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Chapter 6 Rational Expressions § 6.1 Rational Functions and Multiplying and Dividing Rational Expressions Rational Expressions Rational expressions can be written in the form where P and Q are both polynomials and Q 0. P Q Examples of Rational Expressions 3x 2 2 x 4 4x 5 4x 3 y 2 2 2 x 3xy 4 y Martin-Gay, Intermediate Algebra, 5ed 3x 2 4 3 Evaluating Rational Expressions To evaluate a rational expression for a particular value(s), substitute the replacement value(s) into the rational expression and simplify the result. Example: Evaluate the following expression for y = 2. y2 4 4 2 2 5 y 5 (2) 7 7 Martin-Gay, Intermediate Algebra, 5ed 4 Evaluating Rational Expressions In the previous example, what would happen if we tried to evaluate the rational expression for y = 5? y2 3 52 5 y 5 5 0 This expression is undefined! Martin-Gay, Intermediate Algebra, 5ed 5 Undefined Rational Expressions We have to be able to determine when a rational expression is undefined. A rational expression is undefined when the denominator is equal to zero. The numerator being equal to zero is okay (the rational expression simply equals zero). Martin-Gay, Intermediate Algebra, 5ed 6 Undefined Rational Expressions Find any real numbers that make the following rational expression undefined. Example: 9 x3 4 x 15 x 45 The expression is undefined when 15x + 45 = 0. So the expression is undefined when x = 3. Martin-Gay, Intermediate Algebra, 5ed 7 Simplifying Rational Expressions Simplifying a rational expression means writing it in lowest terms or simplest form. Fundamental Principle of Rational Expressions For any rational expression P and any polynomial R, Q where R ≠ 0, PR P R P P 1 QR Q R Q Q PR P . or, simply, QR Q Martin-Gay, Intermediate Algebra, 5ed 8 Simplifying Rational Expressions Simplifying a Rational Expression 1) Completely factor the numerator and denominator of the rational expression. 2) Divide out factors common to the numerator and denominator. (This is the same thing as “removing the factor of 1.”) Warning! Only common FACTORS can be eliminated from the numerator and denominator. Make sure any expression you eliminate is a factor. Martin-Gay, Intermediate Algebra, 5ed 9 Simplifying Rational Expressions Example: Simplify the following expression. 7 x 35 7( x 5) 7 2 x( x 5) x 5x x Martin-Gay, Intermediate Algebra, 5ed 10 Simplifying Rational Expressions Example: Simplify the following expression. x 3x 4 ( x 4)( x 1) x 1 2 x x 20 ( x 5)( x 4) x 5 2 Martin-Gay, Intermediate Algebra, 5ed 11 Simplifying Rational Expressions Example: Simplify the following expression. 7 y 1( y 7) 1 y7 y7 Martin-Gay, Intermediate Algebra, 5ed 12 Multiplying Rational Expressions Multiplying Rational Expressions The rule for multiplying rational expressions is P R PR as long as Q 0 and S 0. Q S QS 1) Completely factor each numerator and denominator. 2) Use the rule above and multiply the numerators and denominators. 3) Simplify the product by dividing the numerator and denominator by their common factors. Martin-Gay, Intermediate Algebra, 5ed 13 Multiplying Rational Expressions Example: Multiply the following rational expressions. 2 3 x x 5 x 6 x 5x 1 3 10 x 12 2 5 x x x 2 2 3 4 2 Martin-Gay, Intermediate Algebra, 5ed 14 Multiplying Rational Expressions Example: Multiply the following rational expressions. ( m n) m (m n)( m n) m 2 m n m mn (m n) m(m n) 2 mn mn Martin-Gay, Intermediate Algebra, 5ed 15 Dividing Rational Expressions Dividing Rational Expressions The rule for dividing rational expressions is P R P S PS Q S Q R QR as long as Q 0 and S 0 and R 0. To divide by a rational expression, use the rule above and multiply by its reciprocal. Then simplify if possible. Martin-Gay, Intermediate Algebra, 5ed 16 Dividing Rational Expressions Example: Divide the following rational expression. 25 ( x 3) 5 x 15 ( x 3) 5 5 x 15 5 25 2 2 ( x 3)( x 3) 5 5 x3 5 5( x 3) Martin-Gay, Intermediate Algebra, 5ed 17 § 6.2 Adding and Subtracting Rational Expressions Martin-Gay, Intermediate Algebra, 5ed 18 Adding and Subtracting with Common Denominators Adding or Subtracting Rational Expressions with Common Denominators The rule for multiplying rational expressions is P R are rational expression, then If and Q Q P R PR P R PR and Q Q Q Q Q Q Martin-Gay, Intermediate Algebra, 5ed 19 Adding Rational Expressions Example: Add the following rational expressions. 4p 3 3p 8 7 p 5 4p 3 3p 8 2p 7 2p 7 2p 7 2p 7 Martin-Gay, Intermediate Algebra, 5ed 20 Subtracting Rational Expressions Example: Subtract the following rational expressions. 8 y 16 8( y 2) 8y 16 8 y2 y2 y2 y2 Martin-Gay, Intermediate Algebra, 5ed 21 Subtracting Rational Expressions Example: Subtract the following rational expressions. 3y 6 3y 6 2 2 2 y 3 y 10 y 3 y 10 y 3 y 10 3( y 2) ( y 5)( y 2) 3 y5 Martin-Gay, Intermediate Algebra, 5ed 22 Least Common Denominators To add or subtract rational expressions with unlike denominators, you have to change them to equivalent forms that have the same denominator (a common denominator). This involves finding the least common denominator of the two original rational expressions. Martin-Gay, Intermediate Algebra, 5ed 23 Least Common Denominators Finding the Least Common Denominator 1) Factor each denominator completely, 2) The LCD is the product of all unique factors each raised to a power equal to the greatest number of times that the factor appears in any factored denominator. Martin-Gay, Intermediate Algebra, 5ed 24 Least Common Denominators Example: Find the LCD of the following rational expressions. 1 3x , 6 y 4 y 12 6 y 2 3y 4 y 12 4( y 3) 2 ( y 3) 2 So the LCD is 2 3 y( y 3) 12 y( y 3) 2 Martin-Gay, Intermediate Algebra, 5ed 25 Least Common Denominators Example: Find the LCD of the following rational expressions. 4 4x 2 , 2 2 x 4 x 3 x 10 x 21 x 4 x 3 ( x 3)( x 1) 2 x 10 x 21 ( x 3)( x 7) 2 So the LCD is (x 3)(x 1)(x 7) Martin-Gay, Intermediate Algebra, 5ed 26 Least Common Denominators Example: Find the LCD of the following rational expressions. 2 3x 4x , 2 2 5x 5 x 2 x 1 5x 5 5( x 1) 5( x 1)( x 1) 2 2 x 2 x 1 ( x 1) 2 2 So the LCD is 5(x 1)(x -1) Martin-Gay, Intermediate Algebra, 5ed 2 27 Least Common Denominators Example: Find the LCD of the following rational expressions. 1 2 , x 3 3 x Both of the denominators are already factored. Since each is the opposite of the other, you can use either x – 3 or 3 – x as the LCD. Martin-Gay, Intermediate Algebra, 5ed 28 Multiplying by 1 To change rational expressions into equivalent forms, we use the principal that multiplying by 1 (or any form of 1), will give you an equivalent expression. P P P R PR 1 Q Q Q R QR Martin-Gay, Intermediate Algebra, 5ed 29 Equivalent Expressions Example: Rewrite the rational expression as an equivalent rational expression with the given denominator. 3 5 9y 72 y 9 4 3 8y 24 y 4 3 4 5 5 9 9y 8y 9y 72 y Martin-Gay, Intermediate Algebra, 5ed 30 Unlike Denominators As stated in the previous section, to add or subtract rational expressions with different denominators, we have to change them to equivalent forms first. Martin-Gay, Intermediate Algebra, 5ed 31 Unlike Denominators Adding or Subtracting Rational Expressions with Unlike Denominators 1) Find the LCD of the rational expressions. 2) Write each rational expression as an equivalent rational expression whose denominator is the LCD found in Step 1. 3) Add or subtract numerators, and write the result over the denominator. 4) Simplify resulting rational expression, if possible. Martin-Gay, Intermediate Algebra, 5ed 32 Adding with Unlike Denominators Example: Add the following rational expressions. 15 8 , 7 a 6a 15 8 6 15 7 8 7 a 6 a 6 7 a 7 6a 73 90 56 146 21a 42a 42a 42a Martin-Gay, Intermediate Algebra, 5ed 33 Subtracting with Unlike Denominators Example: Subtract the following rational expressions. 5 3 , 2x 6 6 2x 5 3 5 3 2x 6 6 2x 2x 6 2x 6 4 222 8 2 x 6 2( x 3) x 3 Martin-Gay, Intermediate Algebra, 5ed 34 Subtracting with Unlike Denominators Example: Subtract the following rational expressions. 7 and 3 2x 3 7 3(2 x 3) 7 3 2x 3 2x 3 2x 3 7 6 x 9 7 6 x 9 16 6 x 2x 3 2x 3 2x 3 2x 3 Martin-Gay, Intermediate Algebra, 5ed 35 Adding with Unlike Denominators Example: Add the following rational expressions. 4 x , 2 2 x x 6 x 5x 6 4 x 4 x 2 2 x x 6 x 5 x 6 ( x 3)( x 2) ( x 3)( x 2) 4( x 3) x( x 3) ( x 3)( x 2)( x 3) ( x 3)( x 2)( x 3) 2 2 x x 12 4 x 12 x 3x ( x 2)( x 3)( x 3) ( x 2)( x 3)( x 3) Martin-Gay, Intermediate Algebra, 5ed 36 § 6.3 Simplifying Complex Fractions Martin-Gay, Intermediate Algebra, 5ed 37 Complex Rational Fractions Complex rational expressions (complex fraction) are rational expressions whose numerator, denominator, or both contain one or more rational expressions. There are two methods that can be used when simplifying complex fractions. Martin-Gay, Intermediate Algebra, 5ed 38 Simplifying Complex Fractions Simplifying a Complex Fraction (Method 1) 1) Simplify the numerator and the denominator of the complex fraction so that each is a single fraction. 2) Perform the indicated division by multiplying the numerator of the complex fraction by the reciprocal of the denominator of the complex fraction. 3) Simplify, if possible. Martin-Gay, Intermediate Algebra, 5ed 39 Simplifying Complex Fractions Example: x 2 2 x 2 2 x 4 x4 2 2 2 x4 2 x4 2 x4 x4 x 4 x4 2 2 2 Martin-Gay, Intermediate Algebra, 5ed 40 Simplifying Complex Fractions Simplifying a Complex Fraction (Method 2) 1) Multiply the numerator and the denominator of the complex fraction by the LCD of the fractions in both the numerator and denominator. 2) Simplify, if possible. Martin-Gay, Intermediate Algebra, 5ed 41 Simplifying Complex Fractions Example: 1 2 2 2 2 y 3 6y 6 4y 2 2 1 5 6y 6y 5y y 6 Martin-Gay, Intermediate Algebra, 5ed 42 Simplifying with Negative Exponents When you have a rational expression where some of the variables have negative exponents, rewrite the expression using positive exponents. The resulting expression is a complex fraction, so use either method to simplify the expression. Martin-Gay, Intermediate Algebra, 5ed 43 Simplifying with Negative Exponents Example: x 2 y 2 Simplify 1 1 . 5x 2 y x 2 y 2 1 1 5x 2 y 1 1 1 2 2 2 2 2 2 x y 1 x y x y 2 2 5 2 x2 y2 1 1 5 xy 2 x y 5 2 x y x y Martin-Gay, Intermediate Algebra, 5ed 44 § 6.4 Dividing Polynomials: Long Division and Synthetic Division Martin-Gay, Intermediate Algebra, 5ed 45 Dividing a Polynomial by a Monomial Dividing a Polynomial by a Monomial Divide each term in the polynomial by the monomial. ab a b , where c 0. c c c Example: Divide 12a 3 36a 15 12a 3 36a 15 3a 3a 3a 3a 5 2 4a 12 a Martin-Gay, Intermediate Algebra, 5ed 46 Dividing a Polynomial by a Monomial 4 3 2 10 t 35 t 5 t Example: Divide 5t 2 10t 4 35t 3 5t 2 10t 4 35t 3 5t 2 5t 2 5t 2 5t 2 5t 2 2t 2 7 t 1 Martin-Gay, Intermediate Algebra, 5ed 47 Dividing Polynomials Dividing a polynomial by a polynomial other than a monomial uses a “long division” technique that is similar to the process known as long division in dividing two numbers, which is reviewed on the next slide. Martin-Gay, Intermediate Algebra, 5ed 48 Dividing Polynomials 168 43 7256 43 295 258 37 6 344 32 Divide 43 into 72. Multiply 1 times 43. Subtract 43 from 72. Bring down 5. Divide 43 into 295. Multiply 6 times 43. Subtract 258 from 295. Bring down 6. Divide 43 into 376. Multiply 8 times 43. Subtract 344 from 376. Nothing to bring down. We then write our result as 32 168 . 43 Martin-Gay, Intermediate Algebra, 5ed 49 Dividing Polynomials As you can see from the previous example, there is a pattern in the long division technique. Divide Multiply Subtract Bring down Then repeat these steps until you can’t bring down or divide any longer. We will incorporate this same repeated technique with dividing polynomials. Martin-Gay, Intermediate Algebra, 5ed 50 Dividing Polynomials 4x 5 2 7 x 3 28x 23x 15 2 28 x 12 x 35 x 15 35x 15 Divide 7x into 28x2. Multiply 4x times 7x+3. Subtract 28x2 + 12x from 28x2 – 23x. Bring down – 15. Divide 7x into –35x. Multiply – 5 times 7x+3. Subtract –35x–15 from –35x–15. Nothing to bring down. So our answer is 4x – 5. Martin-Gay, Intermediate Algebra, 5ed 51 Dividing Polynomials 2 x 10 2 2 x 7 4x 6 x 8 2 4 x 14 x 20x 8 20x 70 78 Divide 2x into 4x2. Multiply 2x times 2x+7. Subtract 4x2 + 14x from 4x2 – 6x. Bring down 8. Divide 2x into –20x. Multiply -10 times 2x+7. Subtract –20x–70 from –20x+8. Nothing to bring down. We write our final answer as 2 x 10 Martin-Gay, Intermediate Algebra, 5ed 78 ( 2 x 7) 52 Dividing Polynomials Using Long Division 2 c 3c 2 Example: Divide: c 1 c c 1 c 2 3c 2 c2 c 2. Multiply c by c + 1. 1. Divide the leading term of the dividend, c2, by the first term of the divisor, x. c c2 c 2c c(c 1) c 2 c 3. Subtract c2 + c from c2 + 3c – 2. (c 2 3c) (c 2 c) 2c Continued. Martin-Gay, Intermediate Algebra, 5ed 53 Dividing Polynomials Using Long Division Example continued: c 2 c 1 c 2 3c 2 c2 c 4. Repeat the process until the degree of the remainder is less than the degree of the binomial divisor. 2c 2 2c 2 4 Bring down the next term to obtain a new polynomial. Remainder 5. Check by verifying that (Quotient)(Divisor) + Remainder = Dividend. (c 2)(c 1) (4) c 2 3c 2 (4) c2 3c 2 c 2 3c 2 c 2 4 c 1 c 1 Martin-Gay, Intermediate Algebra, 5ed 54 Dividing Polynomials Using Long Division Example: Divide: (y2 – 5y + 6) ÷ (y – 2) y –3 y 2 y2 5 y 6 y2 – 2y – 3y + 6 – 3y + 6 0 No remainder Check : (y – 2)(y – 3) = y2 – 5y + 6 (y2 – 5y + 6) ÷ (y – 2) = y – 3 Martin-Gay, Intermediate Algebra, 5ed 55 Synthetic Division To find the quotient and remainder when a polynomial of degree 1 or higher is divided by x – c, a shortened version of long division called synthetic division may be used. Long Division Synthetic Division x 2 x 1 x 2 3x 2 ( x 2 x ) x 2 x 1 x 2 3x 2 x 2x 2 4 2x 2 (2 x 2) 4 Subtraction signs are not necessary. The original terms in red are not needed because they are the same as the term directly above. Continued. Martin-Gay, Intermediate Algebra, 5ed 56 Synthetic Division Continued. x 2 x 1 1 3 2 1 2 2 4 x 2 x 1 x 2 3x 2 x 2x 2 4 x 2 x 1 1 3 2 1 2 2 4 The variables are not necessary. The “boxed” numbers can be aligned horizontally. Martin-Gay, Intermediate Algebra, 5ed Continued. 57 Synthetic Division Continued. x 2 x 1 1 3 2 1 2 1 2 4 The first two numbers in the last row are the coefficients of the quotient, the last number is the remainder. The leading coefficient of the dividend can be brought down. The x + 1 is replaced with a – 1. 1 1 3 2 1 2 1 2 4 Quotient To simplify further, the top row can be removed and instead of subtracting, we can change the sign of each entry and add. Remainder Martin-Gay, Intermediate Algebra, 5ed 58 § 6.5 Solving Equations Containing Rational Expressions Martin-Gay, Intermediate Algebra, 5ed 59 Solving Equations First note that an equation contains an equal sign and an expression does not. To solve EQUATIONS containing rational expressions, clear the fractions by multiplying both sides of the equation by the LCD of all the fractions. Then solve as in previous sections. Note: this works for equations only, not simplifying expressions. Martin-Gay, Intermediate Algebra, 5ed 60 Solving Equations Example: Solve the following rational equation. 5 7 1 3x 6 5 7 6 x 1 6 x 3x 6 10 6x 7 x 10 x Check in the original equation. 5 7 1 3 10 6 5 7 1 30 6 1 7 1 6 6 Martin-Gay, Intermediate Algebra, 5ed true 61 Solving Equations Example: Solve the following rational equation. 1 1 1 2 2 x x 1 3x 3x 1 1 1 6 xx 1 6 xx 1 2 x x 1 3x( x 1) 3x 1 6x 2 3x 3 6x 2 3 3x 2 3x 1 x 1 3 Martin-Gay, Intermediate Algebra, 5ed Continued. 62 Solving Equations Example continued: Substitute the value for x into the original equation, to check the solution. 1 1 1 2 1 1 2 1 3 1 3 1 3 3 3 3 3 3 1 2 4 1 1 3 6 3 3 true 4 4 4 So the solution is x 13 Martin-Gay, Intermediate Algebra, 5ed 63 Solving Equations Example: Solve the following rational equation. x2 1 1 2 x 7 x 10 3x 6 x 5 x2 1 1 3 x 2 x 5 2 3x 2 x 5 x 7 x 10 3 x 6 x 5 3x 2 x 5 3x 2 3x 6 x 5 3x 6 3x x 3x 5 6 6 5x 7 x 7 5 Martin-Gay, Intermediate Algebra, 5ed Continued. 64 Solving Equations Example continued: Substitute the value for x into the original equation, to check the solution. 7 2 1 1 5 2 7 7 7 7 7 10 3 5 6 5 5 5 5 3 1 1 5 49 49 10 21 6 18 5 25 5 5 5 5 5 18 9 18 true So the solution is x 7 5 Martin-Gay, Intermediate Algebra, 5ed 65 Solving Equations Example: Solve the following rational equation. 1 2 x 1 x 1 1 2 x 1x 1 x 1x 1 x 1 x 1 x 1 2x 1 x 1 2x 2 3 x Continued. Martin-Gay, Intermediate Algebra, 5ed 66 Solving Equations Example continued: Substitute the value for x into the original equation, to check the solution. 1 2 3 1 3 1 1 2 2 4 true So the solution is x = 3. Martin-Gay, Intermediate Algebra, 5ed 67 Solving Equations Example: Solve the following rational equation. 12 3 2 2 9a 3 a 3 a 3 2 12 3 a 3 a 3 a 3 a 2 3 a 3 a 9a 12 33 a 23 a 12 9 3a 6 2a 21 3a 6 2a 15 5a 3a Martin-Gay, Intermediate Algebra, 5ed Continued. 68 Solving Equations Example continued: Substitute the value for x into the original equation, to check the solution. 12 3 2 2 33 33 93 12 3 2 0 5 0 Since substituting the suggested value of a into the equation produced undefined expressions, the solution is . Martin-Gay, Intermediate Algebra, 5ed 69 § 6.6 Rational Equations and Problem Solving Martin-Gay, Intermediate Algebra, 5ed 70 Solving Equations with Multiple Variables Solving Equations for a Specified Variable 1) 2) 3) 4) 5) Clear the equation of fractions or rational expressions by multiplying each side of the equation by the LCD of the denominators in the equation. Use the distributive property to remove grouping symbols such as parentheses. Combine like terms on each side of the equation. Use the addition property of equality to rewrite the equation as an equivalent equation with terms containing the specified variable on one side and all other terms on the other side. Use the distributive property and the multiplication property of equality to get the specified variable alone. Martin-Gay, Intermediate Algebra, 5ed 71 Solving Equations with Multiple Variables Example: Solve the following equation for R1 1 1 1 R R1 R2 1 1 1 RR1R2 RR1R2 R R1 R2 R1R2 RR2 RR1 R1R2 RR1 RR2 R1 R2 R RR2 RR 2 R1 R2 R Martin-Gay, Intermediate Algebra, 5ed 72 Finding an Unknown Number Example: The quotient of a number and 9 times its reciprocal is 1. Find the number. 1.) Understand Read and reread the problem. If we let n = the number, then 1 = the reciprocal of the number n Continued Martin-Gay, Intermediate Algebra, 5ed 73 Finding an Unknown Number Example continued: 2.) Translate The quotient of is a number and 9 times its reciprocal n 1 9 n 1 = 1 Continued Martin-Gay, Intermediate Algebra, 5ed 74 Finding an Unknown Number Example continued: 3.) Solve 1 n 9 1 n 9 n 1 n n n 1 9 n2 9 n 3,3 Martin-Gay, Intermediate Algebra, 5ed Continued 75 Finding an Unknown Number Example continued: 4.) Interpret Check: We substitute the values we found from the equation back into the problem. Note that nothing in the problem indicates that we are restricted to positive values. 1 1 3 9 1 3 9 1 3 3 3 3 1 true 3 3 1 true State: The missing number is 3 or –3. Martin-Gay, Intermediate Algebra, 5ed 76 Ratios and Rates Ratio is the quotient of two numbers or two quantities. The ratio of the numbers a and b can also be a written as a:b, or . b The units associated with the ratio are important. The units should match. If the units do not match, it is called a rate, rather than a ratio. Martin-Gay, Intermediate Algebra, 5ed 77 Proportions Proportion is two ratios (or rates) that are equal to each other. a c b d We can rewrite the proportion by multiplying by the LCD, bd. This simplifies the proportion to ad = bc. This is commonly referred to as the cross product. Martin-Gay, Intermediate Algebra, 5ed 78 Solving Proportions Example: Solve the proportion for x. x 1 5 x2 3 3x 1 5x 2 3x 3 5x 10 2x 7 x 7 2 Martin-Gay, Intermediate Algebra, 5ed Continued. 79 Solving Proportions Example continued: Substitute the value for x into the original equation, to check the solution. 7 1 5 2 7 2 3 2 5 25 3 3 2 true So the solution is x 7 2 Martin-Gay, Intermediate Algebra, 5ed 80 Solving Proportions Example: If a 170-pound person weighs approximately 65 pounds on Mars, how much does a 9000-pound satellite weigh? 170 - pound person on Earth 65 - pound person on Mars 9000 - pound satellite on Earth x - pound satellite on Mars 170 x 9000 65 585,000 x 585000 / 170 3441 pounds Martin-Gay, Intermediate Algebra, 5ed 81 Solving Proportions Example: Given the following prices charged for various sizes of picante sauce, find the best buy. • 10 ounces for $0.99 • 16 ounces for $1.69 • 30 ounces for $3.29 Continued. Martin-Gay, Intermediate Algebra, 5ed 82 Solving Proportions Example continued: Size Price Unit Price 10 ounces $0.99 $0.99/10 = $0.099 16 ounces $1.69 $1.69/16 = $0.105625 30 ounces $3.29 $3.29/30 $0.10967 The 10 ounce size has the lower unit price, so it is the best buy. Martin-Gay, Intermediate Algebra, 5ed 83 Solving a Work Problem Example: An experienced roofer can roof a house in 26 hours. A beginner needs 39 hours to do the same job. How long will it take if the two roofers work together? 1.) Understand Read and reread the problem. By using the times for each roofer to complete the job alone, we can figure out their corresponding work rates in portion of the job done per hour. Time in hrs Experienced roofer 26 Beginner roofer 39 Together t Portion job/hr 1/26 /39 1/t Martin-Gay, Intermediate Algebra, 5ed Continued 84 Solving a Work Problem Example continued: 2.) Translate Since the rate of the two roofers working together would be equal to the sum of the rates of the two roofers working independently, 1 1 1 26 39 t Continued Martin-Gay, Intermediate Algebra, 5ed 85 Solving a Work Problem Example continued: 3.) Solve 1 1 1 26 39 t 1 1 1 78t 78t 26 39 t 3t 2t 78 5t 78 t 78 / 5 or 15.6 hours Continued Martin-Gay, Intermediate Algebra, 5ed 86 Solving a Work Problem Example continued: 4.) Interpret Check: We substitute the value we found from the proportion calculation back into the problem. 1 1 1 26 39 78 5 3 2 5 78 78 78 true State: The roofers would take 15.6 hours working together to finish the job. Martin-Gay, Intermediate Algebra, 5ed 87 Solving a Rate Problem Example: The speed of Lazy River’s current is 5 mph. A boat travels 20 miles downstream in the same time as traveling 10 miles upstream. Find the speed of the boat in still water. 1.) Understand Read and reread the problem. By using the formula d=rt, we can rewrite the formula to find that t = d/r. We note that the rate of the boat downstream would be the rate in still water + the water current and the rate of the boat upstream would be the rate in still water – the water current. Distance rate time = d/r Down 20 r + 5 20/(r + 5) Up 10 r – 5 10/(r – 5) Continued Martin-Gay, Intermediate Algebra, 5ed 88 Solving a Rate Problem Example continued: 2.) Translate Since the problem states that the time to travel downstairs was the same as the time to travel upstairs, we get the equation 20 10 r 5 r 5 Continued Martin-Gay, Intermediate Algebra, 5ed 89 Solving a Rate Problem Example continued: 3.) Solve 20 10 r 5 r 5 r 5r 5 20 10 r 5r 5 r 5 r 5 20r 5 10r 5 20r 100 10r 50 10r 150 r 15 mph Martin-Gay, Intermediate Algebra, 5ed Continued 90 Solving a Rate Problem Example continued: 4.) Interpret Check: We substitute the value we found from the proportion calculation back into the problem. 20 10 15 5 15 5 20 10 true 20 10 State: The speed of the boat in still water is 15 mph. Martin-Gay, Intermediate Algebra, 5ed 91 § 6.7 Variation and Problem Solving Martin-Gay, Intermediate Algebra, 5ed 92 Direct Variation y varies directly as x, or y is directly proportional to x, if there is a nonzero constant k such that y = kx. The family of equations of the form y = kx are referred to as direct variation equations. The number k is called the constant of variation or the constant of proportionality. Martin-Gay, Intermediate Algebra, 5ed 93 Direct Variation If y varies directly as x, find the constant of variation k and the direct variation equation, given that y = 5 when x = 30. y = kx 5 = k·30 k = 1/6 1 So the direct variation equation is y = x 6 Martin-Gay, Intermediate Algebra, 5ed 94 Direct Variation Example: If y varies directly as x, and y = 48 when x = 6, then find y when x = 15. y = kx 48 = k·6 8=k So the equation is y = 8x. y = 8·15 y = 120 Martin-Gay, Intermediate Algebra, 5ed 95 Direct Variation Example: At sea, the distance to the horizon is directly proportional to the square root of the elevation of the observer. If a person who is 36 feet above water can see 7.4 miles, find how far a person 64 feet above the water can see. Round your answer to two decimal places. Continued. Martin-Gay, Intermediate Algebra, 5ed 96 Direct Variation Example continued: We substitute our given value for the elevation into the equation. d k e 7.4 k 36 7 .4 d 64 6 7.4 6k 7 .4 k 6 So our equation is 7.4 59.2 d (8) 9.87 miles 6 6 7 .4 d e 6 Martin-Gay, Intermediate Algebra, 5ed 97 Inverse Variation y varies inversely as x, or y is inversely proportional to x, if there is a nonzero constant k such that y = k/x. The family of equations of the form y = k/x are referred to as inverse variation equations. The number k is still called the constant of variation or the constant of proportionality. Martin-Gay, Intermediate Algebra, 5ed 98 Inverse Variation Example: If y varies inversely as x, find the constant of variation k and the inverse variation equation, given that y = 63 when x = 3. y = k/x 63 = k/3 k = 63·3 k = 189 189 So the inverse variation equation is y = x Martin-Gay, Intermediate Algebra, 5ed 99 Powers of x y can vary directly or inversely as powers of x, as well. y varies directly as a power of x if there is a nonzero constant k and a natural number n such that y = kxn y varies inversely as a power of x if there is a nonzero constant k and a natural number n such that y k x n Martin-Gay, Intermediate Algebra, 5ed 100 Powers of x Example: The maximum weight that a circular column can hold is inversely proportional to the square of its height. If an 8-foot column can hold 2 tons, find how much weight a 10-foot column can hold. Continued. Martin-Gay, Intermediate Algebra, 5ed 101 Powers of x Example continued: k w 2 h k k 2 2 8 64 We substitute our given value for the height of the column into the equation. 128 128 w 2 1.28 tons 10 100 k 128 So our equation is 128 w 2 h Martin-Gay, Intermediate Algebra, 5ed 102 Combined Variation y varies jointly as, or is jointly proportional to two (or more) variables x and z, if there is a nonzero constant k such that y = kxz. k is STILL the constant of variation or the constant of proportionality. We can also use combinations of direct, inverse, and joint variation. These variations are referred to as combined variations. Martin-Gay, Intermediate Algebra, 5ed 103 Combined Variation Example: Write the following statements as equations. • a varies jointly as b and c a = kbc • P varies jointly as R and the square of S P = kRS2 • The weight (w) varies jointly with the width (d) and the square of the height (h) and inversely with the length (l) kdh2 w l Martin-Gay, Intermediate Algebra, 5ed 104