction | Excluded Value(s) |
|---|---|---|
| x-1/x+1 | x+1≠ 0 | x ≠ - 1 |
| x^3+7/x^2-x-6 | x^2-x-6≠ 0 | x≠ - 2 and x≠ 3 |
| xy/x(x-3)(y+2) | x(x-3)(y+2)≠ 0 | x≠0, x ≠ 3, and y ≠ - 2 |
| x^4+x^2/x^2+1 | There is no real number that makes x^2+1 zero | None |
Simplifying a rational expression can remove some of the excluded values that appear in the original expression. A rational expression and its simplified form must have the same domain in order for them to be equivalent expressions. This means that the excluded values that are no longer visible in the simplified expression must still be declared.
| Equivalent Expressions | |
|---|---|
| Rational Expression | Simplified Form |
| x-3/(x+2)(x-3), x≠ - 2, 3 | 1/x+2,x≠ - 2, 3 |
| x^2+2x+1/x^2-1, x≠ - 1, 1 | x+1/x-1,x≠ - 1,1 |
| x^3-2x^2+x/x^2,x≠ 0 | x^2-2x+1/x, x≠ 0 |
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Pop Quiz
Determining Values That Make a Rational Expression Undefined
A rational expression is undefined when its denominator is 0. The values that make the denominator of a rational expression equal to 0 are called excluded values because they are excluded from its domain. Determine the excluded values for the indicated rational expressions.
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Discussion
Simplifying Rational Expressions
A rational expression is written in simplest form if the numerator and denominator have no factors in common. Rational expressions that are not in simplest form can be simplified by canceling out the common factors. There are many various methods of factoring polynomials to find any common factors. Consider an example expression. x^2 - 6x + 9/9x - x^3 Rational expressions can be simplified in three steps.
1
Factor the Numerator and the Denominator
expand_more Check first if either the numerator or the denominator — or both — can be factored by the greatest common factor. In this example, the factor x is in both terms of the denominator.
x^2 - 6x + 9/9x - x^3
SplitIntoFactors
Split into factors
x^2 - 6x + 9/x * 9 - x* x^2
FactorOut
Factor out x
x^2 - 6x + 9/x (9 - x^2 )
Determine if it is possible to factor either the numerator or the denominator by using a difference of squares. Look for an expression in the form a^2 - b^2, which can be factored as (a + b)(a - b). In the example, (9 - x)^2 can be factored using this rule.
x^2 - 6x + 9/x(9 - x^2 )
WritePow
Write as a power
x^2 - 6x + 9/x (3^2 - x^2 )
FacDiffSquares
a^2-b^2=(a+b)(a-b)
x^2 - 6x + 9/x (3 + x)(3 - x)
See if either the numerator or the denominator is a perfect square trinomial. Look for an expression in the form a^2 ± 2ab + b^2, which can be factored as (a ± b)^2. Here, the numerator is a perfect square trinomial.
x^2 - 6x + 9/x (3 + x)(3 - x)
SplitIntoFactors
Split into factors
x^2 - 2x(3) + 9/x (3 + x)(3 - x)
WritePow
Write as a power
x^2 - 2* x * 3 + 3^2/x (3 + x)(3 - x)
FacNegPerfectSquare
a^2-2ab+b^2=(a-b)^2
(x - 3)^2/x (3 + x)(3 - x)
In some cases, a negative sign can be factored out for one expression to have the same form as another factor. Here, the factor 3 - x in the denominator is almost identical to the factor x - 3 in the numerator.
(x - 3)^2/x (3 + x)(3 - x)
CommutativePropMult
Commutative Property of Multiplication
(x - 3)^2/(3 - x)x(3 + x)
FactorOutNegSignSwap
a-b=-(b-a)
(x - 3)^2/- (x - 3)x(3 + x)
Now both the numerator and denominator are completely factored.
2
List Restricted Values
expand_more Before simplifying the common factors, check if there are any restrictions on x. (x - 3)^2/- (x - 3)x(3 + x)
Note that the expression is undefined when x=- 3, x=0, or x=3.
3
Cancel Out Common Factors
expand_more In this expression, (x-3) is the common factor of the numerator and the denominator and can be eliminated.
The rational expression is now written in its simplest form. In order for the given expression and the simplified expression to be equivalent, the domain of the simplified expression should be restricted by excluding x=3. x - 3/- x (3 + x), x ≠ 3 Since the restriction x≠ 3 cannot be seen from the simplified expression, it should be written. The other restrictions are evident from the simplified expression.
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Example
Finding the Domain of Rational Expressions
Kevin and Zosia are asked to find the values that make the following rational expression undefined. x^2+12x+35/x^2+2x-35 They disagree about the domain of the rational expression.
Who is correct?
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Hint
A rational expression is undefined for values that make its denominator zero. Therefore, those values should be excluded from the domain.
Solution
A rational expression is undefined when the denominator equals 0. x^2+12x+35/x^2+2x-35
To find the domain of the rational expression, the expression in the denominator is set equal to 0.
x^2+2x-35=0
To solve the resulting equation for x, start by factoring the expression on the left-hand side.
x^2+2x-35=0
(x+7)(x-5)=0
Now the Zero Product Property will be applied to find all excluded values.
(x+7)(x-5)=0
▼
Solve using the Zero Product Property
ZeroProdProp
Use the Zero Product Property
lcx+7=0 & (I) x-5=0 & (II)
SubEqn
(I): LHS-7=RHS-7
lx=-7 x-5=0
AddEqn
(II): LHS+5=RHS+5
lx=-7 x=5
The values x=- 7 and x=5 make the denominator equal to 0. As a result, they make the rational expression undefined. Therefore, Zosia is correct in saying that the domain is all real numbers except - 7 and 5. Kevin may have simplified the expression before identifying the excluded values. This is a common mistake!
x^2+12x+35/(x+7)(x-5)
(x+7)(x+5)/(x+7)(x-5)
▼
Simplify
(x+5)/(x-5)
Note that x=5 is the value that makes the denominator of the simplified expression equal to 0. However, Kevin should have noted that - 7 makes the original expression undefined. Since the original and simplified expressions are equivalent, both should have the same domain.
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Discussion
Multiplying and Dividing Rational Expressions
Operations with rational numbers and rational expressions are similar.
Multiplying Rational Expressions
Multiplying rational expressions works the same way as multiplying fractions. The numerators and denominators are multiplied separately.
P(x)/Q(x) * H(x)/G(x)=P(x)* H(x)/Q(x)* G(x)
As an example, consider the following product. x^2-2x-15/7x+42 * x^2+9x+18/x^2-11x+30 Three steps can be followed to multiply the rational expressions.
1
Factor Each Numerator and Denominator
expand_more Before multiplying, it is helpful to factor the numerators and denominators, if possible. Start with the first rational expression.
Now the numerator and denominator of the second expression will be factored.
x^2+9x+18/x^2-11+30
(x+3)(x+6)/x^2-11x+30
(x+3)(x+6)/(x-5)(x-6)
2
Multiply the Numerators and Multiply the Denominators
expand_more Now that both expressions have been factored, the numerators and denominators can be multiplied.
(x+3)(x-5)/7(x+6) * (x+3)(x+6)/(x-5)(x-6)
MultFrac
Multiply fractions
(x+3)(x-5)(x+3)(x+6)/7(x+6)(x-5)(x-6)
The product is undefined when x=- 6, x=5, and when x=6.
4
Cancel Out Common Factors
expand_more Note that (x−5) and (x+6) are common factors of the numerator and denominator. Finally, the product is simplified by canceling out common factors.
(x+3)(x-5)(x+3)(x+6)/7(x+6)(x-5)(x-6)
CancelCommonFac
Cancel out common factors
(x+3)(x-5) (x+3) (x+6)/7(x+6)(x-5)(x-6)
SimpQuot
Simplify quotient
(x+3)(x+3)/7(x-6)
ProdToPowTwoFac
a* a=a^2
(x+3)^2/7(x-6)
Considering the denominator of the simplified expression and any other denominator used, the values x=- 6, x=5, and x=6 must be excluded from the domain of the simplified expression. In this expression it is clear that x=6 makes the denominator 0. Therefore, the fact that x≠ 6 will not be expressed. (x+3)^2/7(x-6), x ≠ - 6, x ≠ 5
Dividing Rational Expressions
Dividing two rational expressions is the same as multiplying the first expression by the reciprocal of the second expression.
P(x)/Q(x) ÷ H(x)/G(x) = P(x)/Q(x) * G(x)/H(x)
As an example, consider the quotient of two rational expressions. 5-x/x^2+3x ÷ x^2-25/x^2+10x+21 The process of dividing rationals expression can be completed in four steps.
1
Rewrite the Division as the Product of the Dividend and the Reciprocal of the Divisor
expand_more To divide a rational expression by another rational expression, multiply the first expression by the reciprocal of the second. 5-x/x^2+3x ÷ x^2-25/x^2+10x+21 [0.5em] ⇕ [0.6em]
5-x/x^2+3x * x^2+10x+21/x^2-25
Once the quotient is expressed as a product, the remaining steps are the same as those for multiplying rational expressions.
2
Factor Each Numerator and Denominator
expand_more Factor the numerators and the denominators, if possible. The first expression will be factored first.
Now the numerator and the denominator of the second expression will be factored.
x^2+10x+21/x^2-25
(x+3)(x+7)/x^2-25
▼
Factor the denominator
(x+3)(x+7)/(x+5)(x-5)
3
Multiply the Numerators and Multiply the Denominators
expand_more Next, the numerators and denominators can be multiplied.
5-x/x(x+3) * (x+3)(x+7)/(x+5)(x-5) [0.5em] ⇕ [0.6em]
(5-x)(x+3)(x+7)/x(x+3)(x+5)(x-5)
This product is undefined when x= - 5, x=- 3, x=0, and x=5. Also, the values that make the divisor's denominator in the original quotient expression equal 0 should be excluded. These values are x=- 7 and x=- 3.
Excluded Values - 7, - 5, - 3, 0, 5
4
Cancel Out Common Factors
expand_more Finally, the product is simplified by canceling out common factors.
(5-x)(x+3)(x+7)/x(x+3)(x+5)(x-5)
FactorOut
Factor out - 1
- 1(x-5)(x+3)(x+7)/x(x+3)(x+5)(x-5)
CancelCommonFac
Cancel out common factors
- 1 (x-5) (x+3)(x+7)/x(x+3)(x+5)(x-5)
SimpQuot
Simplify quotient
- 1(x+7)/x(x+5)
Distr
Distribute - 1
- x-7/x(x+5)
Considering the denominator of the simplified expression and any other denominator used in this process, the values x=- 7, x=- 5, x=- 3, x=0, and x=5 must be excluded from the domain of the expression. The simplified expression shows that x=0 and x=- 5 must be excluded, so these values are not mentioned. - x-7/x(x+5), x ≠ - 7, x ≠ - 3, x ≠ 5
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Example
Representing Areas with Rational Expressions
Ramsha drew the plan of her house and labeled the sides, measured in meters, as shown.
a Write a simplified rational expression for the area of the house.
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b State any restrictions on x.
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Hint
a Start by factoring the numerator and the denominator of each rational expression.
b The denominator of a rational expression cannot be zero.
Solution
a Start by considering the formula for the area of a rectangle.
A= l w According to the diagram, the length l of the house is represented by 2x^2-6xx^2+18x+81 and the width w by 9x+81x^2-9.
To write an expression for the area, these two expressions will be substituted into the formula for the area of a rectangle. Both factors will then be factored.
A=l w
SubstituteII
l= 2x^2-6x/x^2+18x+81, w= 9x+81/x^2-9
A= 2x^2-6x/x^2+18x+81* 9x+81/x^2-9
▼
Factor the first expression
FactorOut
Factor out 2x
A=2x(x-3)/x^2+18x+81* 9x+81/x^2-9
SplitIntoFactors
Split into factors
A=2x(x-3)/x^2+2(9)x+81* 9x+81/x^2-9
CommutativePropMult
Commutative Property of Multiplication
A=2x(x-3)/x^2+2x(9)+81* 9x+81/x^2-9
WritePow
Write as a power
A=2x(x-3)/x^2+2x(9)+9^2* 9x+81/x^2-9
FacPosPerfectSquare
a^2+2ab+b^2=(a+b)^2
A=2x(x-3)/(x+9)^2* 9x+81/x^2-9
▼
Factor second expression
FactorOut
Factor out 9
A=2x(x-3)/(x+9)^2* 9(x+9)/x^2-9
WritePow
Write as a power
A=2x(x-3)/(x+9)^2* 9(x+9)/x^2-3^2
FacDiffSquares
a^2-b^2=(a+b)(a-b)
A=2x(x-3)/(x+9)^2* 9(x+9)/(x+3)(x-3)
Now that both expressions have been factored, the common factors can be canceled out.
2x(x-3)/(x+9)^2 * 9(x+9)/(x+3)(x-3)
▼
Simplify
MultFrac
Multiply fractions
2x(x-3)* 9(x+9)/(x+9)^2* (x+3)(x-3)
PowToProdTwoFac
a^2=a* a
2x(x-3)* 9(x+9)/(x+9)(x+9)* (x+3)(x-3)
CancelCommonFac
Cancel out common factors
2x(x-3) * 9(x+9)/(x+9)(x+9) * (x+3)(x-3)
SimpQuot
Simplify quotient
2x* 9/(x+9)* (x+3)
Multiply
Multiply
18x/(x+9)(x+3)
This rational expression, which is written in its simplest form, represents the area of Ramsha's house.
b To identify the restrictions, the denominator of the simplified expression and any other denominator used in this process will be analyzed. For simplicity, their factored forms will be used.
| Denominator | Restrictions on the Denominator | Restrictions on the Variable |
|---|---|---|
| (x+9)^2 | (x+9)^2≠ 0 | x≠ - 9 |
| (x+3)(x-3) | x+3≠ 0 and x-3≠ 0 | x≠ -3 and x≠ 3 |
| (x+9)(x+3) | x+9≠ 0 and x+3≠ 0 | x≠ -9 and x≠ -3 |
There are three unique restrictions on the variable x. x≠ - 9, x≠ - 3, x≠ 3
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Example
Determining the Type With the Best Efficiency Ratio
Companies aim to produce packaging using the lowest possible amount of material. They produce their packages in such a way that the ratio of the surface area S of a package to its volume V is as small as possible. Efficiency Ratio: S/V A company is designing two different types of packages. The table shows the expressions for their surface areas and volumes.
| Surface Area, S | Volume, V | |
|---|---|---|
| Type I | x^3-x^2/4x^4+12x^3 | x^2-2x+1/x^3+3x^2 |
| Type II | 2x^2-9x-18/4x^2-28x+24 | 2x^2+x-3 |
a Find the efficiency ratio of each type.
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b Which type is more efficient when x=2?
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Hint
a Start by factoring each denominator and numerator, if possible. Recall that all denominators must be different than 0.
b Substitute the given value for x into each rational expression found in Part A. The smaller the ratio, the more efficient the packaging.
Solution
a To find the efficiency ratios, the ratio of the surface area to the volume will be found for each packaging type. First, the efficiency ratio of Type I will be calculated.
x^3-x^2/4x^4+12x^3 ÷ x^2-2x+1/x^3+3x^2 To divide rational expressions, multiply the first expression by the reciprocal of the second expression. x^3-x^2/4x^4+12x^3 ÷ x^2-2x+1/x^3+3x^2 [0.5em] ⇕ [0.6em] x^3-x^2/4x^4+12x^3* x^3+3x^2/x^2-2x+1 To multiply these expressions, begin by factoring the numerators and the denominators of each expression, if possible. Start with the first rational expression.
x^3-x^2/4x^4+12x^3
FactorOut
Factor out x^2
x^2(x-1)/4x^4+12x^3
FactorOut
Factor out 4x^3
x^2(x-1)/4x^3(x+3)
Now the second rational expression will be factored.
x^3+3x^2/x^2-2x+1
FactorOut
Factor out x^2
x^2(x+3)/x^2-2x+1
▼
Factor denominator
IdPropMult
Identity Property of Multiplication
x^2(x+3)/x^2-2x(1)+1
WritePow
Write as a power
x^2(x+3)/x^2-2x(1)+1^2
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
x^2(x+3)/(x-1)^2
Next the factored rational expressions will be multiplied and the common factors canceled.
x^2(x-1)/4x^3(x+3)* x^2(x+3)/(x-1)^2
MultFrac
Multiply fractions
x^2(x-1)x^2(x+3)/4x^3(x+3)(x-1)^2
▼
Simplify
Write power as a product
x^2(x-1)x * x(x+3)/4x^2 * x(x+3)(x-1)(x-1)
CancelCommonFac
Cancel out common factors
x^2 (x-1) x * x(x+3)/4x^2 * x(x+3) (x-1)(x-1)
SimpQuot
Simplify quotient
x/4 (x-1)
The efficiency ratio of Type I is x4(x-1). The efficiency ratio of the Type II packages can be found by following a similar process. 2x^2-9x-18/4x^2-28x+24 ÷ (2x^2+x-3) To divide a rational expression by polynomial, the first expression is multiplied by the reciprocal of the polynomial. 2x^2-9x-18/4x^2-28x+24 ÷ (2x^2+x-3) [0.6em] ⇕ [0.6em] 2x^2-9x-18/4x^2-28x+24 * 1/2x^2+x-3 The table shows the steps taken to perform the multiplication.
| Multiplying Rational Expressions | |
|---|---|
| Product | 2x^2-9x-18/4x^2-28x+24 * 1/2x^2+x-3 |
| Factor | (2x+3)(x-6)/4(x-1)(x-6) * 1/(2x+3)(x-1) |
| Multiply | (2x+3)(x-6)/4(x-1)(x-6)(2x+3)(x-1) |
| Cancel Out Common Factors | (2x+3)(x-6)/4(x-1)(x-6)(2x+3)(x-1) |
| Simplify | 1/4(x-1)^2 |
The efficiency ratio of Type II is 14(x-1)^2
b In Part A, the efficiency ratio of each type was found. Now, these ratios will be evaluated for x=2.
| Type I | Type II | |
|---|---|---|
| Efficiency Ratio | x/4(x-1) | 1/4(x-1)^2 |
| Substitute | 2/4( 2-1) | 1/4( 2-1)^2 |
| Evaluate | 1/2 | 1/4 |
Recall that the smaller the ratio, the more efficient the packaging. Therefore, Type II is more efficient because 14 < 12.
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Discussion
Complex Fractions
A complex fraction is a rational expression where the numerator, denominator, or both, contain a rational expression.
p(x)r(x)/q(x)m(x)
Here, p(x), r(x), q(x), and m(x) are polynomials. A complex fraction can be simplified by rewriting it as a quotient and then dividing the rational expressions. p(x)r(x)/q(x)m(x) ⇔ p(x)/r(x) ÷ q(x)/m(x) As an example of a complex fraction, consider the following division of rational expressions.
x^2-2x+4x^2-y^2/x-yx+y ⇔ x^2-2x+4/x^2-y^2 ÷ x-y/x+y
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Example
Surface Area to Volume Ratio of Penguins
Animals adapt to their environment. As a result of adaptations, the surface area and volume of animals vary depending on where they live. For example, penguins have a lower surface area to volume ratio to conserve their body heat.
External credits: @macrovector
Suppose that the surface area S and volume V of a penguin are approximated by the following rational expressions. S & = 2π r^3+2π r^2h/3r-6 [1.2em] V & = π r^4h-2π r^3h/3r^2-12r+12 Simplify the complex fraction SV.
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Hint
Start by rewriting the complex fraction as a division expression and then divide the rational expressions.
Solution
The ratio of the surface area S of a penguin to its volume V needs to be simplified.
2π r^3 +2π r^2h3r-6/π r^4h-2π r^3 h3r^2-12r+12
This complex fraction can be rewritten as a division expression.
2π r^3 +2π r^2h/3r-6 ÷ π r^4h-2π r^3 h/3r^2-12r+12
Dividing two rational expressions is the same as multiplying the first expression by the reciprocal of the second expression.
2π r^3 +2π r^2h/3r-6 ÷ π r^4h-2π r^3 h/3r^2-12r+12 [0.6em] ⇕ [0.6em]
2π r^3 +2π r^2h/3r-6 * 3r^2-12r+12/π r^4h-2π r^3 h
Now the multiplication can be performed. To do so, begin by factoring the numerators and denominators of each expression, if possible. The first expression will be factored first.
2π r^3 +2π r^2h/3r-6
FactorOut
Factor out 2π r^2
2π r^2(r +h)/3r-6
FactorOut
Factor out 3
2π r^2(r +h)/3(r-2)
Now the second rational expression will be factored.
3r^2-12r+12/π r^4h-2π r^3 h
▼
Factor the numerator
FactorOut
Factor out 3
3(r^2-4r+4)/π r^4h-2π r^3 h
SplitIntoFactors
Split into factors
3(r^2-2(2)r+4)/π r^4h-2π r^3 h
CommutativePropMult
Commutative Property of Multiplication
3(r^2-2r(2)+4)/π r^4h-2π r^3 h
WritePow
Write as a power
3(r^2-2r(2)+2^2)/π r^4h-2π r^3 h
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
3(r-2)^2/π r^4h-2π r^3 h
FactorOut
Factor out π r^2h
3(r-2)^2/π r^3h(r-2)
Next, the numerators and denominators are multiplied and the common factors are eliminated.
2π r^2(r +h)/3(r-2)* 3(r-2)^2/π r^3h(r-2)
MultFrac
Multiply fractions
2π r^2(r +h)3(r-2)^2/3(r-2) π r^3h(r-2)
▼
Simplify
Write power as a product
2π r^2(r +h) 3(r-2) (r-2)/3(r-2) π r^2 r h(r-2)
CancelCommonFac
Cancel out common factors
2π r^2(r +h) 3(r-2) (r-2)/3(r-2) π r^2 r h(r-2)
SimpQuot
Simplify quotient
2(r+h)/rh
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Closure
Finding Combined Fuel Efficiency of a New Car Model
With the methods seen in the lesson, the challenge given at the beginning can finally be solved. Recall the combined fuel economy formula. C = 1/12 ( 1x+ 1y ) In the formula, x represents miles per gallon (mpg) in the city and y represents miles per gallon on the highway.
a If a car travels 28 mpg in the city and 36 mpg on the highway, what is the combined fuel economy for the car?
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b Let x be the city fuel economy for a new car model. If the highway fuel economy, in terms of x, is 10xx-20, write a simplified expression for the combined fuel economy for such a car.
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Hint
a Substitute the given values and evaluate the formula.
b Substitute the given expression for y and simplify it. When 1 is divided by a rational expression, the result is the reciprocal of the the rational expression.
Solution
a A car travels at 28 mpg in the city and 36 mpg on the highway. The combined fuel economy can be found by substituting these values into the given formula.
C = 1/12 ( 1x+ 1y )
SubstituteII
x= 28, y= 36
C = 1/12 ( 1 28+ 1 36 )
▼
Simplify denominator
ExpandFrac
a/b=a * 9/b * 9
C = 1/12 ( 9252+ 136 )
ExpandFrac
a/b=a * 7/b * 7
C = 1/12 ( 9252 + 7252 )
AddFrac
Add fractions
C = 1/12 ( 16252 )
MultFrac
Multiply fractions
C = 1/16504
DivOneByFracD
1/a/b= b/a
C = 504/16
CalcQuot
Calculate quotient
C = 31.5
The combined fuel economy for the car is 31.5 miles per gallon.
b In this case, the rational expression 10xx-20 will be substituted for y.
C = 1/12 ( 1x + 1 10xx-20 ) Recall that when 1 is divided by a rational expression, the result is the reciprocal of the the rational expression. This fact can be used twice to simplify the right-hand side of the equation.
C = 1/12 ( 1x + 1 10xx-20 )
▼
Simplify denominator
DivOneByFracSL
.1 /a/b.=b/a
C = 1/12 ( 1x + x-2010x )
ExpandFrac
a/b=a * 10/b * 10
C = 1/12 ( 1010x + x-2010x )
AddFrac
Add fractions
C = 1/12 ( x-1010x )
MultFrac
Multiply fractions
C = 1/x-1020x
DivOneByFracSL
.1 /a/b.=b/a
C= 20x/x-10
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Multiplying and Dividing Rational Expressions
Exercise 2.1
Find the expression that makes the following statement true. x-3/x^2+3x-18 ÷ $ $/x^2+4x-21 = x+7/x+6
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We are asked to find the numerator of the divisor in our equation. We will first first solve an equation for the divisor, and then find its numerator. Let's rearrange the equation to isolate the divisor. Dividend/Divisor= Quotient ⇕ Dividend/Quotient = Divisor Let's apply this fact to rearrange our equation. For simplicity, let Y be the expression we are trying to find. x-3x^2+3x-18/Yx^2+4x-21= x+7/x+6 ⇕ x-3x^2+3x-18/x+7x+6 = Y/x^2+4x-21 Let's now divide the rational expressions on the left-hand side of our new equation. To do so, we will start by factoring the denominator of the dividend.
Let's substitute this expression into our equation. 1/x+7 = Y/x^2+4x-21 In order to solve for Y we can use the Cross Products Property. Let's first factor the denominator of the right-hand side of this equation.
Now we will cross multiply our equation to solve for Y.
We see that the expression x-3 makes the given statement true. x-3/x^2+3x-18 ÷ $x-3$/x^2+4x-21 = x+7/x+6
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We want to multiply and divide the given rational expressions. 2x^2 + 5x - 3/2x^2 + x - 10 * (2x + 5) ÷ 2x - 1/3x - 6 We will begin by factoring the numerators and denominators of each expression if possible, starting with the first.
The second expression cannot be factored further. Let's factor the last expression.
Recall that dividing by a rational expression is the same as multiplying by the reciprocal of the expression. (2x - 1)(x + 3)/(2x + 5)(x - 2) * (2x + 5) ÷ 2x - 1/3(x - 2) ⇕ (2x - 1)(x + 3)/(2x + 5)(x - 2) * (2x + 5) * 3(x - 2)/2x - 1 Let's now multiply all expressions and cancel out any common factors.
We simplified the given expression.
We will identify the restrictions on the variable from the denominator of the original expression and from any other denominator used. For simplicity we will use the factored forms.
| Denominator | Restrictions on the Denominator | Restrictions on the Variable |
|---|---|---|
| (2x+5)(x-2) | 2x+5≠ 0 and x - 2 ≠ 0 | x ≠ - 5/2 and x ≠ 2 |
| 3x-6 | 3x-6 ≠ 0 | x ≠ 2 |
| (2x+5)(x-2)(2x-1) | 2x+5≠ 0, x - 2 ≠ 0, and 2x - 1 ≠ 0 | x ≠ - 5/2, x ≠ 2, and x≠ 1/2 |
We found three unique restrictions on the variable. x≠ - 5/2, x ≠ 1/2, x ≠ 2
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Determine whether the given statement is always, sometimes, or never true.
|
Restrictions on variables change when a rational expression is simplified. |
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To determine if the given statement is always, sometimes, or never true, let's recall the definition of a simplified rational expression.
Rational Expression in Simple Form
|-
A rational expression is in simplest form when its numerator and denominator have no common factors.
We factor the numerator and denominator of a rational expression in order to find these common factors. A rational expression and any simplified form must have the same domain in order to be equivalent. Let's see an example.
| Rational Expression | Simplified Form | |
|---|---|---|
| Expression | x^2-4/x-2 | x+2 |
| Domain | x≠ 2 | x ≠ 2 |
As we can see, the restrictions on the variables never change when a rational expression is simplified.
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Zosia performs an operation on rational expressions as shown below.
Describe the error in her work. A. & (x+9)(x-9) is not the factored & form ofx^2-81. B. & The first expression should be multiplied by & the reciprocal of the second expression. C. & The factors(x-3) and(3-x)do not & cancel each other out. D. & The domain should contain -9.
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We are asked to describe the error in finding the product of two rational expressions. To do so, let's pay close attention to Zosia's work.
In the second line, we see that the factors (x-3) and (3-x) are canceled out. However, the signs of these factors are not the same, so we must factor out a - 1 in one of the factors in order to create common factors in the numerator and denominator. (3-x)= -1 (-3+x) ⇕ (3-x)=-1 (x-3) Let's substitute -1 (x-3) for (3-x) into the second line of Zosia's and cancel out any common factors.
The product of the expressions is equal to - x+9. Also, 3 and - 9 are the excluded values. As a result, the factors (x-3) and (3-x) cannot be canceled out. The answer is C.
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Multiplying and Dividing Rational Expressions
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