[(1/9)- (1/n)]
/
[(n+9)/ 9]
[(1/9)- (1/n)]
/
[(n+9)/ 9]
To divide fractions you have to multiply by the reciprocal of the second fraction:
(1/9)-(1/n) * 9/(1+9)
Now, focus on the left side of the multiplication sign. In order to subtract the 2 fractions, you need a common denominator, 9n. Multiply each fraction to make their denominators equal to 9n:
1/9 * n/n = n/9n
1/n * 9/9 = 9/9n
Now it should look like this: (n/9n) - (9/9n)
Subtract the 2 fractions to get: n-9/9n
The full problem should look like this: (n-9/9n) * 9/(1+9)
To help with simplifying later, you can cancel out the 9 in the denominator of the first fraction and the 9 in the numerator of the second fraction because 9/9 = 1.
Now you have: n-9/n * 1/(1+9)
Multiply across:
Numerator: n-9 * 1 = n-9
Denominator (distribute the n): n * 1+9 = n(1+9)
Your answer is: n-9/n(1+9)
When you add or sub fractions you need a common denominator. Just working with the numerator of the whole problem here I will keep the original problem outside of parenthesis and the multipliers inside. Remember, you have to multiply both the top and bottom of the fraction by the same amount if you are altering in any way:
we have a 9 and an n, so our common denominator will be 9n
1 (n) - 1 (9) = n - 9 = n - 9
9 (n) n (9) 9n 9n 9n
Now you have the following whole problem. I am going to represent the division line of the whole fraction with a row of
n - 9
9n
_______
n + 9
9
To divide by a fraction you keep the first(top) fraction the same, and flip the second one(reciprocal) and then multipy the two together
n - 9 x 9 = 9 (n - 9) = 9 (n - 9) = n - 9
9n n + 9 9n(n + 9) 9n(n + 9) n^{2} + 9n
Comments
is the denominator for the whole larger problem [(n+9)/9] or [(1+9)/9]?
Comment