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### Rational Equations

- A
*rational equation*is an equation that contains rational expressions. - To solve a rational equation, multiply each term on both sides by the LCM of all the denominators in the equation. Then solve the resulting equation, which has no fractions.
- An
*extraneous solution*is a value that makes one or more of the denominators in the original equation equal to 0. Always check all potential solutions in the*original*equation. Exclude extraneous values from the solution set. - Here is a better way to deal with extraneous solutions: before solving the equation, determine the values that must be excluded by setting each denominator equal to 0 and solving. Then you will recognize an extraneous solution as soon as it appears as a possible solution.

### Rational Equations

Solve:

[x/(x + 1)] = [(x − 4)/(x − 5)]

[x/(x + 1)] = [(x − 4)/(x − 5)]

- x( x − 5 ) = ( x + 1 )( x − 4 )
- x
^{2}− 5x = x^{2}− 3x − 4 - − 2x = − 4

x = 2

Solve:

[y/(y + 8)] = [(y + 6)/(y − 7)]

[y/(y + 8)] = [(y + 6)/(y − 7)]

- y( y − 7 ) = ( y + 8 )( y + 6 )
- y
^{2}− 7y = y^{2}+ 14y + 48 - − 21y = 48

y = − 2[2/7]

Solve:

[4/(x − 2)] = [(x + 6)/(x − 1)]

[4/(x − 2)] = [(x + 6)/(x − 1)]

- 4( x − 1 ) = ( x − 2 )( x + 6 )
- 4x − 4 = x
^{2}+ 4x − 12 - 8 = x
^{2}

x = ±√8

Solve:

[y/(2y − 1)] + [3y/(y − 5)] = 8

[y/(2y − 1)] + [3y/(y − 5)] = 8

- ( 2y − 1 )( y − 5 )g[y/(2y − 1)] + [3y/(y − 5)]( 2y − 1 )( y − 5 ) = 8( 2y − 1 )( y − 5 )
- y( y − 5 ) + 3y( 2y − 1 ) = 8( 2y
^{2}− 11 + 5 ) - y
^{2}− 5y + 6y^{2}− 3y = 16y^{2}− 88y + 40 - 0 = 9y
^{2}− 80y + 40 - x = [( − b ±√{b
^{2}− 4ac} )/2a] - y = [( − ( − 80 ) ±√{( − 80 )
^{2}− 4( 9 )( 40 )} )/2( 9 )] - y = [(80 ±√{4960} )/18]
- y = [(80 ±4√{310} )/18]
- y = [(40 ±2√{310} )/9]
- Extraneous Solutions: 2y − 1 = 0
- 2y = 1
- y = [1/2]
- y − 5 = 0

y = 5

[n/(n + 3)] + [n/(n − 4)] = 7

- ( n + 3 )( n − 4 )[n/(n + 3)] + ( n + 3 )( n − 4 )[n/(n − 4)] = 7( n + 3 )( n − 4 )
- n(n − 4) + n(n + 3) = 7(n
^{2}− n − 12) - n
^{2}− 4n + n^{2}+ 3n = 7n^{2}− 7n − 84 - 5n
^{2}− 6n − 84 = 0 - n = [( − ( − 6) ±√{( − 6)
^{2}− 4(5)( − 84)} )/2(5)] - n = [(6 ±√{1716} )/10]
- n = [(6 ±2√{429} )/10]
- n = [(3 ±√{429} )/5]
- Extraneous Solutions: n + 3 = 0
- n = − 3
- n − 4 = 0

n = 4

[3r/(r − 2)] + [10r/(2r + 1)] = 1

- 3r(2r + 1) + 10r(r − 2) = (r − 2)(2r + 1)
- 6r
^{2}+ 3r + 10r^{2}− 20r = 2r^{2}− 3r − 2 - 14r
^{2}− 14r + 2 = 0 - r = [( − ( − 14) ±√{( − 14)
^{2}− 4(14)(2)} )/2(14)] - r = [(14 ±√{84} )/28]
- r = [(14 ±2√{21} )/28]
- r = [(7 ±√{21} )/14]
- Extraneous Solutions: r − 2 = 0
- r = 2
- 2r + 1 = 0

r = − [1/2]

Eric finishes a puzzle in 20 minutes. Duy finished the same puzzle in 15 minutes. If they work together, how long does it take them to finish the same puzzle?

- [1/20] + [1/15] = [1/x]
- [1/20](20)(15)(x) + [1/15](20)(15)(x) = [1/x](20)(15)(x)
- 15x + 20x = 300
- 35x = 300

x = 8[4/7] minutes

Irene makes a costume in three hours. Michelle makes the same costume in five hours. If they work together how long does it take them to make the costume?

- [1/3] + [1/5] = [1/x]
- [1/3](5)(3)(x) + [1/5](5)(3)(x) = [1/x](5)(3)(x)
- 5x + 3x = 15
- 8x = 15

x = [15/8] = 1[7/8] hours

[3a/(a + 2)] + [(a − 6)/(a

^{2}− 4)] = 1- LCM = (a − 2)(a + 2)
- [3a/(a + 2)](a − 2)(a + 2) + [(a − b)/(a
^{2}− 4)](a − 2)(a + 2) = (a − 2)(a + 2) - (3a)(a − 2) + (a − 6) = a
^{2}− 4 - 3a
^{2}− 6a + a − 6 = a^{2}− 4 - 2a
^{2}− 5a − 2 = 0 - a = [( − ( − 5 ) ±√{( − 5 )
^{2}− 4( 2 )( 2 )} )/2( 2 )]

a = [(5 ±√{41} )/4]

[p/(3p + 2)] − [5p/(p − 1)] = 5

- LCM = (3p + 2)(p − 1)
- [p/(3p + 2)](3p + 2)(p − 1) − [5p/(p − 1)](3p + 2)(p − 1) = 5(3p + 2)(p − 1)
- p(p − 1) − 5p(3p + 2) = 5(3p
^{2}− p − 2) - p
^{2}− p − 15p^{2}− 10p = 15p^{2}− 5p − 10 - 29p
^{2}+ 6p − 10 = 0 - p = [( − 6 ±√{6
^{2}− 4( 29 )( − 10 )} )/2( 29 )] - p = [( − 6 ±√{1196} )/58]
- p = [( − 6 ±2√{299} )/58]

p = [( − 3 ±√{299} )/29]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Rational Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Definition 0:11
- Example: Cross Multiplication
- Example: Rational Expressions
- Solving Rational Equations 3:12
- Multiply by LCM of Denominators
- Example
- Work Problems 7:19
- Example: Complete a Project
- Extraneous Solutions 12:41
- Check All Solutions
- Example
- Example 1: Solve Rational Equation 17:28
- Example 2: Solve Rational Equation 19:45
- Example 3: Work Problem 27:15
- Example 4: Solve Rational Equation 31:10

1 answer

Last reply by: Dr Carleen Eaton

Tue Jun 18, 2013 7:02 PM

Post by Anwar Alasmari on June 18, 2013

Thank you Dr Carleen for these helpful lessons.

However, there is a typo in Example I that 3+3/3-1 should be 6/2, not 6/3. However, the final solution is correct 3=3.

1 answer

Last reply by: Dr Carleen Eaton

Tue Feb 5, 2013 3:02 PM

Post by Kenneth Montfort on February 1, 2013

I just want to say you have given me many awesome tools in this entire algebra 1 and you made topics so easy to understand with the different strategies that you used...so thank you!

0 answers

Post by Victor Castillo on January 24, 2013

Too fast