lesson 4 homework practice dividing rational numbers

Chapter 3, Lesson 4: Dividing Rational Numbers

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7.1 Multiply and Divide Rational Expressions

Learning objectives.

By the end of this section, you will be able to:

• Determine the values for which a rational expression is undefined
• Simplify rational expressions
• Multiply rational expressions
• Divide rational expressions
• Multiply and divide rational functions

Be Prepared 7.1

Before you get started, take this readiness quiz.

Simplify: 90 y 15 y 2 . 90 y 15 y 2 . If you missed this problem, review Example 5.13 .

Be Prepared 7.2

Multiply: 14 15 · 6 35 . 14 15 · 6 35 . If you missed this problem, review Example 1.25 .

Be Prepared 7.3

Divide: 12 10 ÷ 8 25 . 12 10 ÷ 8 25 . If you missed this problem, review Example 1.26 .

We previously reviewed the properties of fractions and their operations. We introduced rational numbers, which are just fractions where the numerators and denominators are integers. In this chapter, we will work with fractions whose numerators and denominators are polynomials. We call this kind of expression a rational expression .

Rational Expression

A rational expression is an expression of the form p q , p q , where p and q are polynomials and q ≠ 0 . q ≠ 0 .

Here are some examples of rational expressions:

Notice that the first rational expression listed above, − 24 56 − 24 56 , is just a fraction. Since a constant is a polynomial with degree zero, the ratio of two constants is a rational expression, provided the denominator is not zero.

We will do the same operations with rational expressions that we did with fractions. We will simplify, add, subtract, multiply, divide and use them in applications.

Determine the Values for Which a Rational Expression is Undefined

If the denominator is zero, the rational expression is undefined. The numerator of a rational expression may be 0—but not the denominator.

When we work with a numerical fraction, it is easy to avoid dividing by zero because we can see the number in the denominator. In order to avoid dividing by zero in a rational expression, we must not allow values of the variable that will make the denominator be zero.

So before we begin any operation with a rational expression, we examine it first to find the values that would make the denominator zero. That way, when we solve a rational equation for example, we will know whether the algebraic solutions we find are allowed or not.

Determine the values for which a rational expression is undefined.

• Step 1. Set the denominator equal to zero.
• Step 2. Solve the equation.

Example 7.1

Determine the value for which each rational expression is undefined:

ⓐ 8 a 2 b 3 c 8 a 2 b 3 c ⓑ 4 b − 3 2 b + 5 4 b − 3 2 b + 5 ⓒ x + 4 x 2 + 5 x + 6 . x + 4 x 2 + 5 x + 6 .

The expression will be undefined when the denominator is zero.

Determine the value for which each rational expression is undefined.

ⓐ 3 y 2 8 x 3 y 2 8 x ⓑ 8 n − 5 3 n + 1 8 n − 5 3 n + 1 ⓒ a + 10 a 2 + 4 a + 3 a + 10 a 2 + 4 a + 3

ⓐ 4 p 5 q 4 p 5 q ⓑ y − 1 3 y + 2 y − 1 3 y + 2 ⓒ m − 5 m 2 + m − 6 m − 5 m 2 + m − 6

Simplify Rational Expressions

A fraction is considered simplified if there are no common factors, other than 1, in its numerator and denominator. Similarly, a simplified rational expression has no common factors, other than 1, in its numerator and denominator.

Simplified Rational Expression

A rational expression is considered simplified if there are no common factors in its numerator and denominator.

For example,

We use the Equivalent Fractions Property to simplify numerical fractions. We restate it here as we will also use it to simplify rational expressions.

Equivalent Fractions Property

If a , b , and c are numbers where b ≠ 0 , c ≠ 0 , b ≠ 0 , c ≠ 0 ,

Notice that in the Equivalent Fractions Property, the values that would make the denominators zero are specifically disallowed. We see b ≠ 0 , c ≠ 0 b ≠ 0 , c ≠ 0 clearly stated.

To simplify rational expressions, we first write the numerator and denominator in factored form. Then we remove the common factors using the Equivalent Fractions Property.

Be very careful as you remove common factors. Factors are multiplied to make a product. You can remove a factor from a product. You cannot remove a term from a sum.

Removing the x ’s from x + 5 x x + 5 x would be like cancelling the 2’s in the fraction 2 + 5 2 ! 2 + 5 2 !

Example 7.2

How to simplify a rational expression.

Simplify: x 2 + 5 x + 6 x 2 + 8 x + 12 x 2 + 5 x + 6 x 2 + 8 x + 12 .

Simplify: x 2 − x − 2 x 2 − 3 x + 2 . x 2 − x − 2 x 2 − 3 x + 2 .

Simplify: x 2 − 3 x − 10 x 2 + x − 2 . x 2 − 3 x − 10 x 2 + x − 2 .

We now summarize the steps you should follow to simplify rational expressions.

Simplify a rational expression.

• Step 1. Factor the numerator and denominator completely.
• Step 2. Simplify by dividing out common factors.

Usually, we leave the simplified rational expression in factored form. This way, it is easy to check that we have removed all the common factors.

We’ll use the methods we have learned to factor the polynomials in the numerators and denominators in the following examples.

Every time we write a rational expression, we should make a statement disallowing values that would make a denominator zero. However, to let us focus on the work at hand, we will omit writing it in the examples.

Example 7.3

Simplify: 3 a 2 − 12 a b + 12 b 2 6 a 2 − 24 b 2 3 a 2 − 12 a b + 12 b 2 6 a 2 − 24 b 2 .

Simplify: 2 x 2 − 12 x y + 18 y 2 3 x 2 − 27 y 2 2 x 2 − 12 x y + 18 y 2 3 x 2 − 27 y 2 .

Simplify: 5 x 2 − 30 x y + 25 y 2 2 x 2 − 50 y 2 5 x 2 − 30 x y + 25 y 2 2 x 2 − 50 y 2 .

Now we will see how to simplify a rational expression whose numerator and denominator have opposite factors. We previously introduced opposite notation: the opposite of a is − a − a and − a = −1 · a . − a = −1 · a .

The numerical fraction, say 7 −7 7 −7 simplifies to −1 −1 . We also recognize that the numerator and denominator are opposites.

The fraction a − a a − a , whose numerator and denominator are opposites also simplifies to −1 −1 .

This tells us that b − a b − a is the opposite of a − b . a − b .

In general, we could write the opposite of a − b a − b as b − a . b − a . So the rational expression a − b b − a a − b b − a simplifies to −1 . −1 .

Opposites in a Rational Expression

The opposite of a − b a − b is b − a . b − a .

An expression and its opposite divide to −1 . −1 .

We will use this property to simplify rational expressions that contain opposites in their numerators and denominators. Be careful not to treat a + b a + b and b + a b + a as opposites. Recall that in addition, order doesn’t matter so a + b = b + a a + b = b + a . So if a ≠ − b a ≠ − b , then a + b b + a = 1 . a + b b + a = 1 .

Example 7.4

Simplify: x 2 − 4 x − 32 64 − x 2 . x 2 − 4 x − 32 64 − x 2 .

Simplify: x 2 − 4 x − 5 25 − x 2 . x 2 − 4 x − 5 25 − x 2 .

Simplify: x 2 + x − 2 1 − x 2 . x 2 + x − 2 1 − x 2 .

Multiply Rational Expressions

To multiply rational expressions, we do just what we did with numerical fractions. We multiply the numerators and multiply the denominators. Then, if there are any common factors, we remove them to simplify the result.

Multiplication of Rational Expressions

If p , q , r , and s are polynomials where q ≠ 0 , s ≠ 0 , q ≠ 0 , s ≠ 0 , then

To multiply rational expressions, multiply the numerators and multiply the denominators.

Remember, throughout this chapter, we will assume that all numerical values that would make the denominator be zero are excluded. We will not write the restrictions for each rational expression, but keep in mind that the denominator can never be zero. So in this next example, x ≠ 0 , x ≠ 0 , x ≠ 3 , x ≠ 3 , and x ≠ 4 . x ≠ 4 .

Example 7.5

How to multiply rational expressions.

Simplify: 2 x x 2 − 7 x + 12 · x 2 − 9 6 x 2 . 2 x x 2 − 7 x + 12 · x 2 − 9 6 x 2 .

Simplify: 5 x x 2 + 5 x + 6 · x 2 − 4 10 x . 5 x x 2 + 5 x + 6 · x 2 − 4 10 x .

Try It 7.10

Simplify: 9 x 2 x 2 + 11 x + 30 · x 2 − 36 3 x 2 . 9 x 2 x 2 + 11 x + 30 · x 2 − 36 3 x 2 .

Multiply rational expressions.

• Step 1. Factor each numerator and denominator completely.
• Step 2. Multiply the numerators and denominators.
• Step 3. Simplify by dividing out common factors.

Example 7.6

Multiply: 3 a 2 − 8 a − 3 a 2 − 25 · a 2 + 10 a + 25 3 a 2 − 14 a − 5 . 3 a 2 − 8 a − 3 a 2 − 25 · a 2 + 10 a + 25 3 a 2 − 14 a − 5 .

Try It 7.11

Simplify: 2 x 2 + 5 x − 12 x 2 − 16 · x 2 − 8 x + 16 2 x 2 − 13 x + 15 . 2 x 2 + 5 x − 12 x 2 − 16 · x 2 − 8 x + 16 2 x 2 − 13 x + 15 .

Try It 7.12

Simplify: 4 b 2 + 7 b − 2 1 − b 2 · b 2 − 2 b + 1 4 b 2 + 15 b − 4 . 4 b 2 + 7 b − 2 1 − b 2 · b 2 − 2 b + 1 4 b 2 + 15 b − 4 .

Divide Rational Expressions

Just like we did for numerical fractions, to divide rational expressions, we multiply the first fraction by the reciprocal of the second.

Division of Rational Expressions

If p , q , r, and s are polynomials where q ≠ 0 , r ≠ 0 , s ≠ 0 , q ≠ 0 , r ≠ 0 , s ≠ 0 , then

To divide rational expressions, multiply the first fraction by the reciprocal of the second.

Once we rewrite the division as multiplication of the first expression by the reciprocal of the second, we then factor everything and look for common factors.

Example 7.7

How to divide rational expressions.

Divide: p 3 + q 3 2 p 2 + 2 p q + 2 q 2 ÷ p 2 − q 2 6 . p 3 + q 3 2 p 2 + 2 p q + 2 q 2 ÷ p 2 − q 2 6 .

Try It 7.13

Simplify: x 3 + 8 3 x 2 − 6 x + 12 ÷ x 2 − 4 6 . x 3 + 8 3 x 2 − 6 x + 12 ÷ x 2 − 4 6 .

Try It 7.14

Simplify: 2 z 2 z 2 − 1 ÷ z 3 − z 2 + z z 3 + 1 . 2 z 2 z 2 − 1 ÷ z 3 − z 2 + z z 3 + 1 .

Divide rational expressions.

• Step 1. Rewrite the division as the product of the first rational expression and the reciprocal of the second.
• Step 2. Factor the numerators and denominators completely.
• Step 3. Multiply the numerators and denominators together.
• Step 4. Simplify by dividing out common factors.

Recall from Use the Language of Algebra that a complex fraction is a fraction that contains a fraction in the numerator, the denominator or both. Also, remember a fraction bar means division. A complex fraction is another way of writing division of two fractions.

Example 7.8

Divide: 6 x 2 − 7 x + 2 4 x − 8 2 x 2 − 7 x + 3 x 2 − 5 x + 6 . 6 x 2 − 7 x + 2 4 x − 8 2 x 2 − 7 x + 3 x 2 − 5 x + 6 .

Try It 7.15

Simplify: 3 x 2 + 7 x + 2 4 x + 24 3 x 2 − 14 x − 5 x 2 + x − 30 . 3 x 2 + 7 x + 2 4 x + 24 3 x 2 − 14 x − 5 x 2 + x − 30 .

Try It 7.16

Simplify: y 2 − 36 2 y 2 + 11 y − 6 2 y 2 − 2 y − 60 8 y − 4 . y 2 − 36 2 y 2 + 11 y − 6 2 y 2 − 2 y − 60 8 y − 4 .

If we have more than two rational expressions to work with, we still follow the same procedure. The first step will be to rewrite any division as multiplication by the reciprocal. Then, we factor and multiply.

Example 7.9

Perform the indicated operations: 3 x − 6 4 x − 4 · x 2 + 2 x − 3 x 2 − 3 x − 10 ÷ 2 x + 12 8 x + 16 . 3 x − 6 4 x − 4 · x 2 + 2 x − 3 x 2 − 3 x − 10 ÷ 2 x + 12 8 x + 16 .

Try It 7.17

Perform the indicated operations: 4 m + 4 3 m − 15 · m 2 − 3 m − 10 m 2 − 4 m − 32 ÷ 12 m − 36 6 m − 48 . 4 m + 4 3 m − 15 · m 2 − 3 m − 10 m 2 − 4 m − 32 ÷ 12 m − 36 6 m − 48 .

Try It 7.18

Perform the indicated operations: 2 n 2 + 10 n n − 1 ÷ n 2 + 10 n + 24 n 2 + 8 n − 9 · n + 4 8 n 2 + 12 n . 2 n 2 + 10 n n − 1 ÷ n 2 + 10 n + 24 n 2 + 8 n − 9 · n + 4 8 n 2 + 12 n .

Multiply and Divide Rational Functions

We started this section stating that a rational expression is an expression of the form p q , p q , where p and q are polynomials and q ≠ 0 . q ≠ 0 . Similarly, we define a rational function as a function of the form R ( x ) = p ( x ) q ( x ) R ( x ) = p ( x ) q ( x ) where p ( x ) p ( x ) and q ( x ) q ( x ) are polynomial functions and q ( x ) q ( x ) is not zero.

Rational Function

A rational function is a function of the form

where p ( x ) p ( x ) and q ( x ) q ( x ) are polynomial functions and q ( x ) q ( x ) is not zero.

The domain of a rational function is all real numbers except for those values that would cause division by zero. We must eliminate any values that make q ( x ) = 0 . q ( x ) = 0 .

Determine the domain of a rational function.

• Step 3. The domain is all real numbers excluding the values found in Step 2.

Example 7.10

Find the domain of R ( x ) = 2 x 2 − 14 x 4 x 2 − 16 x − 48 . R ( x ) = 2 x 2 − 14 x 4 x 2 − 16 x − 48 .

The domain will be all real numbers except those values that make the denominator zero. We will set the denominator equal to zero , solve that equation, and then exclude those values from the domain.

Try It 7.19

Find the domain of R ( x ) = 2 x 2 − 10 x 4 x 2 − 16 x − 20 . R ( x ) = 2 x 2 − 10 x 4 x 2 − 16 x − 20 .

Try It 7.20

Find the domain of R ( x ) = 4 x 2 − 16 x 8 x 2 − 16 x − 64 . R ( x ) = 4 x 2 − 16 x 8 x 2 − 16 x − 64 .

To multiply rational functions, we multiply the resulting rational expressions on the right side of the equation using the same techniques we used to multiply rational expressions.

Example 7.11

Find R ( x ) = f ( x ) · g ( x ) R ( x ) = f ( x ) · g ( x ) where f ( x ) = 2 x − 6 x 2 − 8 x + 15 f ( x ) = 2 x − 6 x 2 − 8 x + 15 and g ( x ) = x 2 − 25 2 x + 10 . g ( x ) = x 2 − 25 2 x + 10 .

Try It 7.21

Find R ( x ) = f ( x ) · g ( x ) R ( x ) = f ( x ) · g ( x ) where f ( x ) = 3 x − 21 x 2 − 9 x + 14 f ( x ) = 3 x − 21 x 2 − 9 x + 14 and g ( x ) = 2 x 2 − 8 3 x + 6 . g ( x ) = 2 x 2 − 8 3 x + 6 .

Try It 7.22

Find R ( x ) = f ( x ) · g ( x ) R ( x ) = f ( x ) · g ( x ) where f ( x ) = x 2 − x 3 x 2 + 27 x − 30 f ( x ) = x 2 − x 3 x 2 + 27 x − 30 and g ( x ) = x 2 − 100 x 2 − 10 x . g ( x ) = x 2 − 100 x 2 − 10 x .

To divide rational functions, we divide the resulting rational expressions on the right side of the equation using the same techniques we used to divide rational expressions.

Example 7.12

Find R ( x ) = f ( x ) g ( x ) R ( x ) = f ( x ) g ( x ) where f ( x ) = 3 x 2 x 2 − 4 x f ( x ) = 3 x 2 x 2 − 4 x and g ( x ) = 9 x 2 − 45 x x 2 − 7 x + 10 . g ( x ) = 9 x 2 − 45 x x 2 − 7 x + 10 .

Try It 7.23

Find R ( x ) = f ( x ) g ( x ) R ( x ) = f ( x ) g ( x ) where f ( x ) = 2 x 2 x 2 − 8 x f ( x ) = 2 x 2 x 2 − 8 x and g ( x ) = 8 x 2 + 24 x x 2 + x − 6 . g ( x ) = 8 x 2 + 24 x x 2 + x − 6 .

Try It 7.24

Find R ( x ) = f ( x ) g ( x ) R ( x ) = f ( x ) g ( x ) where f ( x ) = 15 x 2 3 x 2 + 33 x f ( x ) = 15 x 2 3 x 2 + 33 x and g ( x ) = 5 x − 5 x 2 + 9 x − 22 . g ( x ) = 5 x − 5 x 2 + 9 x − 22 .

Section 7.1 Exercises

Practice makes perfect.

In the following exercises, determine the values for which the rational expression is undefined.

ⓐ 2 x 2 z 2 x 2 z , ⓑ 4 p − 1 6 p − 5 4 p − 1 6 p − 5 , ⓒ n − 3 n 2 + 2 n − 8 n − 3 n 2 + 2 n − 8

ⓐ 10 m 11 n 10 m 11 n , ⓑ 6 y + 13 4 y − 9 6 y + 13 4 y − 9 , ⓒ b − 8 b 2 − 36 b − 8 b 2 − 36

ⓐ 4 x 2 y 3 y 4 x 2 y 3 y , ⓑ 3 x − 2 2 x + 1 3 x − 2 2 x + 1 , ⓒ u − 1 u 2 − 3 u − 28 u − 1 u 2 − 3 u − 28

ⓐ 5 p q 2 9 q 5 p q 2 9 q , ⓑ 7 a − 4 3 a + 5 7 a − 4 3 a + 5 , ⓒ 1 x 2 − 4 1 x 2 − 4

In the following exercises, simplify each rational expression.

− 44 55 − 44 55

56 63 56 63

8 m 3 n 12 m n 2 8 m 3 n 12 m n 2

36 v 3 w 2 27 v w 3 36 v 3 w 2 27 v w 3

8 n − 96 3 n − 36 8 n − 96 3 n − 36

12 p − 240 5 p − 100 12 p − 240 5 p − 100

x 2 + 4 x − 5 x 2 − 2 x + 1 x 2 + 4 x − 5 x 2 − 2 x + 1

y 2 + 3 y − 4 y 2 − 6 y + 5 y 2 + 3 y − 4 y 2 − 6 y + 5

a 2 − 4 a 2 + 6 a − 16 a 2 − 4 a 2 + 6 a − 16

y 2 − 2 y − 3 y 2 − 9 y 2 − 2 y − 3 y 2 − 9

p 3 + 3 p 2 + 4 p + 12 p 2 + p − 6 p 3 + 3 p 2 + 4 p + 12 p 2 + p − 6

x 3 − 2 x 2 − 25 x + 50 x 2 − 25 x 3 − 2 x 2 − 25 x + 50 x 2 − 25

8 b 2 − 32 b 2 b 2 − 6 b − 80 8 b 2 − 32 b 2 b 2 − 6 b − 80

−5 c 2 − 10 c −10 c 2 + 30 c + 100 −5 c 2 − 10 c −10 c 2 + 30 c + 100

3 m 2 + 30 m n + 75 n 2 4 m 2 − 100 n 2 3 m 2 + 30 m n + 75 n 2 4 m 2 − 100 n 2

5 r 2 + 30 r s − 35 s 2 r 2 − 49 s 2 5 r 2 + 30 r s − 35 s 2 r 2 − 49 s 2

a − 5 5 − a a − 5 5 − a

5 − d d − 5 5 − d d − 5

20 − 5 y y 2 − 16 20 − 5 y y 2 − 16

4 v − 32 64 − v 2 4 v − 32 64 − v 2

w 3 + 216 w 2 − 36 w 3 + 216 w 2 − 36

v 3 + 125 v 2 − 25 v 3 + 125 v 2 − 25

z 2 − 9 z + 20 16 − z 2 z 2 − 9 z + 20 16 − z 2

a 2 − 5 a − 36 81 − a 2 a 2 − 5 a − 36 81 − a 2

In the following exercises, multiply the rational expressions.

12 16 · 4 10 12 16 · 4 10

32 5 · 16 24 32 5 · 16 24

5 x 2 y 4 12 x y 3 · 6 x 2 20 y 2 5 x 2 y 4 12 x y 3 · 6 x 2 20 y 2

12 a 3 b b 2 · 2 a b 2 9 b 3 12 a 3 b b 2 · 2 a b 2 9 b 3

5 p 2 p 2 − 5 p − 36 · p 2 − 16 10 p 5 p 2 p 2 − 5 p − 36 · p 2 − 16 10 p

3 q 2 q 2 + q − 6 · q 2 − 9 9 q 3 q 2 q 2 + q − 6 · q 2 − 9 9 q

2 y 2 − 10 y y 2 + 10 y + 25 · y + 5 6 y 2 y 2 − 10 y y 2 + 10 y + 25 · y + 5 6 y

z 2 + 3 z z 2 − 3 z − 4 · z − 4 z 2 z 2 + 3 z z 2 − 3 z − 4 · z − 4 z 2

28 − 4 b 3 b − 3 · b 2 + 8 b − 9 b 2 − 49 28 − 4 b 3 b − 3 · b 2 + 8 b − 9 b 2 − 49

72 m − 12 m 2 8 m + 32 · m 2 + 10 m + 24 m 2 − 36 72 m − 12 m 2 8 m + 32 · m 2 + 10 m + 24 m 2 − 36

3 c 2 − 16 c + 5 c 2 − 25 · c 2 + 10 c + 25 3 c 2 − 14 c − 5 3 c 2 − 16 c + 5 c 2 − 25 · c 2 + 10 c + 25 3 c 2 − 14 c − 5

2 d 2 + d − 3 d 2 − 16 · d 2 − 8 d + 16 2 d 2 − 9 d − 18 2 d 2 + d − 3 d 2 − 16 · d 2 − 8 d + 16 2 d 2 − 9 d − 18

6 m 2 − 13 m + 2 9 − m 2 · m 2 − 6 m + 9 6 m 2 + 23 m − 4 6 m 2 − 13 m + 2 9 − m 2 · m 2 − 6 m + 9 6 m 2 + 23 m − 4

2 n 2 − 3 n − 14 25 − n 2 · n 2 − 10 n + 25 2 n 2 − 13 n + 21 2 n 2 − 3 n − 14 25 − n 2 · n 2 − 10 n + 25 2 n 2 − 13 n + 21

In the following exercises, divide the rational expressions.

v − 5 11 − v ÷ v 2 − 25 v − 11 v − 5 11 − v ÷ v 2 − 25 v − 11

10 + w w − 8 ÷ 100 − w 2 8 − w 10 + w w − 8 ÷ 100 − w 2 8 − w

3 s 2 s 2 − 16 ÷ s 3 + 4 s 2 + 16 s s 3 − 64 3 s 2 s 2 − 16 ÷ s 3 + 4 s 2 + 16 s s 3 − 64

r 2 − 9 15 ÷ r 3 − 27 5 r 2 + 15 r + 45 r 2 − 9 15 ÷ r 3 − 27 5 r 2 + 15 r + 45

p 3 + q 3 3 p 2 + 3 p q + 3 q 2 ÷ p 2 − q 2 12 p 3 + q 3 3 p 2 + 3 p q + 3 q 2 ÷ p 2 − q 2 12

v 3 − 8 w 3 2 v 2 + 4 v w + 8 w 2 ÷ v 2 − 4 w 2 4 v 3 − 8 w 3 2 v 2 + 4 v w + 8 w 2 ÷ v 2 − 4 w 2 4

x 2 + 3 x − 10 4 x ÷ ( 2 x 2 + 20 x + 50 ) x 2 + 3 x − 10 4 x ÷ ( 2 x 2 + 20 x + 50 )

2 y 2 − 10 y z − 48 z 2 2 y − 1 ÷ ( 4 y 2 − 32 y z ) 2 y 2 − 10 y z − 48 z 2 2 y − 1 ÷ ( 4 y 2 − 32 y z )

2 a 2 − a − 21 5 a + 20 a 2 + 7 a + 12 a 2 + 8 a + 16 2 a 2 − a − 21 5 a + 20 a 2 + 7 a + 12 a 2 + 8 a + 16

3 b 2 + 2 b − 8 12 b + 18 3 b 2 + 2 b − 8 2 b 2 − 7 b − 15 3 b 2 + 2 b − 8 12 b + 18 3 b 2 + 2 b − 8 2 b 2 − 7 b − 15

12 c 2 − 12 2 c 2 − 3 c + 1 4 c + 4 6 c 2 − 13 c + 5 12 c 2 − 12 2 c 2 − 3 c + 1 4 c + 4 6 c 2 − 13 c + 5

4 d 2 + 7 d − 2 35 d + 10 d 2 − 4 7 d 2 − 12 d − 4 4 d 2 + 7 d − 2 35 d + 10 d 2 − 4 7 d 2 − 12 d − 4

For the following exercises, perform the indicated operations.

10 m 2 + 80 m 3 m − 9 · m 2 + 4 m − 21 m 2 − 9 m + 20 ÷ 5 m 2 + 10 m 2 m − 10 10 m 2 + 80 m 3 m − 9 · m 2 + 4 m − 21 m 2 − 9 m + 20 ÷ 5 m 2 + 10 m 2 m − 10

4 n 2 + 32 n 3 n + 2 · 3 n 2 − n − 2 n 2 + n − 30 ÷ 108 n 2 − 24 n n + 6 4 n 2 + 32 n 3 n + 2 · 3 n 2 − n − 2 n 2 + n − 30 ÷ 108 n 2 − 24 n n + 6

12 p 2 + 3 p p + 3 ÷ p 2 + 2 p − 63 p 2 − p − 12 · p − 7 9 p 3 − 9 p 2 12 p 2 + 3 p p + 3 ÷ p 2 + 2 p − 63 p 2 − p − 12 · p − 7 9 p 3 − 9 p 2

6 q + 3 9 q 2 − 9 q ÷ q 2 + 14 q + 33 q 2 + 4 q − 5 · 4 q 2 + 12 q 12 q + 6 6 q + 3 9 q 2 − 9 q ÷ q 2 + 14 q + 33 q 2 + 4 q − 5 · 4 q 2 + 12 q 12 q + 6

In the following exercises, find the domain of each function.

R ( x ) = x 3 − 2 x 2 − 25 x + 50 x 2 − 25 R ( x ) = x 3 − 2 x 2 − 25 x + 50 x 2 − 25

R ( x ) = x 3 + 3 x 2 − 4 x − 12 x 2 − 4 R ( x ) = x 3 + 3 x 2 − 4 x − 12 x 2 − 4

R ( x ) = 3 x 2 + 15 x 6 x 2 + 6 x − 36 R ( x ) = 3 x 2 + 15 x 6 x 2 + 6 x − 36

R ( x ) = 8 x 2 − 32 x 2 x 2 − 6 x − 80 R ( x ) = 8 x 2 − 32 x 2 x 2 − 6 x − 80

For the following exercises, find R ( x ) = f ( x ) · g ( x ) R ( x ) = f ( x ) · g ( x ) where f ( x ) f ( x ) and g ( x ) g ( x ) are given.

f ( x ) = 6 x 2 − 12 x x 2 + 7 x − 18 f ( x ) = 6 x 2 − 12 x x 2 + 7 x − 18 g ( x ) = x 2 − 81 3 x 2 − 27 x g ( x ) = x 2 − 81 3 x 2 − 27 x

f ( x ) = x 2 − 2 x x 2 + 6 x − 16 f ( x ) = x 2 − 2 x x 2 + 6 x − 16 g ( x ) = x 2 − 64 x 2 − 8 x g ( x ) = x 2 − 64 x 2 − 8 x

f ( x ) = 4 x x 2 − 3 x − 10 f ( x ) = 4 x x 2 − 3 x − 10 g ( x ) = x 2 − 25 8 x 2 g ( x ) = x 2 − 25 8 x 2

f ( x ) = 2 x 2 + 8 x x 2 − 9 x + 20 f ( x ) = 2 x 2 + 8 x x 2 − 9 x + 20 g ( x ) = x − 5 x 2 g ( x ) = x − 5 x 2

For the following exercises, find R ( x ) = f ( x ) g ( x ) R ( x ) = f ( x ) g ( x ) where f ( x ) f ( x ) and g ( x ) g ( x ) are given.

f ( x ) = 27 x 2 3 x − 21 f ( x ) = 27 x 2 3 x − 21 g ( x ) = 3 x 2 + 18 x x 2 + 13 x + 42 g ( x ) = 3 x 2 + 18 x x 2 + 13 x + 42

f ( x ) = 24 x 2 2 x − 8 f ( x ) = 24 x 2 2 x − 8 g ( x ) = 4 x 3 + 28 x 2 x 2 + 11 x + 28 g ( x ) = 4 x 3 + 28 x 2 x 2 + 11 x + 28

f ( x ) = 16 x 2 4 x + 36 f ( x ) = 16 x 2 4 x + 36 g ( x ) = 4 x 2 − 24 x x 2 + 4 x − 45 g ( x ) = 4 x 2 − 24 x x 2 + 4 x − 45

f ( x ) = 24 x 2 2 x − 4 f ( x ) = 24 x 2 2 x − 4 g ( x ) = 12 x 2 + 36 x x 2 − 11 x + 18 g ( x ) = 12 x 2 + 36 x x 2 − 11 x + 18

Writing Exercises

Explain how you find the values of x for which the rational expression x 2 − x − 20 x 2 − 4 x 2 − x − 20 x 2 − 4 is undefined.

Explain all the steps you take to simplify the rational expression p 2 + 4 p − 21 9 − p 2 . p 2 + 4 p − 21 9 − p 2 .

ⓐ Multiply 7 4 · 9 10 7 4 · 9 10 and explain all your steps. ⓑ Multiply n n − 3 · 9 n + 3 n n − 3 · 9 n + 3 and explain all your steps. ⓒ Evaluate your answer to part ⓑ when n = 7 n = 7 . Did you get the same answer you got in part ⓐ ? Why or why not?

ⓐ Divide 24 5 ÷ 6 24 5 ÷ 6 and explain all your steps. ⓑ Divide x 2 − 1 x ÷ ( x + 1 ) x 2 − 1 x ÷ ( x + 1 ) and explain all your steps. ⓒ Evaluate your answer to part ⓑ when x = 5 . x = 5 . Did you get the same answer you got in part ⓐ ? Why or why not?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved your goals in this section! Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific!

…with some help. This must be addressed quickly as topics you do not master become potholes in your road to success. Math is sequential - every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

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Lesson Video: Dividing Rational Numbers Mathematics • First Year of Preparatory School

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In this video, we will learn how to divide rational numbers, including fractions and decimals.

Video Transcript

In this lesson, what we’ll be looking at is dividing rational numbers. And this will include fractions and decimals. So by the end of the lesson, what we should be able to do is divide a rational decimal by a rational decimal, divide a fraction by a fraction, divide rational numbers in various different forms, and, finally, solve word problems involving the division of rational numbers.

But before we start to do any of that, what is a rational number? Well, in fact, a rational number is a real number that can be written as 𝑎 over 𝑏, where 𝑎 and 𝑏 are integers and 𝑏 is not equal to zero. Another way of saying this is that it is any number that can be represented as the ratio between two integers. So what we have here are some examples. So first of all, we’ve got two-fifths which we can see is written in the form 𝑎 over 𝑏. Well, then we have 0.3 recurring. But we think, “hold on! This isn’t written as 𝑎 over 𝑏. So how come this is a rational number?” Well, 0.3 recurring can be written as one over three or one-third. So it is also worth noting at this point that, in fact, any recurring decimal is in fact a rational number. Then we have three.

But, again, we’re thinking, well, that’s not in the form 𝑎 over 𝑏. Well, in fact, it could be written as three over one or even nine over three; they would both give us three. Then we have another recurring decimal, which is 0.142857 recurring. This time, I’ve just put this in here just to show there’s a different way of showing recurring here. And that’s with a straight line, not just a dot. We could have a dot over the one and a dot over the seven to mean the same thing. And as we said before, all recurring decimals are rational numbers. And this one is the same as one over seven or one-seventh. Then, finally, for our examples, we have 0.125, which you might already know is an eighth. And it’s also worth pointing out here is that this is something called a terminating decimal. And any terminating decimal is also a rational number.

So therefore, what we can also surmise is that the converse of this, anything that doesn’t satisfy this rule, is not going to be a rational number. It’s going to be an irrational number. Okay, great. So we’ve looked at what rational numbers are. So now what we’re going to do is move on to our questions.

Evaluate 0.8 divided by 0.4.

So what we have here are two terminating decimals, and we’re going to divide them. And we have a couple of methods to do this. So what we’re gonna do is have a look at both of them. So for method one, what we’re gonna do is we’re gonna multiply both of our decimals. And we’re gonna multiply them both by 10. And that’s because what we’re gonna do is make it so, in fact, we’re not dividing decimals at all. We’re gonna be dividing units. So if we multiply 0.8 and 0.4 by 10, what’s gonna happen is that each of the numbers is going to move one place value to the left. So what we’re gonna get is eight divided by four. And we can do this because we’ve done it to both terms. So therefore, it’s going to give us the same result. Well, this is nice and straightforward. And that’s because eight divided by four is equal to two. So therefore, we can say that 0.8 divided by 0.4 is two.

So now we’re gonna take a look at method two. And for method two, what we’re going to do is convert both of our decimals to fractions. And we can do that because they’re both terminating decimals. And we know that terminating decimals are rational numbers, so therefore can be converted to a fraction with an integer as the numerator and an integer as the denominator. Well, if we start with 0.8, what this means is eight-tenths. Well, in turn, we can simplify eight-tenths by dividing the numerator and denominator by two, which will give us four-fifths. Then we have 0.4, which is gonna be equal to four-tenths, which again we can simplify by dividing the numerator and denominator by two to give us two-fifths.

Okay, great. We now got our two fractions. So we now have four-fifths divided by two-fifths. And we’ve got a method for dividing fractions. And what we can do is use our memory aid to remind us how to do that. And that is KCF, which is keep it, change it, flip it. And this means we keep the first fraction the same, we change the sign from a divide to a multiply, and we flip the second fraction. And it’s worth noting that if we flip a fraction, this is in fact the reciprocal of that fraction. And then if we multiply two fractions, all we do is multiply the numerators and multiply the denominators. So this is gonna be equal to 20 over 10, which once again gives us an answer of two.

Now this is quite straightforward because we’re multiplying two easy fractions. However, there is a quick tip, which can be useful. If you’re ever multiplying fractions, have a look at the numerators and denominators and see if there’s a common factor. So here we can see that five is a common factor. So if we divide both the numerator and denominator by five, we’re just left with four multiplied by one over one multiplied by two, which is four over two, which again would’ve given us two.

So that was our first example. So now what we’re gonna do is have a look at an example where we’re going to evaluate an expression using multiplication and division of rational numbers. And all of these are gonna be in fractional form.

Evaluate three-quarters multiplied by negative two over three divided by a fifth giving the answer in its simplest form.

So to help us evaluate this expression, what we’re going to use is PEMDAS. And what PEMDAS is, is a way of remembering the order of operations. So here it says that the P — we’re gonna deal with the parentheses first, then exponents, multiplication, division, addition, then subtraction. So we can see that, in our expression, what we have are parentheses, so we’ll deal with these first. So what we have is three-quarters multiplied by negative two over three. So to multiply fractions, what we do is multiply numerators then multiply denominators. So this is gonna give us negative six over 12. Well, we can simplify this by dividing both the numerator and the denominator by six. So we’re gonna get negative one over two. So now we’ve dealt with our parentheses. So what we can do is put this value back into our expression.

Well, next we have a division. And that division is negative one over two, which is the result of the parentheses calculation, divided by one over five or one-fifth. Now, to help us remember what we’re gonna do with the division of fractions, we use our memory aid KCF, keep it, change it, flip it. So we keep the first fraction the same. We change our divide to a multiply. And we flip our second fraction. So we get negative one over two multiplied by five over one. So then what we do is multiply our numerators and denominators. So we can say that if we evaluate three-quarters multiplied by negative two over three divided by a fifth, we’re gonna get negative five over two.

So what we had a look at here was a problem involving fractions. What we’re gonna have a look at now is a problem that involves recurring decimals and the modulus or absolute value.

Evaluate 0.8 recurring divided by the modulus or absolute value of negative five over four giving the answer in its simplest form.

So in this question, the first thing we’re gonna have a look at is the recurring decimal. So we got 0.8 recurring. Now, this is, in fact, a rational number. And it’s a rational number because we can represent it as a fraction with an integer as the numerator and an integer as the denominator. And this is something we know about all recurring decimals. So to change this into a fraction, what we’re gonna do is let 𝑥 be equal to 0.8 recurring. And then what we’re gonna do is multiply 𝑥 by 10 to give us 10𝑥 and multiply 0.8 recurring by 10 to give us 8.8 recurring. So we now know that 10𝑥 is equal to 8.8 recurring.

So you might have thought, “Well, why have we just done that?” But it’s actually rather clever because what we’re gonna do now is eliminate the recurring part of our decimal. So let’s label them equation one and equation two. So what we can do is we can subtract equation one from equation two. So when we do that, what we’re gonna get on the left-hand side is nine 𝑥, cause we’ve got 10𝑥 minus 𝑥 which is nine 𝑥. And then on the right-hand side, we’re just gonna have eight, and that’s because if we have 8.8 recurring minus 0.8 recurring, the recurring parts cancel each other out and we’re left with just eight. So then what we’re gonna do is just divide through by nine. And what we get is 𝑥 is equal to eight over nine or eight-ninths. And as 𝑥 is equal to 0.8 recurring, we can say that 0.8 recurring is equal to eight over nine or eight-ninths. And this is a fraction with an integer numerator and an integer denominator.

Okay, great. So we converted that into a fraction. Well, now what about our absolute value or modulus of negative five over four? Well, these vertical lines mean the absolute value or modulus. And what this means is that we’re only interested in the positive value because what the absolute value or modulus means is the distance from zero or magnitude of a value. So therefore, we’re not interested in the negative part. So therefore, what we can do is just write our modulus or absolute value of negative five over four as five over four. So now what our calculation has become is eight over nine or eight-ninths divided by five over four. So now what we’re gonna do is divide our fractions.

And to remember how to do that, we can use our trusty memory aid, KCF: keep it, change it, flip it. So, to use this, we keep the first fraction the same. Then we change the sign from a divide to a multiply. And now we flip the second fraction. So we’ve now got eight over nine multiplied by four over five. So then if we multiply our numerators and denominators, we’re gonna get 32 over 45. And we can see this can’t be canceled down any further. So it is, in fact, in its simplest form. So then we can say the answer is 32 over 45.

Okay, great. So we’ve looked at a number of different skills so far, but what we’re gonna have look at now is to see if we can use these skills to solve problems. And what we’re gonna try and do is find our missing value.

The product of two rational numbers is negative 16 over nine. If one of the numbers is negative four over three, find the other number.

So the first thing we’re gonna do is look at a couple of key terms. So we’ve got product, which means multiply. So if we find the product of two numbers, that means we’re multiplying them together. And then we’re also looking at the term, rational. And what this means is a number that can be written as a fraction with an integer as the numerator and an integer as the denominator, which is gonna help us when we’re gonna try and find the number that we’re looking for. So taking the information we’ve got from the question, what we can do is write it down. And we’ve got negative four over three multiplied by 𝑎 over 𝑏 equals negative 16 over nine. And it’s this 𝑎 over 𝑏 that we’re trying to find.

Well, there are, in fact, a couple of ways we could solve this. So we’re gonna have a look at both of those. So first of all, what we could do is divide both sides by negative four over three. So when we do that, we’ll have 𝑎 over 𝑏 equals negative 16 over nine divided by negative four over three. So then what we can do is divide our fractions. And to do that, we can use our memory aid, KCF — keep it, change it, flip it — which is gonna give us 𝑎 over 𝑏 is equal to negative 16 over nine multiplied by negative three over four. So now, before we multiply, what we can do is divide through by any common factors. Well, first of all, we can divide numerators and denominators by four and then by three.

So now what we’ve got is negative four over three multiplied by negative one over one. Well, a negative multiplied by a negative is a positive. So therefore, what we’re gonna get is 𝑎 over 𝑏 is equal to four over three. So therefore, we’ve found our missing number. And what we can do is check this by using the alternate method. And the alternate method is equating the numerators and denominators. Well, as we know, we’ve got negative four over three in the left-hand side and the result is negative 16 over nine. We know that a negative has to be multiplied by a positive to give us a negative result. So therefore, we know that 𝑎 over 𝑏 will be positive. So we could ignore the signs when we’re gonna equate the numerators and denominators.

Well, if you equate the numerators, we’ve got four 𝑎 cause four multiplied by 𝑎 is equal to 16. So therefore, 𝑎 will be equal to four. So then if we equate the denominators, we’re gonna get three 𝑏 is equal to nine. So 𝑏 is equal to three. So therefore, 𝑎 over 𝑏 is gonna be equal to four over three, which is what we’ve got with the first method.

Okay, great. So we’ve just looked at a problem where we had to find a missing value. So for our final example, we’re gonna take a look at a worded problem, so a problem where we’re gonna use these skills in context.

Noah constructs three-quarters of a wall in one and two-thirds days. How many days will he need to construct the wall?

So what we’re gonna do is use a diagram to help us visualize the problem. So what we can see is it takes one and two-thirds days to build three-quarters of the wall. So therefore, if we want to find out how long it takes to build one-quarter of the wall, what we can do is divide one and two-thirds by three. Well, to complete this calculation, what we want to do is convert one and two-thirds, a mixed number, into a top heavy or improper fraction. So to do that, what we do is we see that there are three-thirds in a whole one add two gives us five over three. So it’s one multiplied by three add two over three. So we’ve got five-thirds divided by three.

Well, if we want to divide five-thirds by three, we can think of this as five-thirds divided by three over one. Well then what we can use is our memory aid for dividing by a fraction, which is keep it, change it, flip it, KCF, which will give us five-thirds multiplied by one over three, cause we keep the first fraction, change the sign, flip the second fraction. So therefore, what we’ve got is the time taken for one-quarter of the wall to be built is five over nine or five-ninths because five multiplied by one is five and three multiplied by three is nine.

So now what’s the next stage? Now what do we need to do? Well, we can see that the whole wall is four-quarters. So therefore, we need to multiply the time taken for a quarter of the wall to be built by four. So this means five-ninths multiplied by four. Well, to help us think what we’ll do with this calculation, we can think of four as four over one. So then we’re gonna do five multiplied by four over nine, which is equal to 20 over nine. Well, this is an improper fraction. So what we could do now is convert this back into a mixed number. Well, to do that, what we do is we see how many nines go into 20, which is two with a remainder of two. So therefore, we know that it takes two and two-ninths days to construct the wall.

In fact, it is worth noting that we could’ve completed this question with one calculation. And that would’ve been one and two-thirds multiplied by four over three because if we divide by three and multiply by four, it’s the same as multiplying by four over three. And it could also have been solved by the calculation one and two-thirds divided by three over four because if we go backwards from keep it, change it, flip it, we could see that one and two-thirds divided by three over four is the same as one and two-thirds multiplied by four over three.

So we’ve taken a look at a number of examples and covered the key objectives for the lesson, so now let’s take a look at the key points. So the first key point is that a rational number is in fact a real number that can be written as 𝑎 over 𝑏, where 𝑎 and 𝑏 are integers and 𝑏 is not equal to zero. So therefore, it can be written as a fraction. Another way we can think of that is that a rational number is a number that can be represented as the ratio of two integers. And we also know that both recurring decimals and terminating decimals are also rational numbers because these can be written as fractions.

For example, on the left, we have our recurring decimal, which can be written as a seventh. And on the right, we have a terminating decimal which could be written as an eighth. And for our final key point, if we’re gonna divide two fractions, so 𝑎 over 𝑏 divided by 𝑐 over 𝑑, then this is equal to 𝑎 over 𝑏 multiplied by 𝑑 over 𝑐. So you multiply it by the reciprocal with the second fraction. And we have a memory aid to help us remember this. And that is KCF: keep it, change it, flip it. Keep the first fraction the same, change the sign from a divide to a multiply, and flip the second fraction.

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Multiplying and Dividing Rational Expressions

Multiplication

To Multiply a rational expression:

1. Factor all numerators and denominators.

2. Cancel all common factors.

3. Either multiply the denominators and numerators together or leave the solution in factored form.

Multiply and then simplify the product

$\frac{{2x + 4}}{x} \cdot \frac{3}{{6x + 12}}$

Multiply the following rational expressions:

$\frac{{{x^2} + 6x + 9}}{{{x^2} - 9}} \cdot \frac{{3x - 9}}{{{x^2} + 2x - 3}}$

1 : Factor all numerators and denominators:

$\frac{{{x^2} + 6x + 9}}{{{x^2} - 9}} \cdot \frac{{3x - 9}}{{{x^2} + 2x - 3}} = \frac{{(x + 3)(x + 3)}}{{(x - 3)(x + 3)}} \cdot \frac{{3(x - 3)}}{{(x + 3)(x - 1)}}$

2 : Cancel all common factors:

$\frac{{{x^2} + 6x + 9}}{{{x^2} - 9}} \cdot \frac{{3x - 9}}{{{x^2} + 2x - 3}} = \frac{{\cancel{{(x + 3)}}\cancel{{(x + 3)}}}}{{\cancel{{(x - 3)}}\cancel{{(x + 3)}}}} \cdot \frac{{3\cancel{{(x - 3)}}}}{{\cancel{{(x + 3)}}(x - 1)}}$

3 : Multiply the denominators and numerators:

$\frac{{{x^2} + 6x + 9}}{{{x^2} - 9}} \cdot \frac{{3x - 9}}{{{x^2} + 2x - 3}} = \frac{{\cancel{{(x + 3)}}\cancel{{(x + 3)}}}}{{\cancel{{(x - 3)}}\cancel{{(x + 3)}}}} \cdot \frac{{3\cancel{{(x - 3)}}}}{{\cancel{{(x + 3)}}(x - 1)}} = \frac{1}{1} \cdot \frac{3}{{x - 1}} = \frac{3}{{x - 1}}$

Division of rational expressions

When we divide rational functions we multiply by the reciprocal.

Perform the indicated operations:

$\frac{{2{x^2} + x - 6}}{{{x^2} - 2x - 8}}:\frac{{2{x^2} - x - 3}}{{{x^2} - 3x - 4}}$

Solution 3:

$$\frac{{2{x^2} + x - 6}}{{{x^2} - 2x - 8}}:\frac{{2{x^2} - x - 3}}{{{x^2} + 3x - 4}} = $$ $$ = \frac{{2{x^2} + x - 6}}{{{x^2} - 2x - 8}} \cdot \frac{{{x^2} - 3x - 4}}{{2{x^2} - x - 3}} = $$ $$ = \frac{{2\left( {x - \frac{3}{2}} \right)(x + 2)}}{{(x + 2)(x - 4)}} \cdot \frac{{(x - 4)(x + 1)}}{{2\left( {x - \frac{3}{2}} \right)(x + 1)}} = $$ $$ = \frac{{\bcancel{{(2x - 3)}}\cancel{{(x + 2)}}}}{{\cancel{{(x + 2)}}\bcancel{{(x - 4)}}}} \cdot \frac{{\bcancel{{(x - 4)}}\cancel{{(x + 1)}}}}{{\bcancel{{(2x - 3)}}\cancel{{(x + 1)}}}} = 1$$

$\frac{{\frac{{x + 4}}{{2x - 6}}}}{{\frac{{3x + 12}}{{4x - 12}}}}$

Solution 4:

$$\frac{{\frac{{x + 4}}{{2x - 6}}}}{{\frac{{3x + 12}}{{4x - 12}}}} = \frac{{x + 4}}{{2x - 6}} \cdot \frac{{4x - 12}}{{3x + 12}} = $$ $$ = \frac{{\cancel{{x + 4}}}}{{2\cancel{{(x - 3)}}}} \cdot \frac{{4\cancel{{(x - 3)}}}}{{3\cancel{{(x + 4)}}}} = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}$$

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Basic Algebra/Working with Numbers/Dividing Rational Numbers

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Dividing rational numbers covers a general area of equations. For an equation that is such has only to have a numerator and denominator that are both rational numbers. In turn, one will come out with a quotient that fits the terms applied to a "rational number". Given the fact that you already understand rational numbers, you will understand this unit. If, on the other hand, you have no clue what a rational number is, then you should do some research concerning this subject so that you can understand the explanation of dividing such numbers that follows this text.

Anyway, dividing rational numbers, sometimes worded "quotients of rational expressions", is simply dividing a rational number by a rational number. For instance, look at the example problems, dividing rational numbers is very easy. If you have a fraction dividing another fraction then you simply flip the dividend and, by multiplying, one will come out with exactly the same number. The knowledge of expressing how this works is beyond the scope of this lesson. But, it works every time. You are still dividing, but you have switched your means of doing so. When you come to more complicated problems that have unknown variables the same method works. So if you have a fraction of 7 over 5 divided by 3 over 4, you will simply flip the 3 over 4 and multiply the fractions instead of dividing. This is a method that will be used again and again in math, so know it well. Look at the examples given and, although this is easy, make sure you know it.

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Dividing Rational Numbers

Lesson Narrative

In this lesson, students complete their work extending all four operations to signed numbers by studying division. They use the relationship between multiplication and division to develop rules for dividing signed numbers. In preparation for the next lesson on negative rates of change, students look at a context, drilling a well, that is modeled by an equation \(y = kx\) where \(k\) is a negative number. This builds on their previous work with proportional relationships.

Learning Goals

Teacher Facing

• Apply multiplication and division of signed numbers to solve problems involving constant speed with direction, and explain (orally) the reasoning.
• Generalize (orally) a method for determining the quotient of two signed numbers.
• Generate a division equation that represents the same relationship as a given multiplication equation with signed numbers.

Student Facing

Let's divide signed numbers.

Learning Targets

• I can divide rational numbers.

CCSS Standards

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Glossary Entries

A solution to an equation is a number that can be used in place of the variable to make the equation true.

For example, 7 is the solution to the equation  \(m+1=8\) , because it is true that  \(7+1=8\) . The solution to  \(m+1=8\)  is not 9, because  \(9+1 \ne 8\) . 

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