Study Guide Help

Question

Having some trouble on these three questions. Appreciate the help!

 

Find the missing numerator: 2/c-d=?/c^2-d^2

 

 

Rewrite each group of fractions with their LCD

 

x+1/8 and x-1/28

 

5/4bc^2 and 7/10c^3 and 4/b^2c

Answer 1

Hi Jack, if I understand your fractions correctly, here is what I would do:

 

For 2/(c - d) = x/(c² - d²)

 

c² - d² = (c - d) * (c + d)

 

Use this information to replace c² - d²

 

2/(c - d) = x/((c - d) * (c + d))

 

Cross-multiply:

 

2 * ((c - d) * (c + d)) = x * (c - d)

 

cancel (c - d) from both sides:

 

2 * (c + d) = x

 

The missing numerator is 2 * (c + d)

 

To find the LCD for 8 and 28, find their prime factorization:

 

Prime factorization for 8: 2 * 2 * 2, or 2³

 

Prime factorization for 28: 2 * 2 * 7, or 2² * 7

 

The LCD for 8 and 28 is composed of each prime number raised to its highest power. Here there are two prime numbers, 2 and 7. The highest power of 2 is 3 and the highest power of 7 is 1.

 

LCD = 2³ * 7, or 56

 

The two fractions rewritten with a denominator of 56 is:

 

(7 * (x + 1)) / 56 and (2 * (x - 1)) / 56

 

To rewrite the three fractions, you need to find the LCD for all three. There are three parts to consider. Part 1, the LCD of 4 and 10. Part 2, highest power for b, and part 3, highest power for c.

 

Part 1:

 

LCD for 4 and 10 is 20. 4 divides into 20, 5 times. 10 divides into 20, 2 times.

 

Part 2:

 

The highest power for b is 2.

 

Part 3:

 

The highest power for c is 3.

 

The LCD for the three terms is 20b²c³.

 

5/4bc² becomes 25bc/20b²c³ - multiply the fraction by 5bc/5bc

 

7/10c³ becomes 14b²/20b²c³ - multiply the fraction by 2b²/2b²

 

4/b²c becomes 80c²/20b²c³ - multiply the fraction by 20c²/20c²

 

Questions?