please help me i dont remember how to do these anymore
1. add (1/x+3) + (5/x-4)
2. subtract (3/x) - (7/x-2)
3. subract (3/x-3) - (4/x+5)
please help me i dont remember how to do these anymore
1. add (1/x+3) + (5/x-4)
2. subtract (3/x) - (7/x-2)
3. subract (3/x-3) - (4/x+5)
Fair enough!
What you need to do for these is to find a common denominator, which is most easily done by multiplying the two denominators. Here goes:
$\displaystyle \frac{1}{x+3}\cdot \frac{x-4}{x-4} + \frac{5}{x-4} \cdot \frac{x+3}{x+3} = \frac{x-4}{(x+3)(x-4)} + \frac{5(x+3)}{(x+3)(x-4)}$
$\displaystyle = \frac{x-4+5x+15}{(x+3)(x-4)} = \frac{6x+11}{(x+3)(x-4)}$
Well, by the same method. You find common denominators, and just subtract instead of adding. Here's how to start the second one:
$\displaystyle \frac{3}{x}\cdot\frac{x-2}{x-2} - \frac{7}{x-2}\cdot\frac{x}{x} = \frac{3(x-2)-7x}{x(x-2)}.$
Now simplify, and you should have your answer.
Same method for #3.
Let us know what you get, and we'll check your answers.
By the way, and not to be a stick-in-the-mud, but you're still not expressing these in a mathematically correct way, which is important because it's hard for us to tell what you mean. The correct way to use the parentheses would be
1. 1/(x+3) + 5/(x-4)
2. 3/x - 7/(x-2)
3. 3/(x-3) - 4/(x+5)
In other words, you need to group all the numerators and denominators together, so we know what is being divided into what. I hope that makes sense.
Unfortunately, no. For the second one, I have shown you how to get to
$\displaystyle \frac{3(x-2)-7x}{x(x-2)}.$
What you need to do now is carry out the multiplication in the numerator and combine like terms. You won't have to do anything to the denominator.
Give it a shot.