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MATH 121, Lesson 2.5 Notes

by: Mallory McClurg

MATH 121, Lesson 2.5 Notes Math 121

Marketplace > University of Mississippi > Math > Math 121 > MATH 121 Lesson 2 5 Notes
Mallory McClurg
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About this Document

These notes are on rational expressions and equations. I've included details and step-by-step explanations on every type of problem we'll need to know how to solve!
College Algebra
Class Notes
Math, rational, equations, college, Algebra
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This 4 page Class Notes was uploaded by Mallory McClurg on Sunday September 11, 2016. The Class Notes belongs to Math 121 at University of Mississippi taught by Dirle in Fall 2016. Since its upload, it has received 62 views. For similar materials see College Algebra in Math at University of Mississippi.


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Date Created: 09/11/16
Math 121 Chapter 2 Notes Lesson 2.5 – Rational Expressions and Equations Example 1. 4 / (6x-6) (To find the variable restrictions, set the denominator to zero and solve. Why? Because we know that if the numerator equals zero, then the fraction is undefined.) 6x – 6 = 0 x ≠ 1 Ex2mple 2. 2 (x + x + 15)/(8x – 3x) (To find the restricted values of x, or the values that make this equation undefined, we can set the denominator equal to zero, because we know a zero denominator will make this undefined.) 2 8x – 3x = 0 (First, we can factor out an x.) x(8x – 3) = 0 (Set “8x – 3” equal to zero and solve for x.) x ≠ 0 x ≠ 3/8 (These are the restricted values of x.) Example 3. (3x – 10x – 8)/(x – 6x + 8) (We can see that there are to quadratic expressions, so we can try to factor each of them out.) (3x + 2)(x – 4)/(x – 4)(x – 2) (Like the last problem, the values of x in the denominator will help us determine what values of the variable must be excluded. So, set “x – 4” and “x – 2” equal to zero and solve for x.) x ≠ 4 x ≠ 2 (These are the variable that must be excluded.) Exam2le 4. (5x +18x – 8)/(x + 4) (To simplify this expression, first factor out the numerator.) (5x – 2)(x + 4) / (x + 4) (Cancel out the “x + 4” in the numerator and denominator. What’s left?) 5x – 2 (This is our simplified answer. What value for the variable must be excluded?) x ≠ (2/5) Example 5. (20y + 40)/(5 – (20/y )) 2 (We’re asked to simplify this expression. To get rid of the fraction in the denominator, we need to multi2ly everything in the equation by 20y .) (20y(20y ) + 40(20y ))/ (5(20y ) – (20/y )(20y )) 2 (Simplify.) 3 2 2 (400y + 800y ) / (100y – 400) (Now, what similar values do you see in the numerator? Factor out “400y ” 2 from the numerator. What similar values do you see in the denominator? Factor out “100” from the denominator.) (400y (y + 2)) / (100 (y – 4)) (We should notice that “y – 4” is the difference of two squares, so we can factor that into “y + 2” and “y – 2”.) (400y (y + 2)) / (100 (y + 2) (y – 2)) (To simplify, cross out the “y + 2” in the numer2tor and denominator. Then, simplify the 400y over 100.) 4y / (y – 2) Example 6. (4z – 9y ) / (2y + 3z) / z 2 (We are asked to simplify this complex rational equation. We know that a value with negative exponents can be rewritten as a fraction. The first fraction would 2 be “4/z ” and the second would be “-9/y . To combine the two fractions in the numerator, we must give them a common 2 2 denominator by multiplying both2by “y2z ”. Your numerator should now read “4y – 9z / y z ”. We should realize now that this numerator is the difference of two squares, so factor it out.) 2 2 2 ((2y + 3z) (2y – 3z)) / y z / (2y + 3z) / z (Now that we have the top numerator factored out, we can deal with the fraction on the bottom of the equation. Flip the fraction on the bottom so it reads (z / 2y + 3z) and then multiply it by the top fraction.) 2 2 2 ((2y + 3z)(2y – 3z)/ y z ) * (z / 2y + 3z) (Cross out all of the common values you see in the numerator and denominator across the multiplication problem to simplify.) 2 (2y – 3z)/y (This is as simplified as it gets!) Exam2le 7. 2 2 ((z + 4z + 4) / (5z +13z +6)) * ((5z – 17z -12)/ (z + 2)) (We’re asked to multiply the rational equations. First, we can factor out the numerator of the first fraction, which is (z + 2)(z + 2). Then we can factor out the denominator, which is (5z + 3)(z + 2). Cross out the similar “z + 2” in the numerator and denominator.) ((z + 2)/(5z + 3)) * ((5z – 17x – 12)/(z + 2)) (Now, factor out the numerator of the second fraction. It should read (5z + 3) and (z – 4).) ((z + 2)/(5z + 3)) * (5z + 3)(z – 4)/(z + 2) (Since we’re multiplying fractions, we can combine the numerators and denominators.) (z + 2)(5z + 3)(z – 4)/(5z + 3)(z + 2) (Cross out similar groups you see in the numerator and denominator to simplify.) z – 4 Example 8. (z – 13z + 40)/(z + 5) ÷ (z + 3z – 40)/(z + 5) (We are asked to divide the two fractions. First, factor out the numerators of both expressions. The first is (z – 5)(z – 8). The second is (z + 8)(z – 5).) (z – 5)(z – 8)/(z + 5) ÷ (z + 8)(z – 5)/(z + 5) (To divide fractions, we can choose one, flip the numerator and denominator, and then multiply the two fractions.) (z – 5)(z – 8)/(z + 5) * (z + 5)/(z + 8)(z – 5) (Now cross out the similar groups from the numerator and denominator of both equations to simplify.) (z – 8)/(z + 8) Example 9. ((y + 4)/(y – 6)) – ((y – 2)/(y + 5)) (We’re asked to subtract the fractions. So first we need to get a common denominator by multiplying the first fraction by (y + 5) and multiplying the second fraction by (y – 6).) ((y + 4)(y + 5)/(y – 6)(y + 5)) – ((y – 2)(y – 6)/(y + 5)(y – 6)) (Now that we have the same denominator in both fra2tions, we should FOIL both 2 numerators.) ((y + 9y + 20)/(y – 6)(y + 5)) – ((y – 8y + 12)/(y + 5)(y – 6)) (Since the denominators are the same, we just subtract numerator from numerator.) (17y + 8)/(y – 6)(y + 5) (This is your answer!) Example 10. (3/2x) – (7/12) = (1/8x) (To solve for x, we must first get the least common multiple of the denominators, which is 24x here. That is the smallest number that all of the denominators will fit into. Now, multiply the numerators of all three fractions by “24x”.) (72x/2x) – (168x/12) = (24x/8x) (Simplify each of the fractions.) 36 – 14x = 3 (Now, solve for x. Add 36 to both sides, then divide by -14.) x = 33/14 Example 11. (x/(x + 1)) – (x/(x +3x +2)) = (5/(x + 2)) (To solve this equation, we need the fractions to have the same denominator. First, factor out the denominator of the second fraction. It will be (x +1)(x +2), which shares values with the other two denominators, so this is good.) (x/(x + 1)) – (x/(x+1)(x +2)) = (5/(x + 2)) (Now, multiply the first fraction by (x +2), and the last fraction by (x+1). Distribute the “x” with (x+2), and distribute the “5” with (x +1). You should now have three fractions with (x +1)(x+2) as the denominators, which means we can now solve with the numerators.) 2 x + 2x – x = 5x +5 (Combine like terms.) x – 4x – 5 = 0 (Factor it out, then solve for “x”.) x = 1 x = 5 (We need to check both of these with the original equation, to make sure none of the values for x is restricted.) x ≠ 1 x = 5 (There is only one solution to this problem.) Example 12. ((3z/10) – (10/3z)) / (3 – (10/z) (To be clear, this is a fraction minus a fraction, over a 3 minus a fraction.. To simplify, we must first determine the least common multiple of the top of the equation, which is “30z”. Multiply the numerators of both fractions by 30z, which then makes this 2 ((90z /10) – (300z/3z)). Simplify this fraction to (9z – 100), then factor it out because it’s the difference of two squares.) ((3z +10)(3z – 10) / 30z) * (z/(3z – 10)) (Cross out the common terms to simplify.) (3z + 10)/30 (This is as simplified as it gets!)


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