To solve the problem, I knew I had to subtract the distance Jill jumped from the distance her friend jumped. Since 6 - 6 = 0 I knew I needed to subtract the fraction in Jill's long-jump distance from the fraction in her friend's long-jump distance. When subtracting fractions, the first step is to find a common denominator. Since 21 is a common multiple of 3 and 7, I decided to make the common denominator 21. I multiplied the numerator and the denominator by the sarne number to write an equivalent fraction with a denominator of 21. Finally, I subtracted 7 from 15 to find the numerator, and I kept the denominator the same, 21.

Question

To solve the problem, I knew I had to subtract the distance Jill jumped from the distance her friend jumped. Since 6 - 6 = 0 I knew I needed to subtract the fraction in Jill's long-jump distance from the fraction in her friend's long-jump distance. When subtracting fractions, the first step is to find a common denominator. Since 21 is a common multiple of 3 and 7, I decided to make the common denominator 21. I multiplied the numerator and the denominator by the sarne number to write an equivalent fraction with a denominator of 21. Finally, I subtracted 7 from 15 to find the numerator, and I kept the denominator the same, 21.

Change the wording but still answer the question Jill jumped feet in the long-jump event. Jill's best friend jumped feet. How much farther did Jill's best friend jump? Describe in words the process you used to solve the problem.

Answer 1

To determine the extent to which Jill's best friend outperformed her in the long-jump event, I subtracted the distance Jill jumped from the distance her friend jumped. Since 6 - 6 equals 0, I recognized that I needed to calculate the difference between the fractions representing their long-jump distances. Subtracting fractions involves finding a common denominator, so I identified 21 as a common multiple of 3 and 7. Consequently, I aimed to establish a common denominator of 21 by multiplying the numerator and denominator of each fraction by the same number to achieve equivalent fractions. Finally, by subtracting 7 from 15, I computed the numerator, while maintaining the denominator as 21.