Iwouldhighly appreciate it if someone could solve these forme.I think I typed themout correctly. What I'm looking at,for instance,the first one,is displayed 5n over 3 times 4n over 3

1.(5n/3)- (4n/3)
2.(3x/10)+(9x/10)
3.(n-2/n+4)+(8/[n+4])
4.(5m+8/[m-1]) - (m-3/[m-1])
5.(x^2/[x-1]) - (1/[x-1])
6.(4x+y/[2x+3y])-(2x -2y/[2x+3y])
7.(x^2+3x+2/[x^2-16] + (3x++6/[x^2-16])
8.7/(x-5) - (2+x/[x-5])

First, "(5n/3)- (4n/3)" is not "5n over 3 times 4n over 3."

If the fractions have the same denominator, just do the required operations in the numerator.

(5n/3)- (4n/3) = n/3

Use the same principle for the remaining problems.

I hope this helps. Thanks for asking.

To solve the given expressions, I will go through each one and simplify using the principles mentioned earlier. Here are the solutions:

1. (5n/3) - (4n/3) = n/3

2. (3x/10) + (9x/10) = (3x + 9x)/10 = 12x/10 = 6x/5

3. (n-2)/(n+4) + 8/(n+4) = (n-2+8)/(n+4) = (n+6)/(n+4)

4. (5m+8)/(m-1) - (m-3)/(m-1) = (5m+8-(m-3))/(m-1) = (5m+8-m+3)/(m-1) = (4m+11)/(m-1)

5. (x^2/(x-1)) - (1/(x-1)) = (x^2-1)/(x-1)

6. (4x+y)/(2x+3y) - (2x - 2y)/(2x+3y) = ((4x+y)-(2x-2y))/(2x+3y) = (4x+y-2x+2y)/(2x+3y) = (2x+3y)/(2x+3y) = 1

7. (x^2+3x+2)/(x^2-16) + (3x+6)/(x^2-16) = ((x^2+3x+2)+(3x+6))/(x^2-16) = (x^2+3x+2+3x+6)/(x^2-16) = (x^2+6x+8)/(x^2-16)

8. 7/(x-5) - (2+x)/(x-5) = (7-(2+x))/(x-5) = (7-2-x)/(x-5) = (5-x)/(x-5)

These are the simplified expressions for each problem. If you have any more questions, feel free to ask!