 ### Simplify Rational Expressions

After studying this section, you will be able to:

1. Simplify an algebraic fraction by factoring.

Recall that a rational number is a number that can be written as one integer divided by another integer such as 3 ÷ 4 or 3/4. We usually use the word fraction to mean 3/4 We can extend this idea to algebraic expressions. A rational expression is an algebraic expression divided by another algebraic expression such as

(3x + 2) ÷ (x +4)   or  (3x+2)/(x+4)

The last fraction is sometimes called a fractional algebraic expression. There is a special restriction for all fractions, including fractional algebraic expressions. The denominator of the fraction cannot be 0. For example, in the expression (3x+2)/(x+4) the denominator cannot be 0. Therefore, the value of x cannot be -4. The following important restriction will apply throughout this lesson. We state it here to avoid having to mention it repeatedly throughout this lesson.

We have discovered that fractions can be simplified (or reduced) in the following way.

15/25=(3*5)/(5*5)=3/5

This is sometimes referred to as the basic rule of fractions and can be stated as follows:

We will examine several examples where a, b, c are real numbers, as well as more involved examples where a, b, and c are polynomials. In either case we shall make extensive use of our factoring skills in this section.

There is one essential property that is revealed by the basic rule of fractions. It is this: If the numerator and denominator of a given fraction are multiplied by the same quantity, an equivalent fraction is obtained. The rule can be used two ways. You can start with (ac)/(bc) and end with the equivalent fraction a/b. Or, you can start with a/b and end with the equivalent fraction (ac)/(bc).

EXAMPLE 1

(a) Write a fraction equivalent to 3/5 with a denominator of 10.

(b) Reduce 21/39.

(a) 3/5=(39*2)/(5*2)=6/10   Use the rule a/b=(ac)/(bc). Let c = 2 since 5*2 = 10.

(b) 21/39 = (7*3)/(13*3)=7/13   Use the rule (ac)/(bc)=a/b. Let c = 3 because 3 is the greatest common factor of 21 and 39.

Simplifying Algebraic Fractions

The process of reducing the fraction shown above is sometimes called dividing out common factors. Remember, only factors of both the numerator and the denominator can be divided out. Now, to apply this rule, it is usually necessary that the numerator and denominator of the fraction be completely factored. You will need to use your factoring skills from earlier tutorial to accomplish this step. When you are applying this rule you are simplifying the fraction .

EXAMPLE 2. Simplify, (4x+12)/(5x+15)

(4x+12)/(5x+15)=(4(x+3))/(5(x+3))   Factor 4 from the numerator. Factor 5 from the denominator.

= 4/5        Apply the basic rule of fractions

EXAMPLE 3 Simplify. (x^2+9x+14)/(x^2-4)

= ((x+7)(x+2))/((x-2)(x+2))   Factor both the numerator and the denominator.

= (x+7)/(x-2)   Apply the basic rule of fractions.

Some problems may involve more than one step of factoring. Always remember to remove any common factors as the first step, if it is possible to do so.

EXAMPLE 4 Simplify. (9x-x^3)/(x^3+x^2-6x)

= (x(9-x^2))/(x(x^2+x-6))

Remove a common factor from the polynomial in the numerator and in the denominator.

= (x(3+x)(3-x))/(x(x+3)(x-2))

Factor each polynomial and apply the basic rule of fractions. Note that (3 + x) is equivalent to (x + 3) since addition is commutative.

= (3-x)/(x-2)

Be on the lookout for a special simplifying situation. Watch for a situation where each term in the denominator is opposite in sign from each term in the numerator. In such cases you should factor -1 or another negative number from one polynomial so that the expression in each set of parentheses is equivalent. Look carefully at the following two examples.

EXAMPLE 5 Simplify. (5x-15)/(6-2x)

Notice that the variable term in the numerator, 5x, and the variable term in the denominator, -2x, are opposite in sign. Likewise, the numerical terms -15 and 6 are opposite in sign. Factor out a negative number from the denominator.

= (5(x-3))/(-2(-3+x))

Factor 5 from the numerator. Factor -2 from the denominator. Note that (x-3) and (-3 + x) are equivalent since +x -3 = -3 + x.

5/2   Apply the basic rule of fractions.

Note that 5/-2 is not considered to be in simple form. We usually avoid leaving a negative number in the denominator. Therefore, to simplify, give the result as 5/2 or -5/2.

EXAMPLE 6 Simplify. (2x^2-11x+12)/(16-x^2)

= ((x-4)(2x-3))/((4-x)(4+x))

Factor numerator and denominator. Observe that (x-4) and (4-x) are opposite in sign.

= ((x-4)(2x-3))/(-1(-4+x)(4+x))

Factor -1 out of (+4-x) to obtain -1(-4 + x)

= (2x-3)/(-1(4+x))

Apply the basic rule of fractions, since (x-4) and (-4 + x) are equivalent.

(2x-3)/(4+x)

After doing Examples 5 and 6, you will notice a pattern. Whenever the factor in the numerator and the factor in the denominator are exactly opposite in sign, the value -1 results. We could actually make this a definition of that property.

You may use this definition in reducing fractions if it is helpful to you. Otherwise, you may use the factoring method discussed earlier.

Some problems will involve two or more variables. In such cases, you will need to factor carefully and make sure that each set of parentheses contains the correct letters.

EXAMPLE 7 Simplify, (x^2-7xy+12y^2)/(2x^2-7xy-4y^2

Factor both the numerator and the denominator.

= ((x-4y)(x-3y))/((2x+y)(x-4y))

Apply the basic rule of fractions.

= (x-3y)/(2x+y)

### Multiplication and Division of Rational Expressions

After studying this section, you will be able to:

1. Multiply algebraic fractions and write the answer in simplest form.

2. Divide algebraic fractions and write the answer in simplest form.

Multiplying Algebraic Fractions

To multiply two fractional expressions, we multiply the numerators and we multiply the denominators. As before, the denominators cannot equal zero.

Simplifying or reducing fractions prior to multiplying them usually makes the problem easier to do. To do the problem the long way would certainly increase the chance for error! This long approach should be avoided.
As an example, let’s do the same problem two ways to see which way seems easier. First let’s multiply by the ‘‘long way’’:

5/7 × 49/125

5/7 × 49/125 = 245/875

Multiply the numerators and multiply the denominators

= 7/25

Reduce the fraction. (Note: It takes a bit of trial and error to discover how to reduce it.)

Compare this with the method of reducing the fractions prior to multiplying them. Let’s multiply by simplifying before multiplication:

5/7 × 49/125

Step 1. It is easier to factor first. We factor the numerator and denominator of the second fraction.

5/7 × (7*7)(5*5*5)

Step 2. We express the product as one fraction (by the definition of multiplication of fractions).

(5*7*7)/(7*5*5*5)

Step 3. Then we apply the basic rule of fractions to divide common factors of 5 and 7 that appear in the denominator and in the numerator.

(5*7*7)/(7*5*5*5)=7/25

A similar approach can be used with the multiplication of rational algebraic expressions. We first factor the numerator and denominator of each fraction wherever possible. Then we cancel any factor that is common to a numerator and a denominator. Finally, we multiply the remaining numerators and the remaining denominators.

EXAMPLE 1 Multiply. (x^2-x-12)/(x^2-16)*(2x^2+7x-4)/(x^2-4x-21)

Factor wherever possible is always the first step.

= ((x-4)(x+3))/((x-4)(x+4))*((x+4)(2x-1))/((x+3)(x-7))

Apply the basic rule of fractions. (Three pairs of factors divide out.)

= (2x-1)/(x-7)  The final answer.

In some cases, a given numerator can be factored more than once. You should always be sure that you remove the common factor first wherever possible.

EXAMPLE 2 Multiply. (x^4-16)/(x^3+4x)*(2x^2-8x)/(4x^2+2x-12)

Factor each numerator and denominator. Factoring out the common factor first is very important.

= ((x^2+4)(x^2-4))/(x(x^2+4))*(2x(x-4))/(2(2x^2+x-6))

Factor again where possible.

= ((x^2 + 4)(x + 2)(x-2))/(x(x^2+4))*(2x(x-4))/(2(x+2)(2x-3))

Remove factors that appear in both numerator and denominator. (There are four such pairs to be removed.)

= ((x-2)(x-4))/(2x-3)  or   (x^2 -6x+8)/(2x-3)

Write the answer as one fraction. (Usually, if there is more than one set of parentheses in a numerator, the answer is left in factored form.)

Dividing Algebraic Fractions

For any two fractions a/b and c/d the operation of division can be performed by inverting the second fraction and multiplying it by the first fraction.

This property holds whether a, b, c, and d are polynomials or numerical values. (It is assumed, of course, that no denominator is zero.)

In the first step for dividing two algebraic fractions, you should invert the second fraction and write the problem as a multiplication. Then you follow the procedure for multiplying algebraic fractions.

EXAMPLE 3. Divide. (6x+12y)/(2x-6y) ÷ (9x^2-36y^2)/(4x^2-36y^2)

Invert the second fraction and write the problem as the product of two fractions.

(6x+12y)/(2x-6y)* (4x^2-36y^2)/(9x^2-36y^2)

Factor each numerator and denominator.

= (6(x+2y))/(2(x-3y))*(4(x^2-9y^2))/(9(x^2-4y^2))

Factor again where possible

= ((3)(2)(x+2y))/(2(x-3y))*((2)(2)(x+3y)(x-3y))/((3)(3)(x+2y)(x-2y))

Remove factors that appear in both numerator and denominator.

((2)(2)(x+3y))/(3(x-2y))

Write the result as one fraction and simplify.

= (4(x+3y))/(3(x-2y))

Usually, answers are left in this form.

Although it is correct to write this answer as (4x+12y)/(3x-6y) it is customary to leave the answer in factored form in order to ensure that the final answer is simplified.

A polynomial that is not in fraction form can be written as a fraction if you write a denominator of 1.

EXAMPLE 4  Divide. (15-3x)/(x+6) ÷ (x^2-9x + 20)

Note that x^2-9x + 20 can be written as (x^2-9x + 20)/1

Invert and multiply.

= (15-3x)/(x+6)*1/(x^2-9x + 20)

Factor where possible. Note we had to factor -3 from the first numerator so that we would have a common factor with the second denominator.

= (-3(-5+x))/(x+6)*1/((x-5)(x-4))

Divide out the common factor. (-5 + x) is equivalent to (x-5) and the final answer

= -3/((x+6)(x-4))

Note that the answer can be written in several equivalent forms.

-3/((x+6)(x-4)) or

3/((x+6)(x-4)) or

3/((x+6)(4-x))

Take a minute to study each. Each is equivalent to the others.

A word of caution:

It is logical to assume that the types of problems encountered in Section 6.2 have at least one common factor that can be divided out. Therefore if after factoring you do not observe any common factors, you should be somewhat suspicious. In such cases, it would be wise to double check your factoring steps to see if an error has been made.

### Addition and Subtraction of Rational Expressions

After studying this section, you will be able to:

1. Add and subtract algebraic fractions with the same denominator.

2. Determine the LCD for two or more algebraic fractions with different denominators.

3. Add or subtract algebraic fractions with different denominators.

Adding and Subtracting Algebraic Fractions with the Same Denominator

If fractional algebraic expressions have the same denominator, they can be combined in a fashion similar to that used in arithmetic. The numerators are added or subtracted and the denominator remains the same.

EXAMPLE 1 Add. (5a)/(a+2b)+(6a)/(a+2b)

Note the denominators are the same. Add the numerators.

(5a)/(a+2b)+(6a)/(a+2b)=(5a+6a)/(a+2b)=(11a)/(a+2b)

EXAMPLE 2 Subtract, (3x)/((x+y)(x-2y))-(8x)/((x+y)(x-2y))

Write as one fraction and simplify.

(3x)/((x+y)(x-2y))-(8x)/((x+y)(x-2y))=(3x-8x)/((x+y)(x-2y))

= (-5x)/((x+y)(x-2y))

Determining the LCD

How do we add or subtract fractional expressions when the denominators are not the same? First we must find the least common denominator (LCD). You need to be clear on how to find a least common denominator and how to add and subtract fractions from arithmetic before you attempt this lesson.

EXAMPLE 3 Find the LCD of the following algebraic fractions. 5/(2x-4), 6/(3x-6)

Factor each denominator.

2x-4 = 2(x-2)

3x-6 = 3(x-2)

The different factors are 2, 3, and (x-2). Since no factor is repeated more than once in any one denominator, the LCD is the product of these three factors.

LCD = (2)(3)(x-2) = 6(x-2)

EXAMPLE 4 Find the LCD.

(a) 5/(12ab^2c), 13/(18a^3bc^4)  (b) 7/(x^2-2x), 9/(x^2-4x+4)  (c) 8/(x^2-5x+4), 12/(x^2+2x-3)

If a factor occurs more than once in any one denominator, the LCD will contain that factor repeated the greatest number of times that it occurs in any One denominator.

Adding and Subtracting Fractions with Different Denominators

If two fractions have different denominators, we first change them to equivalent fractions with the least common denominator. Then we add or subtract the numerators and keep the common denominator.

EXAMPLE 5 Add. 5/(xy)+2/y

The denominators are different. Find the LCD. The two factors are x and y. We observe that the LCD is xy.

5/(xy)+2/y=5/(xy)+2/y*x/x   Multiply the second fraction by x/x.

= 5/(xy)+(2x)/(xy)  Now each fraction has a common denominator of xy.

= (5+2x)/(xy)  Write the sum as one fraction.

EXAMPLE 6 Add.  (3x)/((x+y)(x-y))+5/(x+y)

The two factors are (x + y) and (x-y). We observe that the LCD = (x + y)(x-y). Multiply the second fraction by (x-y)/(x-y).

= (3x)/((x+y)(x-y))+5/(x+y)*(x-y)/(x-y)

Now each fraction has a common denominator of (x + y)(x-y).

= (3x)/((x+y)(x-y))+(5x-5y)/((x+y)(x-y))

Write the sum of the numerators over one common denominator.

= (3x+5x-5y)/((x+y)(x-y))

Collect like terms.

= (8x-5y)/(x+y)(x-y)

It is important to remember that the LCD is the smallest algebraic expression into which each denominator can be divided. For algebraic expressions the LCD must contain each factor that appears in any denominator. If the factor is repeated, the LCD must contain each factor the greatest number of times that it appears in any one denominator.

EXAMPLE 7  Add. 5/(xy^2)+3/(x^2y)+2/y^3

The x factor is squared in one fraction. The y factor is cubed in one fraction. Therefore, the LCD is x^2y^3.
Multiply each fraction by the appropriate value to obtain x^2y^3 in each denominator. (Note: Here we always multiply the numerator and denominator of any one fraction by the same value to obtain an equivalent fraction.)

= 5/(xy^2)*(xy)/(xy)+3/(x^2y)*y^2/y^2+2/y^3*x^2/x^2

Now all fractions have a common denominator.

= (5xy)/(x^2y^3)+(3y^2)/(x^2y^3)+(2x^2)/(x^2y^3)

Write the sum as one fraction.

= (5xy+3y^2+2x^2)/(x^2y^3)

In many cases, the denominators in an addition or subtraction problem are not in factored form. You must factor each denominator in order to determine the LCD. Collect like terms in the numerator; then look to see if that final numerator can be factored. If so, the fraction can sometimes be simplified.

EXAMPLE 8 Add. 5/(x^2-y^2)+(3x)/(x^3+x^2y)

5/(x^2-y^2)+(3x)/(x^3+x^2y)=5/((x+y)(x-y))+(3x)/(x^2(x+y))

Factor the two denominators. Observe that the LCD is x^2(x + y)(x-y).

Multiply each fraction by the appropriate value to obtain a common denominator of x^2(x + y)(x-y).

= 5/((x+y)(x-y))*x^2/x^2+(3x)/(x^2(x+y))*(x-y)/(x-y)

= (5x^2)/(x^2(x+y)(x-y))+(3x^2-3xy)/(x^2(x+y)(x-y))

Write the sum of the numerators over one common denominator.

= (5x^2+3x^2-3xy)/(x^2(x+y)(x-y))

Collect like terms.

= 8x^2-3xy)/(x^2(x+y)(x-y))

Remove the common factor x in the numerator and denominator and simplify.

= (x(8x-3y))/(x^2(x+y)(x-y))

= (8x-3y)/(x(x+y)(x-y))

It is very easy to make a mistake in sign when subtracting two fractions. You will find it helpful to place parentheses around the numerator of the second fraction so that you will not forget to subtract the entire numerator.

EXAMPLE 9 Subtract. (3x+4)/(x-2)-(x-3)/(2x-4)

Factor the second denominator and observe that the LCD is 2(x-2).

= (3x+4)/(x-2)-(x-3)/(2(x-2))

Multiply the first fraction by 2/2 so that the resulting fraction will have the common denominator.

= 2/2*(3x+4)/(x-2)-(x-3)/(2(x-2))

Write the indicated subtraction as one fraction. Note the parentheses around (x-3).

= (2(3x+4)-(x-3))/(2(x-2)

Remove the parentheses in the numerator.

= (6x+8-x+3)/(2(x-2))

Collect like terms.

= (5x+11)/(2(x-2))

To avoid making errors when subtracting two fractions, some students find it helpful to change subtraction to addition of the opposite of the second fraction.

EXAMPLE 10 Subtract. (8x)/(x^2-16)-4/(x-4)

Factor the first denominator. Use the property that a/b-c/b=a/b+(-c)/b

(8x)/(x^2-16)-4/(x-4)=(8x)/((x+4)(x-4))+(-4)/(x-4)

Multiply the second fraction by (x+4)/(x+4)

= (8x)/((x+4)(x-4))+(-4)/(x-4)*(x+4)/(x+4)

Write the addition of the numerators over one common denominator.

= (8x+(-4)(x+4))/((x+4)(x-4))

Remove parentheses.

= (8x-4x-16)/((x+4)(x-4))

Collect like terms. Note that the numerator can be factored.

= (4x-16)/((x+4)(x-4))

= (4(x-4))/((x+4)(x-4))

Divide out. Since (x-4) is a factor of the numerator and the denominator, we may divide out the common factor.

= 4/(x+4)

### Simplify Complex Rational Expressions

After studying this section, you will be able to:

1. Simplify complex fractions by adding or subtracting in the numerator and denominator.

2. Simplify complex fractions by multiplying by the LCD of all the denominators.

Simplifying Complex Fractions by Adding or Subtracting in the Numerator and the Denominator

A complex fraction has a fraction in the numerator or in the denominator, or both.

(3+2/x)/(x/7+2)  (x/y+1)/2  ((a+b)/3)/((x-2y)/4)

The bar in a complex fraction is both a grouping symbol and a symbol for division.

((a+b)/3)/((x-2y)/4) is equivalent to ((a+b)/3) ÷ ((x-2y)/4)

We need a procedure for simplifying complex fractions.

EXAMPLE 1 Simplify. (1/x)/(2/y^2+1/y)

Step 1 Add the two fractions in the denominator.

= (1/x)/(2/y^2+1/y*y/y)=(1/x)/((2+y)/y^2)

Step 2 Divide the fraction in the numerator by the fraction in the denominator.

= 1/x ÷ (2+y)/y^2=1/x*y^2/(2+y)=y^2/(x(2+y)

In some problems there may be two or more fractions in the numerator and the denominator.

EXAMPLE 2 Simplify. (1/x+1/y)/(3/a-2/b)

We observe that the LCD of the fractions in the numerator is xy. The LCD of the fractions in the denominator is ab.

Multiply each fraction by the appropriate value to obtain common denominators.

= (1/x*y/y+1/y*x/x)/(3/a*b/b-2/b*a/a

Add the two fractions in the numerator and subtract the two fractions in the denominator.

= (y+x)/(xy))/((3b-2a)/(ab))

Invert the fraction in the denominator and multiply it by the numerator.

= (y+x)/(xy)*(ab)/(3b-2a)

Write the answer as one fraction.

= (ab(y+x))/(xy(3b-2a))

In some problems, factoring may be necessary to determine the LCD and to combine fractions.

EXAMPLE 3 Simplify. (1/(x^2-1)+2/(x+1))/x

We observe the need for factoring x^2-1.

The LCD for the fractions in the numerator is (x + 1)(x-1).

= (1/((x+1)(x-1))+2/(x+1)*(x-1)/(x-1))/x

Add the two fractions in the numerator.

= ((1+2x-2)/((x+1)(x-1)))/x

Simplify the numerator. Invert the fraction in the denominator and multiply.

= (2x-1)/((x+1)(x-1))*1/x

Write the answer as one fraction.

(2x-1)/(x(x+1)(x-1))

In simplifying complex fractions, always be alert to see if the final fraction can be reduced or simplified.

EXAMPLE 4 Simplify. (3/(a+b)-3/(a-b))/(5/(a^2-b^2))

The LCD of the two fractions in the numerator is (a + b)(a-b).

(3/(a+b)*(a-b)/(a-b)-3/(a-b)*(a+b)/(a+b))/(5/(a^2-b^2))=((3a-3b)/((a+b)(a-b))-(3a+3b)/((a+b)(a-b)))/(5/(a^2-b^2))

Study carefully how we combine the two fractions in the numerator. Do you see how we obtain -6b?

= ((-6b)/((a+b)(a-b)))/(5/((a+b)(a-b)))    Factor a^2-b^2 as (a+b)(a-b)

= (-6b)/((a+b)(a-b))*((a+b)(a-b))/5

Note that since (a + b)(a-b) are factors in both numerator and denominator they may be divided out.

= (-6b)/5

Simplifying Complex Fractions by Multiplying by the LCD of All the Denominators

There is another way to simplify complex fractions: Multiply the numerator and denominator of the complex fraction by the least common denominator of all the denominators appearing in the complex fraction.

EXAMPLE 5 Simplify by multiplying by the LCD. (5/(ab^2)-2/(ab))/(3-5/(2a^2b))

The LCD of all the denominators in the complex fraction is 2a^2b^2. Multiply each term by 2a^2b^2.

(2a^2b^2(5/(ab^2)-2/(ab)))/(2a^2b^2(3-5/(2a^2b)))=(2a^2b^2(5/(ab^2))-2a^2b^2(2/(ab)))/(2a^2b^2(3)-2a^2b^2(5/(2a^2b)))

Simplify.

= (10a-4b)/(6a^2b^2-5b)

So that you can compare the two methods, we will redo Example 4 by the alternative method.

EXAMPLE 6  Simplify.  (3/(a+b)-3/(a-b))/(5/(a^2-b^2))  Use alternative method.

The LCD of all individual fractions contained in the complex fraction is (a+b)(a-b). Multiply each term by the LCD.

((a+b)(a-b)(3/(a+b))-(a+b)(a-b)(3/(a-b)))/((a+b)(a-b)(5/((a+b)(a-b))

Simplify.

= (3(a-b)-3(a+b))/5

Remove parentheses.

= (3a-3b-3a-3b)/5

Simplify.

= (-6b)/5

### Equations Involving Rational Expressions

After studying this section, you will be able to:

1. Solve equations involving algebraic fractions that have solutions.

2. Determine if an equation involving algebraic fractions has no solution.

Solving Equations Involving Algebraic Fractions

In earlier lesson we developed procedures to solve linear equations containing fractions whose denominators were numerical values. In this section we use a similar approach to solve equations containing fractions whose denominators are polynomials.

EXAMPLE 1 Solve for x and check your solution. 5/x+2/3=2-2/x-1/6

We observe that the LCD is 6x. Multiply each term by 6x.

6x(5/x)+6x(2/3)=6x(2)-6x(2/x)-6x(1/6)

Simplify. Do you see how each term is obtained?

30+4x=12x-12-x

Collect like terms.

30+4x=11x-12

Subtract 4x from both sides.

30=7x-12

Add 12 to both sides.

42=7x

Divide both sides by 7.

6=x

Check 5/6+2/3 ≟ 2-2/6-1/6   Replace each x by 6.

5/6+4/6 ≟ 12/6-2/6-1/6

9/6=9/6  ✔  It checks

It is sometimes necessary to factor denominators before the correct LCD can be determined.

EXAMPLE 2 Solve and check your solution. 3/(x+5)-1=(4-x)/(2x+10)

We observe a need to factor the 2x + 10. We determine that the LCD is 2(x + 5).

3/(x+5)-1=(4-x)/(2(x+5))

Multiply each term by the LCD.

2(x+5)(3/(x+5))-2(x+5)(1)=2(x+5)[(4-x)/(2(x+5)]

Simplify.

2(3)-2(x+5)=4-x

Remove parentheses.

6-2x-10=4-x

Collect like terms.

-2x-4=4-x

Add 2x to both sides.

-4=4+x

Subtract 4 from both sides.

-8=x

Check :  Replace each x in the original equation by -8.

3/(-8+5)-1 ≟ (4-(-8))/(2(-8)+10

-3/3-1 ≟ (4+8)/(-16+10)

-1-1 ≟ 12/-6

-2 = -2  ✔  It checks.   x = -8 is the solution.

Determining If an Equation Involving Algebraic Fractions Has No Solution

Linear equations containing fractions that have variables in denominators sometimes appear to have solutions when in fact they do not. By this we mean that the "solutions" we get by using completely correct methods are, in actuality, not solutions.

In the case where a value makes a denominator in the equation equal to zero, we say there is no solution to the equation. Such a value is called an extraneous solution. An extraneous solution is an apparent solution that does not satisfy the original equation. It is important that you check the number in the original equation.

EXAMPLE 3 Solve and check. y/(y-2)-4=2/(y-2)

We observe that the LCD is y-2. Multiply each term by (y-2).

(y-2)(y/(y-2))-(y-2)(4)=(y-2)(2/(y-2))

Simplify. Do you see how this is done?

y-4(y-2)=2

Remove parentheses.

y-4y+8=2

Collect like terms.

-3y+8=2

Subtract 8 from both sides.

-3y=-6

Divide both sides by negative 3.

(-3y)/-3=-6/-3

y=2

y=2 is only an apparent solution.

There is no solution to this problem.

Why? We can see immediately that y = 2 is not a possible solution for the original equation. The use of the value y = 2 in a denominator would make the denominator equal to zero, and the expression is undefined.

Check.  Suppose that you try to check the apparent solution by substituting y = 2.

2/(2-2)-4 ≟ 2/(2-2)

This does not check since you do not obtain a real number when you divide by zero.

There is no such number as 2 ÷ 0. We see that y = 2 does not check. There is no solution to this problem.

### Ratio, Proportion, and Other Applied Problems

After studying this section, you will be able to:

1. Solve problems involving ratio and proportion.

2. Solve problems involving similar triangles.

3. Solve certain distance problems.

4. Solve work problems.

Solving Problems Involving Ratio and Proportion

A ratio is a comparison of two quantities. You may be familiar with ratios that compare miles to hours or miles to gallons. A ratio is often written as a quotient in the form of a fraction. For example, the ratio of 7 to 9 can be written as 7/9.

A proportion is an equation that states that two ratios are equal. For example,

7/9=21/27    2/3=10/15    a/b=c/d

Let’s take a closer look at the last proportion. We can see that the LCD of the fractional equation is bd.

(bd)a/b =(bd)c/d   Multiply each side by the LCD.

da = bc

ad = bc   Since multiplication is commutative, da = ad.

Thus we have proved the following:

This is sometimes called cross-multiplying. It can be applied only if you have one fraction and nothing else on each side of the equation.
We can use a proportion and the technique of cross-multiplying to solve a variety of
applied problems involving two ratios.

EXAMPLE 1 Michael took 5 hours to drive 245 miles on the turnpike. If he continues at the same rate, how many hours will it take him to drive a distance of 392 miles?

1. Understand the problem.

Let x = the number of hours it will take to drive 392 miles. If 5 hours are needed to drive 245 miles, then x hours is needed to drive 392.

2. Write an equation.

We can write this as a proportion. Compare time to distance in each ratio.

3. Solve the equation and state the answer.

5(392) = 245x   Cross-multiply.

1960/245=x   Divide both sides by 245

8=x
It would take Michael 8 hours to drive 392 miles.

EXAMPLE 2 If 3/4 inch on a map represents an actual distance of 20 miles, how long is the distance represented by 41/8 inches on the same map?

Let x = the distance represented by 41/8 inches.

Cross-multiply.

(3/4)(x) = (20)(41/8)

Write 41/8 as 33/8 and simplify.

(3/4)(x)=(20)(33/8)

Multiplication of fractions.

(3x)/4=165/2

Multiply each side by 4.

3x=330

Divide both sides by 3.

x=110

41/8 inches on the map represents an actual distance of 110 miles.

Solving Problems Involving Similar Triangles

Similar triangles are triangles that have the same shape, but may be a different size. For example, if you draw a triangle on a sheet of paper and place the paper in a photocopy machine and make a copy that is reduced by 25%, you would create a triangle that was similar to the original triangle. The two triangles would have the same shape. The corresponding sides of the triangles would be proportional.

You can use the proportion equation to show that the corresponding sides of the above triangles are proportional. In fact, you can use the proportion equation to find the length of a side of a triangle if it is not known.

EXAMPLE 3 Find the length of side x in the two similar triangles pictured below.

32x = (9)(15)   Cross-multiply.

32x = 135   Multiply.

x= 135/32   Divide both sides by 32.

or x = 47/32

We can also use similar triangles for indirect measurement, for instance, to find the measure of an object that is too tall to measure using standard measuring devices. When the sun shines on two vertical objects at the same time, the shadows and the objects form similar triangles.

EXAMPLE 4 A woman who is 5 feet tall casts a shadow that is 8 feet long. At the same time of day, a building casts a shadow that is 72 feet long. How tall is the building?

1. Understand the problem.

First we draw a sketch.

We do not know the height of the building, so we call it x.

2. Write an equation and solve.

5/8=x/72

(5)(72) = 8x   Cross-multiply.

360 = 8x

45 =x

The height of the building is 45 feet.

In problems such as Example 4 we are assuming that the building and the person are standing exactly perpendicular to the ground. In other words, each triangle is assumed to be a right triangle. In some similar problems of this type, if the triangle is not a right triangle you must be careful to be sure that the angle between the object and the ground is the same in each case.

Solving Distance Problems Involving Algebraic Fractions

Some distance problems are solved using rational equations. We will need the formula

Distance = Rate × Time, D = RT, which we can write in the form T = D/R

EXAMPLE 5 Plane A flies at a speed that is 50 kilometers per hour faster than plane B. Plane A flies 500 kilometers in the amount of time that plane B flies 400 kilometers. Find the speed of each plane.

1. Understand the problem.

Let s = the speed of plane B in kilometers per hour.

Let s + 50 = the speed of plane A in kilometers per hour.

Make a simple table for D, R, and T.

Since T = D/R we divide the expression for D by the expression for R and write it in the table in the column for time.

2. Write an equation and solve.

Each plane flies for the same amount of time. That is, the time for plane A equals the time for plane B.

500/(s+50)=400/s

You can solve this equation using the methods you learnt in earlier lesson or you may cross- multiply. Here we will cross-multiply.

500s = (s + 50)(400)   Cross-multiply.

500s = 400s + 20,000   Remove parentheses.

100s = 20,000   Subtract 400s from each side.

s = 200   Divide each side by 100.

Plane B travels 200 kilometers per hour. Since

s + 50 = 200 + 50 = 250

plane A travels 250 kilometers per hour.

Solving Work Problems

Some applied problems involve the length of time needed to do a job. These problems are often referred to as work problems.

EXAMPLE 6 Reynaldo can sort a huge stack of mail on an old sorting machine in 9 hours. His brother Carlos can sort the same amount of mail using a newer sorting machine in 8 hours. How long would it take them to do the job working together? Express your answer in hours and minutes. Round to the nearest minute.

1. Understand the problem.

Let’s do a little reasoning.
If Reynaldo can do the job in 9 hours, then in 1 hour he could do 1/9 of the job.
If Carlos can do the job in 8 hours, then in 1 hour he could do 1/8 of the job.
Let x = the number of hours it takes Reynaldo and Carlos to do the job together.
In 1 hour together they could do 1/x of the job.

2. Write an equation and solve.

The amount of work Reynaldo can do in 1 hour plus the amount of work Carlos can do in 1 hour must be equal to the amount of work they could do together in 1 hour.

We observe the LCD is 72x. Multiply each term by the LCD.

72x(1/9)+72x(1/8)=72x(1/x)

Simplify.

8x+9x=72

Collect like terms.

17x=72

Divide each side by 17.

x=72/17

x = 44/17

To change 4/17 of an hour to minutes we multiply.

To the nearest minute this is 14 minutes. Thus doing the job together will take 4 hours and 14 minutes.