Fractional equations

[holdyourhorsiieees]

3/1 + 1/x+2
--------------- * 4(x+2)/4(x+2) = 12(x+2) + 4/8 + (x+2)
2/x+2 + 1/4


I can cancel the (x+2) correct, leaving me with 16/8=2?

 

[Denis]

3/1 + 1/x+2
--------------- * 4(x+2)/4(x+2) = 12(x+2) + 4/8 + (x+2)
2/x+2 + 1/4

VERY unclear; do you mean:

3/1 + 1/(x+2)
---------------- * [4(x+2)/4(x+2)] = 12(x+2) + 4/8 + (x+2)
2/(x+2) + 1/4

If that's what you mean, then [4(x+2)/4(x+2)] = 1, so cancel it out.
NO: you cannot cancel out the (x+2)'s in the division: you should know that :shock:

To show us where you're at, simplify the right side: 12(x+2) + 4/8 + (x+2)
What do you get?

 

[holdyourhorsiieees]

3/1 + 1/x+2
--------------- <----- That is the problem I am given.
2/x+2 + 1/4


I then multiply the numerator and denominator by 4(x+2)/4(x+2), the common denominator.


so, I get: 12(x+2) + 4
----------------
8+(x+2)

I don't understand what you mean stating I cannot cancel out the (x+2)s in the division. I think I see why the (x+2)s woudn't cancel out in this problem, but are you saying I cannot cancel out the (x-3)s in this type of situation- (x+3)(x-3) ?
--------------
(x-3)(x-2)


You say to simplify the right side, so I have 12(x+2) + 4/(x+10)

Is that my final answer?

 

[Loren]

3/1 + 1/x+2
--------------- <----- That is the problem I am given.
2/x+2 + 1/4

\(\displaystyle \frac{\frac{3}{1}+\frac{1}{x}+2}{\frac{2}{x}+2+\frac{1}{4}}\)

If this is not what you mean, you need to use parenthesis to clarify the problem.

If you mean \(\displaystyle \frac{\frac{3}{1}+\frac{1}{x+2}}{\frac{2}{x+2}+\frac{1}{4}}\), it is written as ...

3/1 + 1/(x+2)
----------------- <----- That is the problem I am given.
2/(x+2) + 1/4

In either case, I have found the best approach is to find the least common denominator of the numerator and denominator (of the main fraction), then multiply both numerator and denominator by that. That process will leave you with a simple fraction which can then be simplified if possible.

 

[Denis]

so, I get:
12(x+2) + 4
----------------
8+(x+2)

I don't understand what you mean stating I cannot cancel out the (x+2)s in the division.

If you don't understand that the (x+2)'s CANNOT, repeat CANNOT, be
cancelled out, then you need teacher/classroom help. Roger.

 

[holdyourhorsiieees]

I stated that I understand why the (x+2)s do not cancel out in this problem. I just didn't know if you meant you can NEVER cancel out identical binomials in a division problem. I was taught that you could.



I apologize that I obviously was not clear in asking my question.


I truly appreciate everyone's replies, but I'm not getting anywhere here.

I'll just ask my teacher.

 

[Denis]

I'll just ask my teacher.

Good idea; much easier to explain face to face than by typing out loud :wink:

(x+2)(x-3)
=======
7(x+2)
cancel (x+2)'s ? YES.

(x+2) + (x-3)
=========
7(x+2)
cancel (x+2)'s ? NO.

 

[stapel]

I don't understand what you mean stating I cannot cancel out the (x+2)s in the division.

You can only cancel factors; you can never cancel terms! :!:

For instance:

. . . . .\(\displaystyle \L \frac{12}{24}\, =\, \frac{1\,\sout{2}}{\sout{2}\,4}\, =\, \frac{1}{4}\)

...is nonsense (and incorrect), and:

. . . . .\(\displaystyle \L \frac{12}{24}\, =\, \frac{2\, +\, 10}{2(10)\, +\, 4}\, =\, \frac{2\, +\, \sout{10}}{2\,\sout{(10)}\,+\,4}\, =\, \frac{2}{2\, + 6}\, =\, \frac{2}{8}\, =\, \frac{1}{4}\)

...is invalid (and incorrect), but:

. . . . .\(\displaystyle \L \frac{12}{24}\, =\, \frac{(1)(12)}{(2)(12)}\, =\, \frac{(1)\sout{(12)}}{(2)\sout{(12)}}\, =\, \frac{1}{2}\)

...is fine. :wink:

Hope that helps!

Eliz.