University:
California State University, Northridge
Course:
MATH 320 | Foundations of Higher Mathematics
Academic year:

2024
Views:

324

Pages:

16
Author:

zufriedendyu

§ 14.4 Adding and Subtracting Rational Expressions with Different Denominators 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. 38 Unlike Denominators Adding or Subtracting Rational Expressions with Unlike Denominators 1) Find the LCD of all the rational expressions. 2) Rewrite each rational expression as an equivalent one with the LCD as the denominator. 3) Add or subtract numerators and write result over the LCD. 4) Simplify rational expression, if possible. 39 Adding with Unlike Denominators Example Add the following rational expressions. 15 8 , 7 a 6a 15 8 6 ⋅15 7 ⋅ 8 + = + = 7 a 6a 6 ⋅ 7 a 7 ⋅ 6a 73 90 56 146 + = = 42a 42a 42a 21a 40 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 2⋅2⋅2 8 = = 2 x − 6 2( x − 3) x − 3 41 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 42 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 − 3 x = ( x + 2)( x − 3)( x + 3) ( x + 2)( x − 3)( x + 3) 43 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. 73 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 Continued 74 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 75 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 76 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 3 2 5 + = 78 78 78 5 true State: The roofers would take 15.6 hours working together to finish the job. 77 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 78 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 79 Solving a Rate Problem Example continued 3.) Solve 20 10 = r +5 r −5  20   10  (r + 5)(r − 5) (r + 5)(r − 5) =  r +5  r −5 20(r − 5) = 10(r + 5) 20r − 100 = 10r + 50 10r = 150 r = 15 mph Continued 80 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. 81