college algebra

Module 5: Rational Expressions, Equations, and Functions

Rational Equation

A  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

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.

Module 5: Rational Expressions, Equations, and Functions

Rational Equation

A  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

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.