Can someone tell me if I am doing this right, I'm not sure?!

C

CalebsMomma

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Jul 13, 2009
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Add and express the result in the simplest form:
1/(x+4)+ 2/x

I need to get common denominators, so I should add 4 to 2/x. And since I add to the denominator I have to also add to the numerator, correct?
Which gives me...
1/(x+4) + 6/(x+4)=
(1+2+4)/x+4=
3/x, because the 4's cancel each other out, correct?
 
CalebsMomma said:
Add and express the result in the simplest form:
1/(x+4)+ 2/x

3/x, because the 4's cancel each other out, correct?
Click to expand...

No you can't, is thier multiplication between the 1 , 2 and 4 ?? - So you can't cancel them out
 
Ok, so now I have another question since I got that part down...
If I have an equation that reads:
2/(x+4)-1/(x-3)

Do I add 7 to 1/(x-3) to make it equal to the other denominator (x+4) or do I subtract one from (x+4) to make it equal to (x-3)? Or does it matter???
 
CalebsMomma said:
Ok, so now I have another question since I got that part down...
If I have an equation that reads:
2/(x+4)-1/(x-3)

Do I add 7 to 1/(x-3) to make it equal to the other denominator (x+4) or do I subtract one from (x+4) to make it equal to (x-3)? Or does it matter???
Click to expand...
\(\displaystyle \frac{(2)(x-3)-(x+4)}{(x+4)(x-3)}\)

\(\displaystyle \frac{2x-6-x-4}{(x+4)(x-3)}\)

\(\displaystyle \frac{x-10}{(x+4)(x-3)}\)
 
I appreciate the help but if you would could you please explain why you added 4 on to the x on the first equation yet multiplied both denominators on the second equation?
 
Calebsmoma

Can you reduce the following into a single fraction - without using calculator:

\(\displaystyle \frac{4}{5} \, + \, \frac{3}{4} \, = \, ???\)

show your work - step by step. (This will be useful in explaining the question you had asked)
 
4/5+3/4= 20/20 + 15/20= 35/20= 7/4

I can do just normal numbers but when you throw a variable "x" in the mix, it's when it messes me up.

For instance, I got 2/x/(x+4) - 1/ (x-3)=0 I can work this problem down to 2x(x-3)-1(x+4)/(x+4)(x-3)=0
2x^2-7x-4/x^2+1x-12=
(2x+1)(x-4)/(x+4)(x-3)

But this is where I get lost, i am not sure if now that this is multiplication if I can cancel or not?!
 
CalebsMomma said:
4/5+3/4= 20/20 + 15/20= 35/20= 7/4 <<< How did you get that? should it be

(4*4)/(5*4) + (3*5)/(4*5) = 16/20 + 15/20 = 31/20

Take time and go through EVERY step


I can do just normal numbers but when you throw a variable "x" in the mix, it's when it messes me up.

For instance, I got 2/x/(x+4) - 1/ (x-3)=0 <<< Are you sure your problem looks like this? (Anyway, you don't tell us what did you want to do with it)

If it did, then,

It is equivalent to

2/[x*(x+4)] - 1/(x-3) = 0

The LCM of the denominator is x*(x+4)*(x-3) - we need to make all the fractions have the same denominator - lke the numerical problem above. Then we have

{2*(x-3)}/[x*(x+4)(x-3)] - {x*(x+4)}/[x*(x+4)*(x-3)] = 0

[{2*(x-3)} - {x*(x+4)}] / [x*(x+4)*(x-3)] = 0

[-x[sup:357rlwed]2[/sup:357rlwed] - 2x - 6] / [{x*(x+4)(x-3)} = 0

For a fraction to be equal to zero - the numerator must equal to zero

then

-x[sup:357rlwed]2[/sup:357rlwed] - 2x - 6 = 0

x[sup:357rlwed]2[/sup:357rlwed] + 2x + 6 = 0

I can work this problem down to 2x(x-3)-1(x+4)/(x+4)(x-3)=0 <<<What happened to the parenthesis on the numerator?

2x^2-7x-4/x^2+1x-12= <<< what happened to the zero?? What happened to the parenthesis on the numerator?

(2x+1)(x-4)/(x+4)(x-3) <<< what happened to the zero??


For a fraction to be equal to zero - the numerator must equal to zero

I find that your biggest trouble is that you are in too big of a hurry and careless - all your mistakes stem from that fact

But this is where I get lost, i am not sure if now that this is multiplication if I can cancel or not?!
Click to expand...
 
I did think that we would cross multiply on the problem that you gave me and since they have no common denominator we need to make them common which would be 20, so that is how I got that.

And the problem I typed was 2x/(x+4) - 1(x-3)=0 It was typed that way on purpose, it wasn't a mistake.
What i need to do is solve for x. And I can get the equation down to
2x/(x+4) - 1(x-3)=0
2x(x-3)-1(x+4)/(x+4)(x-3)=0
2x^2-7x-4/x^2+1x-12=0
(2x+1)(x-4)/(x+4)(x-3) =0

I just usually drop the zero until I get everything worked out so it's not in my way.
 
CalebsMomma said:
I did think that we would cross multiply on the problem that you gave me and since they have no common denominator we need to make them common which would be 20, so that is how I got that.

And the problem I typed was 2x/(x+4) - 1(x-3)=0 It was typed that way on purpose, it wasn't a mistake.

You mean you typed the above problem as

2/x/(x+4) - 1/ (x-3)=0
Click to expand...

with a purpose? You changed a multiplication (2x) to a division (2/x).

Whatever that purpose was - it escapes me.


What i need to do is solve for x. And I can get the equation down to

2x/(x+4) - 1/(x-3)=0

[2x(x-3)-1(x+4)]/[(x+4)(x-3)] = 0

[2x^2-7x-4]/[x^2+1x-12] = 0

(2x+1)(x-4)/[(x+4)(x-3)] = 0

I just usually drop the zero until I get everything worked out so it's not in my way. It is not a good idea - may create confusion
Click to expand...
 
CalebsMomma said:
4/5+3/4= 20/20 + 15/20= 35/20= 7/4

I can do just normal numbers but when you throw a variable "x" in the mix, it's when it messes me up.

For instance, I got 2/x/(x+4) - 1/ (x-3)=0 I can work this problem down to 2x(x-3)-1(x+4)/(x+4)(x-3)=0
2x^2-7x-4/x^2+1x-12=
(2x+1)(x-4)/(x+4)(x-3)

But this is where I get lost, i am not sure if now that this is multiplication if I can cancel or not?!
Click to expand...

It was???
 
Thanks, sorry, guys, math is not my subject, thankfully after Monday, I'm done with math classes!! lol
 
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