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Simplify: n 4 n n + 5 1 n + 5 + 1 n 5 .

Solution

.
Simplify the numerator and denominator.
Find the LCD and add the fractions in the numerator.
Find the LCD and add the fractions in the denominator.
.
Simplify the numerators. .
Subtract the rational expressions in the numerator and add in the denominator.

Simplify.
.
Rewrite as fraction division. .
Multiply the first times the reciprocal of the second. .
Factor any expressions if possible. .
Remove common factors. .
Simplify. .

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Simplify: b 3 b b + 5 2 b + 5 + 1 b 5 .

b ( b + 2 )

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Simplify: 1 3 c + 4 1 c + 4 + c 3 .

3 c + 3

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Simplify a complex rational expression by using the lcd

We “cleared” the fractions by multiplying by the LCD when we solved equations with fractions. We can use that strategy here to simplify complex rational expressions. We will multiply the numerator and denominator by LCD of all the rational expressions.

Let’s look at the complex rational expression we simplified one way in [link] . We will simplify it here by multiplying the numerator and denominator by the LCD. When we multiply by LCD LCD we are multiplying by 1, so the value stays the same.

Simplify: 1 3 + 1 6 1 2 1 3 .

Solution

.
The LCD of all the fractions in the whole expression is 6.
Clear the fractions by multiplying the numerator and denominator by that LCD. .
Distribute. .
Simplify. .
.
.

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Simplify: 1 2 + 1 5 1 10 + 1 5 .

10 3

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Simplify: 1 4 + 3 8 1 2 5 16 .

7 3

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How to simplify a complex rational expression by using the lcd

Simplify: 1 x + 1 y x y y x .

Solution

The above image has 3 columns. It shows the steps on how to simplify a complex rational expression using the LCD for 1 divided by x plus 1 divided by y divided by x divided by y minus y divided by x. Step one is to find the LCD of all fractions in the complex rational expression. The LCD of all the fractions is x y. Multiply the numerator and denominator by the LCD. Step two is to multiply both the numerator and denominator by x y to get x y times 1 divided by x plus 1 divided by y divided x y times x divided by y minus y divided by x. Step three is to simplify the expression. Distribute to get x y times 1 divided by x plus x y times 1 divided y divided by x y times x divided by y minus x y times y divided by x. Simplify to get y plus x divided by x squared minus y squared. Remove common factors. Cross out y plus x in the numerator. Cross out x plus y in the numerator. Simplify to get 1 divided by x minus y.
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Simplify: 1 a + 1 b a b + b a .

b + a a 2 + b 2

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Simplify: 1 x 2 1 y 2 1 x + 1 y .

y x x y

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Simplify a complex rational expression by using the lcd.

  1. Find the LCD of all fractions in the complex rational expression.
  2. Multiply the numerator and denominator by the LCD.
  3. Simplify the expression.

Be sure to start by factoring all the denominators so you can find the LCD.

Simplify: 2 x + 6 4 x 6 4 x 2 36 .

Solution

.
Find the LCD of all fractions in the complex rational expression. The LCD is ( x + 6 ) ( x 6 ) .
Multiply the numerator and denominator by the LCD. .
Simplify the expression.
Distribute in the denominator. .
Simplify. .
Simplify. .
To simplify the denominator, distribute and combine like terms. .
Remove common factors. .
Simplify. .
Notice that there are no more factors common to the numerator and denominator.

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Simplify: 3 x + 2 5 x 2 3 x 2 4 .

3 x 6 5 x + 7

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Simplify: 2 x 7 1 x + 7 6 x + 7 1 x 2 49 .

x + 21 6 x + 43

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Simplify: 4 m 2 7 m + 12 3 m 3 2 m 4 .

Solution

.
Find the LCD of all fractions in the complex rational expression. The LCD is ( m 3 ) ( m 4 ) .
Multiply the numerator and denominator by the LCD. .
Simplify. .
Simplify. .
Distribute. .
Combine like terms. .

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Simplify: 3 x 2 + 7 x + 10 4 x + 2 + 1 x + 5 .

3 5 x + 22

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Simplify: 4 y y + 5 + 2 y + 6 3 y y 2 + 11 y + 30 .

6 y + 34 3 y

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Simplify: y y + 1 1 + 1 y 1 .

Solution

.
Find the LCD of all fractions in the complex rational expression.
The LCD is ( y + 1 ) ( y 1 ) .
Multiply the numerator and denominator by the LCD. .
Distribute in the denominator and simplify. .
Simplify. .
Simplify the denominator, and leave the numerator factored. .
.
Factor the denominator, and remove factors common with the numerator. .
Simplify. .

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Simplify: x x + 3 1 + 1 x + 3 .

x x + 4

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Simplify: 1 + 1 x 1 3 x + 1 .

x ( x + 1 ) 3 ( x 1 )

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Key concepts

  • To Simplify a Rational Expression by Writing it as Division
    1. Simplify the numerator and denominator.
    2. Rewrite the complex rational expression as a division problem.
    3. Divide the expressions.
  • To Simplify a Complex Rational Expression by Using the LCD
    1. Find the LCD of all fractions in the complex rational expression.
    2. Multiply the numerator and denominator by the LCD.
    3. Simplify the expression.

Practice makes perfect

Simplify a Complex Rational Expression by Writing It as Division

In the following exercises, simplify.

2 a a + 4 4 a 2 a 2 16

a 4 2

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3 b b 5 b 2 b 2 25

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5 c 2 + 5 c 14 10 c + 7

3 2 ( c 2 )

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8 d 2 + 9 d + 18 12 d + 6

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1 2 + 5 6 2 3 + 7 9

24 26

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2 3 1 9 3 4 + 5 6

20 57

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n m + 1 n 1 n n m

n 2 + m m n 2

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1 r + 1 t 1 r 2 1 t 2

r t t r

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2 v + 2 w 1 v 2 1 w 2

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x 2 x x + 3 1 x + 3 + 1 x 3

( x + 1 ) ( x 3 ) 2

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y 2 y y 4 2 y 4 2 y + 4

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2 2 a + 3 1 a + 3 + a 2

4 a + 1

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4 4 b 5 1 b 5 + b 4

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Simplify a Complex Rational Expression by Using the LCD

In the following exercises, simplify.

1 3 + 1 8 1 4 + 1 12

11 8

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5 6 + 2 9 7 18 1 3

19

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c d + 1 d 1 d d c

c 2 + c c d

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1 p + 1 q 1 p 2 1 q 2

p q q p

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2 r + 2 t 1 r 2 1 t 2

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2 x + 5 3 x 5 + 1 x 2 25

2 x 10 3 x + 16

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5 y 4 3 y + 4 + 2 y 2 16

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5 z 2 64 + 3 z + 8 1 z + 8 + 2 z 8

3 z 19 3 z + 8

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3 s + 6 + 5 s 6 1 s 2 36 + 4 s + 6

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4 a 2 2 a 15 1 a 5 + 2 a + 3

4 3 a 2

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5 b 2 6 b 27 3 b 9 + 1 b + 3

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5 c + 2 3 c + 7 5 c c 2 + 9 c + 14

2 c + 29 5 c

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6 d 4 2 d + 7 2 d d 2 + 3 d 28

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2 + 1 p 3 5 p 3

( 2 p 5 ) 5

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m m + 5 4 + 1 m 5

m ( m 5 ) 4 m 2 + m 95

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Simplify

In the following exercises, use either method.

3 4 2 7 1 2 + 5 14

13 24

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2 a + 4 1 a 2 16

2 ( a 4 )

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3 b 2 3 b 40 5 b + 5 2 b 8

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3 m + 3 n 1 m 2 1 n 2

3 m n n m

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2 r 9 1 r + 9 + 3 r 2 81

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x 3 x x + 2 3 x + 2 + 3 x 2

( x 1 ) ( x 2 ) 6

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Everyday math

Electronics The resistance of a circuit formed by connecting two resistors in parallel is 1 1 R 1 + 1 R 2 .

  1. Simplify the complex fraction 1 1 R 1 + 1 R 2 .
  2. Find the resistance of the circuit when R 1 = 8 and R 2 = 12 .

R 1 R 2 R 2 + R 1
24 5

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Ironing Lenore can do the ironing for her family’s business in h hours. Her daughter would take h + 2 hours to get the ironing done. If Lenore and her daughter work together, using 2 irons, the number of hours it would take them to do all the ironing is 1 1 h + 1 h + 2 .

  1. Simplify the complex fraction 1 1 h + 1 h + 2 .
  2. Find the number of hours it would take Lenore and her daughter, working together, to get the ironing done if h = 4 .
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Writing exercises

In this section, you learned to simplify the complex fraction 3 x + 2 x x 2 4 two ways:

rewriting it as a division problem

multiplying the numerator and denominator by the LCD

Which method do you prefer? Why?

Answers will vary.

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Efraim wants to start simplifying the complex fraction 1 a + 1 b 1 a 1 b by cancelling the variables from the numerator and denominator. Explain what is wrong with Efraim’s plan.

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Self check

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

The above image is four columns and three rows. The first row is the header row. The first header is labeled “I can…”, the second “Confidently”, the third, “With some help”, and the fourth “No – I don’t get it!”. In the first column under “I can”, the next row reads “simplify a complex rational expression by writing it as division.”, the next row reads “simplify a complex rational expression by using the LCD.” The remaining columns are blank.

After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
Eunice Reply
what is a capacitor?
Raymond Reply
Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
Maria Reply
please solve
Sharon
8m/s²
Aishat
What is Thermodynamics
Muordit
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
Saheed Reply
50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
Ramon Reply
I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
Someone
about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
Someone
definitely of physics
Haryormhidey Reply
how many start and codon
Esrael Reply
what is field
Felix Reply
physics, biology and chemistry this is my Field
ALIYU
field is a region of space under the influence of some physical properties
Collete
what is ogarnic chemistry
WISDOM Reply
determine the slope giving that 3y+ 2x-14=0
WISDOM
Another formula for Acceleration
Belty Reply
a=v/t. a=f/m a
IHUMA
innocent
Adah
pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
Nassze Reply
how do lnternal energy measures
Esrael
Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
JALLAH Reply
No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Robert
like charges repel while unlike charges atttact
Raymond
What is specific heat capacity
Destiny Reply
Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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