Chapter 7 Class Notes. Intermediate Algebra, MAT1033C. SI Leader Joe Brownlee. Palm Beach State College

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1 Chapter 7 Class Notes Intermediate Algebra, MAT033C Palm Beach State College

2 Class Notes 7. Professor Burkett 7. Rational Expressions and Functions; Multiplying and Dividing Chapter 7 takes factoring to the next level, so make sure you re set on the basics! In 7. we re dealing with rational expressions and functions, which is a fancy way of saying fractions. Example : Find all numbers that are not in the domain. f(x) = x + 4 x 6 When we re asked to find the numbers that are not in the domain, all we do is set the denominator equal to zero and solve. x 6 = 0 x = 6 Add six to both sides. x 6 x cannot equal six. The reason x cannot equal six is because if it did, it would make the denominator 0 (6-6=0); as we know, the denominator cannot equal 0, otherwise it is undefined. Example : Find all numbers that are not in the domain. f(x) = 3 + x 5 In this case, since there is no variable in the denominator, nothing we plug-in for x will make the denominator 0. There Example 3: Find all numbers that are not in the domain. f(x) = x x 6 Factor the denominator into (x + )(x 3) and set each factor equal to zero. x + = 0 x 3 = 0 x = x = 3 Therefore, x cannot equal - or 3, otherwise the denominator will equal 0, which will make it undefined.

3 Class Notes 7. Professor Burkett Next we ll be writing rational expressions in lowest terms. Let s practice a couple easier problems before tackling more difficult problems. Example 4: Write each rational function in lowest terms. 8k 6 We see that the 8 and 6 can be reduced: 8k 6 So we re left with k. Example 5: Write the rational function in lowest terms. y + y 4 y+ (y+)(y ) Factor the denominator (difference of two squares). y+ (y+)(y ) Cancel. = (y ) Lowest terms. Example 6: y y When the numerator and denominator are exact opposites, the rational expression is reduced to -. = -

4 Class Notes 7. Professor Burkett Example 7: m 6 m + m + 4 (m+4)(m 4) m+ m+4 Factor each numerator and denominator into lowest terms. (m+4)(m 4) m+ m+4 Cancel like terms. (m 4) Multiply numerators, multiply denominators. m+ = m 4 m + Example 8: y y 3 y + 4y + 4 y y + y When dividing fractions, we flip the second fraction and turn the division sign into multiplication. Sometimes this is referred to as keep, change, flip. y y 3 y + 4y + 4 y + y y (y+)(y 3) (y+)(y ) (y+)(y+) (y+)(y ) Factor each numerator and denominator into lowest terms. (y+)(y 3) (y+)(y ) (y+)(y+) (y+)(y ) Cancel. = (y 3) (y+) 3

5 Class Notes 7. Professor Burkett 7. Adding and Subtracting Rational Expressions When adding and subtracting fractions, they must have the same (or common ) denominator. If they start with the same denominator, great; just deal with the numerators. If the denominators are different, you must make them the same, by multiplying as necessary, then deal with the numerators. Example : Basic Review Example : Notice the denominator was the same in start, so we rewrite the problem with the numerators combined over one denominator. Remove the parenthesis in the numerator, ensuring to distribute the negative. Combine like terms in the numerator. Factor the GCF out of the numerator and cancel.

6 Class Notes 7. Professor Burkett Example 3: Notice the denominators in the original are different. We must not find a common denominator in order to continue. In this case, the common denominator will be each of the original denominators, (x+) (x-4). Once we find the common denominator, we multiply the numerator by what was missing in the original denominator. In this case, the 8 x+ was missing the (x-4) in its denominator, so when we put it in the denominator, we must multiply the numerator by that same binomial, because whatever we do to the top we must also do to the bottom. Therefore, we get 8(x-4) in the numerator. Repeat this process for the 5 x 4 to get 5(x+) in the new numerator. Example 4: Rewrite the fractions, factoring both denominators. Determine the common denominator, then multiply the numerators as necessary. Now that the denominators are equal, combine and simplify the numerator. In this case, when we simplified the numerator, nothing could cancel.

7 Class Notes 7.4 Professor Burkett 7.4 Equations with Rational Expressions and Graphs In this section, we ll build on the skills learned in 7. and 7.. Example : Solve the equation. 3 x + = 5 6x 3(6x) x + (6x) = 5(6x) 6x Find your common denominator and multiply each numerator by it to clear your fractions x = 5 Clear your fractions and simplify. 3x = 5 x = 5 Solve for x. Example : Solve the equation. x x x 4 = 8 x 6 x x x 4 = 8 (x + 4)(x 4) Factor the denominators to better determine a common denominator. (x 4)(x + 4) + x + 4 x(x + 4)(x 4) x 4 = 8(x + 4)(x 4) (x + 4)(x 4) Multiply each fraction by the common denominator to clear the fractions. x 4 + x + 4x = 8

8 Class Notes 7.4 Professor Burkett x + 5x + 4 = 0 Combine like terms and set the equation equal to zero. (x + )(x + 4) = 0 Factor. x + = 0 x + 4 = 0 Set each factor to 0. x = 4, x = You must check to see if plugging the solution into the original equation will make one of the denominators 0. In this case, if we plugged -4 into the original equation, we get a zero for two of the denominators. Therefore, we must drop that as an answer and only use -.

9 Class Notes 7.5 Professor Burkett Example : Solve for p, if f=5 and q=5. f = p + q 7.5 Applications of Rational Expressions 5 = p + 5 Plug in the given values where indicated. (75p) 5 = (75p) p + (75p) 5 Multiply each fraction by the common denominator to clear the fractions. 5p = p Simplify. p = 75 p = 75 Solve for p. Example : Solve for the specified variable. A = Rr R + r for R. A(R + r) = Rr(R + r) R + r Multiply both sides of the equation by the common denominator to clear the fraction. AR + Ar = Rr Simplify. Ar = Rr AR Since we re solving for R, bring all terms with R to one side of the equation.

10 Class Notes 7.5 Professor Burkett Ar = R(r A) Factor out R from the right side. Then divide to get R by itself. R = Ar r A Example 3: Lauren s car uses 5 gal. of gasoline to drive 495 miles. She has 6 gal. of gas in the car, and she wants to know how much more gasoline she will need to drive 600 miles. 5 gallons 495 miles = x gallons 600 miles Set up a simple proportion. 5(600) = 495x Cross multiply = 495x x = 8. gallons Simplify. Therefore, if Lauren already has 6 gallons, she will need an additional. gallons of gas to make her 600 mile journey.