Lesson 8: Rational Expressions Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 1 of 8

Rational Equations Activity

Solve.

1. Each weekday a mother drives 5 miles to pick up her son from soccer

practice. She then drives 5 miles to the gym to pick up her daughter. Due to heavy traffic, her speed going to the gym is 5 mph less than her speed

going to the soccer field. After picking her daughter up, she takes a 6-mile shortcut home at the same speed that she began with. If the whole

trip takes 34 minutes, how fast does she drive on each leg of her trip?

D r t= ⋅ D

tr

=

5

soccert

r=

5

5gymt

r=

− hom

6e

tr

=

Total time = 34 minutes = 34

hour60

Total time = homsoccer gym et t t+ +

Home 5 Soccer

Gym

5 6

Lesson 8: Rational Expressions Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 2 of 8

( ) ( )

( ) ( ) ( )

hom

2

2

2

2

34

60

5 5 6 34

5 60

5 5 6 3460 5 60 5

5 60

300 5 300 360 5 34 5

300 1500 300 360 1800 34 170

960 3300 34 170

34 1130 3300 0

17 565 1650 0

17

soccer gym et t t

r r r

r r r rr r r

r r r r r

r r r r r

r r r

r r

r r

r

+ + =

+ + =−

− + + = ⋅ − −

− + + − = −

− + + − = −

− = −

− + =

− + =

( ) ( )55 30 0r− − =

17 55 0 or 30 0

55 or 30

17

r r

r r

− = − =

= =

If 55

17r = , then

305

17r

−− = �

If 30r = , then her speed from home to soccer and from

the gym to home was 30 mph. Her speed from soccer to the gym was 25 mph.

Check:

5 1

hr 10 min30 6

soccert = = =

5 1

hr 12 min25 5

gymt = = =

hom

6 1 hr 12 min

30 5et = = =

Total time = 10 min + 12 min + 12 min = 34 minutes �

Lesson 8: Rational Expressions Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 3 of 8

2. Fill in the numerators for the rational expression below so that the equation has one integer solution and one extraneous solution. Hint: In

order for an equation to have two solutions, it must be a quadratic

equation.

25 2 3 10b b b b

+ =− + − −

There are many ways to approach this problem and many correct answers. The solution given below is one strategy for

solving this problem.

To have an extraneous solution, we need for one of our solutions

to make the denominator zero. We could have either 5b = or

2b = . I chose 5b = to be one solution and b=-3 to be the other

solution.

( ) ( )

2

5 or 3

5 0 or 3 0

5 3 0

2 15 0

b b

b b

b b

b b

= = −

− = + =

− + =

− − =

After multiplying both sides of the above equation by the LCD

and simplifying, I want the equation to be 2 2 15 0b b− − = .

I let the first numerator be b, and the second numerator be 7. (Note: These were chosen randomly.) I’ll temporarily let K be

the third numerator.

2

7

5 2 3 10

b K

b b b b+ =

− + − −

Multiplying both sides by the LCD and simplifying gives:

Lesson 8: Rational Expressions Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 4 of 8

( ) ( )

( ) ( )( ) ( )

( ) ( )

( ) ( )

2

2

2

7

5 2 3 10

7

5 2 5 2

75 2 5 2

5 2 5 2

2 7 5

2 7 35

9 35 0

b K

b b b b

b K

b b b b

b Kb b b b

b b b b

b b b K

b b b K

b b K

+ =− + − −

+ =− + − +

− + + = − + − + − +

+ + − =

+ + − =

+ − − =

We want this equation to be 2 2 15 0b b− − = . If we set the

equations equal to each other, we get:

2 29 35 2 15

11 20 0

11 20

b b K b b

b K

b K

+ − − = − −

− − =

− =

So, one equation that will work is:

2

7 11 20

5 2 3 10

b b

b b b b

−+ =

− + − −

Check:

( ) ( )

( ) ( )( ) ( )

( ) ( )

( ) ( )

( ) ( )

2

2

2

7 11 20

5 2 3 10

7 11 20

5 2 5 2

7 11 205 2 5 2

5 2 5 2

2 7 5 11 20

2 7 35 11 20

2 15 0

5 3 0

b b

b b b b

b b

b b b b

b bb b b b

b b b b

b b b b

b b b b

b b

b b

−+ =

− + − −−

+ =− + − +

− − + + = − + − + − +

+ + − = −

+ + − = −

− − =

− + =

5 0 or 3 0

5 or 3

b b

b b

− = + =

= = −

Lesson 8: Rational Expressions Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 5 of 8

When we substitute 5b = , we get zero in the denominator. This

is an extraneous solution.

When we substitute 3b = − , we get:

( )( ) ( )

?

2

?

2

?

?

?

7 11 20

5 2 3 10

11 3 203 7

3 5 3 2 3 3 3 10

3 7 33 20

8 1 9 9 10

3 56 53

8 8 8

53 53

8 8

b b

b b b b

−+ =

− + − −− −−

+ =− − − + − − − −

− −− =

+ −−

− =

− −= �

So 3b = − is an integer solution.

3. If 5x y+ = and 8xy = , find 1 1

x y+ .

We could try to solve this system of equations for x and y, using substitution:

5

8

x y

xy

+ =

=

8

xy

=

Substituting into the first equation gives:

( )2

2

85

85

8 5

5 8 0

yy

y y yy

y y

y y

+ =

⋅ + = ⋅

+ =

− + =

Lesson 8: Rational Expressions Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 6 of 8

Unfortunately, this won’t factor. (In Lesson 10, you’ll learn how to solve these using the quadratic formula.)

We can try the “working backwards” strategy. Start with 1 1

x y+

and simplify:

1 1 1 1y x

x y y x x y

y x

xy xy

x y

xy

+ = ⋅ + ⋅

= +

+=

Since we know that 5x y+ = and 8xy = , we can substitute to

get:

1 1 5

8

x y

x y xy

++ = =

4. You hire a moving company to pack and move your business. One employee, Joseph, could do the entire job in 6 hours. Another employee,

Logan, could do the entire job in 8 hours. Logan works alone for an hour,

then Joseph shows up, and they work together. Joseph leaves after two hours of work, and Logan finishes the rest of the job.

a) How much longer would Logan have to work to complete the job?

Lesson 8: Rational Expressions Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 7 of 8

In one hour, Logan can do 1

8 of the job. In x hours, he can do

8

x of the job. In one hour, Joseph can do

1

6 of the job.

Logan worked for 1 hour and did 1

8 of the job. Joseph joined

him, and they worked for 2 hours and together did 1 1

28 6

+

of the job. After Joseph left, Logan worked x hours and did 8

x

of the job.

The complete job is represented by:

1 1 1

2 18 8 6 8

x + + + =

Solving for x gives:

( )

( )

1 1 12 1

8 8 6 8

1 2 21

8 8 6 8

3 11

8 3

3 124 1 24

8 3

3 3 8 24

9 3 8 24

3 7

7 12 2 hrs 20 min

3 3

x

x

x

x

x

x

x

x

+ + + =

+ + + =

++ =

+ ⋅ + = ⋅

+ + =

+ + =

=

= = =

Lesson 8: Rational Expressions Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 8 of 8

b) If the company bills $50 per hour each employee worked, what would the total bill be?

Logan worked 1 hr + 2 hrs + 2 hrs 20 min = 5 hrs 20 min. Joseph worked 2 hrs. The total number of billable hours is

7 hrs 20 min, which is 1

73 hours. The total bill would be:

1$50 7 $366.67

3

⋅ =