College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.1 - Real Numbers - R.1 Assess Your Understanding - Page 16: 78



Work Step by Step

Find the LCD of the two fractions: $15 = 5(3) \\9=3(3)$ The LCM is $5(3)(3) =45$ Thus, the LCD is also $45$ Make the fractions similar using their LCD to obtain: $=\dfrac{2(3)}{15(3)} + \dfrac{8(5)}{9(5)} \\=\dfrac{6}{45}+\dfrac{40}{45} \\=\dfrac{6+40}{45} \\=\dfrac{46}{45}$
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