(3/x-3)-(18/x^2-9)=5/x+3?

Question

Help

Answer 1

x^2 - 9 = (x-3) (x+3)

So 3 / (x-3) - 18 / (x^2 - 9)
= 3 / (x-3) - 18 / [(x-3)(x+3)]
= [3(x+3)] / [(x-3)(x+3)] - 18 / [(x-3)(x+3)]
= [3x + 9 - 18] / [(x-3)(x+3)]
= [3x - 9] / [(x-3)(x+3)]
= [3(x-3)] / [(x-3)(x+3)]
= 3 / (x+3) for x ≠ 3, -3.

Danny L's answer is incorrect - if this is supposed to be an equation to solve, there is no solution as written. I made the assumption that you're trying to simplify the LHS, but if it is an equation we can see from my simplification above that we get 3 / (x+3) = 5 / (x+3) and hence 3 = 5 for any value of x for which the original equation is defined, so there is no solution.

If x = 3 then Danny L is multiplying by 0 on both sides, so the result does come out equal. But to check your answers you always have to put them into the ORIGINAL equation, and if you do that here with x = 3 you will get
3/0 - 18/0 = 5/6
which is obviously invalid.

Answer 2

I'm going to assume what you really mean is

3/(x-3)-18/(x^2-9)=5/(x+3), right? Remember, in order of operations, multiplications and divisions come before +/-, so as you stated it, it would actually be (3/x)-3-(18/x^2)-9=3+5/x...
anyway,


3/(x-3)-18/(x^2-9)=5/(x+3) ==

(factor that (x^2-9) )
3/(x-3)-18/[(x-3)(x+3)]=5/(x+3) ==

(now find common denominators-)

3(x+3)/[(x-3)(x+3)] - 18/[(x-3)(x+3)] = 5(x-3)/[(x-3)(x+3)] ==

(now toss the common denominators)

3(x+3) - 18 = 5(x-3) ==

3x + 9 - 18 = 5x -15 ==

3x - 9 = 5x -15 ==

6 = 2x ==

x = 3

Damn! Above is correct, when you substitute 3 for x back into the original equation, you have zero for 2 denominators - no solution. Wazzup?

Answer 3

the common denomator is (x-3)(x+3) thus multiplying on each side gives

3(x+3) - 18 = 5(x-3)

distribute

3x+9 - 18 = 5x - 15
-2x = -6
x=3

check

3(x+3) - 18 = 5(x-3) for x=3

3(6) -18 = 5(3-3)
18-18 =5(0)

0 = 0

Answer 4

(3/x) - 3 - (18/x^2) + 9 = (-18/x^2) + (3/x) + 6.

However,

(3/x) - 3 + (2/x) +6 = (5/x) + 3.

You should use parenthesis to indicate whether you mean (3/x) - 3 or 3/(x-3) etc.

Answer 5

3/(x-3) -18/(x^2-9) = 5/(x+3)
3(x+3)/(x-3)(x+3) - 18/(x+3)(x-3) = 5(x-3)/(x+3)(x-3)
3(x+3) - 18 = 5(x-3)
3x -9 = 5x - 15
6 = 2x
x = 3
Unfortunately, (3/x-3)-(18/x^2-9)=5/x+3 is undefined at x=3, because (x-3) = 0 at that point and you can't divide by zero.

Answer 6

Well if I knew what I'm looking for. Are we trying to find the LCD? Which would be (x-3)(x+3) because you foil out the x^2 -9.

Are we looking for the domain? Which is all real numbers except x cannot equal + or - 3. Which you find by setting the parts of the LCD above equal to zero.

Or are we solving for x??? In which you set the entire equation equal to zero.

Answer 7

(3/x) / (3/(x-2)) is the comparable as multiplying (3/x) by the reciprocal of (3/(x-2)), that's (x-2)/3 (3/x) ((x-2)/3) the three's cancel (x-2)/x separate the each words to simplify (x/x) - (2/x) = a million - (2/x)