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CAmodule5lecturediscussion.docx

Module 5:  Solving rational equations

Discussion Instructions:

Please review  the class discussion directions . They tell you how to prepare and reflect before responding to the prompt below. They also include detailed guidelines how you must respond to classmates and then reply to the responses you receive. Please pay close attention to the response directions.  To earn your discussion points, you must post, respond, and reply as directed.

Discussion writing prompt

Please create a new forum thread and address the 3 parts of the initial post below:

1. Find a math problem as an example that demonstrates the following, “When adding or subtracting rational expressions, we write the fractions with the LCD then add or subtract the numerators while keeping the common denominator.”

2. Include an example of a rational equation and provide step-by-step explanation as to how to solve it.

3. Share two resources that can provide extra practice in solving rational explanations

 

Remember to review the  academic expectations  for your submission.

 

Submission reminders:

· Submit your initial discussion post by 11:59 pm ET on Wednesday.  Respond to two of your classmates' discussion posts, and reply to the responses you received by 11:59 pm ET on Sunday.

· Contribute a minimum of 250 words to the initial post.

· Follow established  netiquette (Links to an external site.)  guidelines when participating in forums.

Reply

CAmodule5discussion.docx

Module 5: Rational Expressions, Equations, and Functions

Rational Equation

rational equation is an equation that contains at least one rational expression.

 Steps to solve Rational Equations:

1. Determine the LCD of all rational expressions in the equation.

2. Multiply both sides of the equations by the LCD. This will result in every term in the equation being multiplied by the LCD. This will eliminate all the fractions from the equation.

3. Remove any parentheses and combine like terms on each side of the equation.

4. Solve the equation using the properties discussed in previous modules.

5. Check the solution in the original equation.

Example:

solve 3x4+12=2x−34

LCD: 4, then multiply both sides of the equation by 4

4(3x4+12)=(2x−34)4

Distributive property:  4(3x4)+4(12)=2x−3

group terms: 3x + 2 = 2x - 3  🡪  3x -2x = -3 -2

    x = -5

 

Proportions

A proportion is an equation of the form ab=cd,b≠0,d≠0

Proportions are rational equations and therefore may be solved by multiplying both sides by the LCD.

Example:

Solve x2x−3=9x−3

LCD: x - 3, therefore, multiply both sides by x - 3

          (x−3)x2x−3=9x−3(x−3)

          x2=9

          x2−9=0

Factor the difference of two squares: (x−3)(x+3)=0

x=±3

Check:

a) x=−3(−3)2−3−3=9−3−3=9−6=−32(true)

b) x=3323−3=93−3=90?90(undefined)

 

Solve Problems Involving Rational Functions

     Consider the functionf(x)=x−2x

     Find all  a for whichf(a)=1

Solution:

f(a)=a−2a=1, in which a≠0

LCD = a

Multiply both sides for the LCD: a(a−2a)=1.a

a2−2=a

a2−2−a=0ora2−a−2=0

(a−2)(a+1)=0

a−2=0ora+1=0

 a=2ora=−1

Check:

f(x)=x−2x

f(2)=2−22=2−1=1

f(−1)=−1−2−1=−1+2=1

Solve a Variable in a Formula Containing Rational

Expressions:

If, while solving for a variable in a formula, the variable appears in more than one term, factoring is needed to solve for the variable.

Example:

A formula used in banking is A = P + Prt, where A represents the amount that must be repaid to the bank when P dollars are borrowed at simple interest rate,  r, for time,  t, in years. Solve this equation for P.

Since both sides of the equation contains P

A = P + Prt

Factor out P : A = P (1+ rt)

Divide both sides by (1 + rt) to isolate P: A1+rt=P(1+rt)1+rt

Solution:  P =A1+rt

Rational Equations: Applications and Problem Solving

Example:

James Wan turns on the faucet to his tub, but forgets to close the drain. The faucet can fill the tub in 10 minutes and the drain can empty the tub in 18 minutes. If the faucet is on and the drain remains open, how long will it take for the tube to fill?

Solution:

Let x = amount of time needed to fill the tub

As water from the faucet is filling the tub, water going down the drain is emptying the tub (the faucet and drain are working against each other

Rate of work

Time worked

Part of tub

filled or emptied

Faucet filling tub

                  110

                       x

x10

Drain emptying tub

                  118

                       x

x18

 

part of tub filled in x minutes – part of tub emptied in x minutes = 1 (whole tub filled)

                                          x10−x18=1

LCD: 90

Multiply both sides by 90    ⟶90(x10−x18)=90(1)

90(x10)−90(x18)=90

9x−5x=90

4x=90

x=22.5   the tub will fill in 22.5 minutes

 

Variation

Variation equations show how one quantity changes in relation to another quantity or quantities.

Direct Variation: If a variable  y varies directly as variable x, then

y = k*x  (k: constant of proportionality)

Example:

The circumference of a circle,  C, is directly proportional to its radius,  r. Write the equation for the circumference of a circle if the constant of proportionality, k, is 2π

C = k*r

C = 2π*r = 2πr

Inverse Variation: Involve two variables in which one variable increases as the other decreases and vice versa.

If a variable  y varies inversely as a variable  x, then:

y=kx     (k: constant of proportionality)

Example:

The amount of time, t, it takes a block of ice to melt in water is inversely proportional to water’s temperature, T.

Write the variation equation”

t=KT

Now, if a block of ice takes 15 minutes to melt in 60°F water, determine the constant proportionality:

k = t*T (from the variation equation) = 15(60) = 900

References

Angel, A., & Rundle, D. (2018).  Intermediate algebra for college students. (10th edition). Pearson.