Solving Complex Fractions: [ 3/(x+2) + 2/3 ] / [ (2x)/(x+2) - 1/x ]

F

FritoTaco

New member
Joined
Nov 14, 2015
Messages
42
Hello, I'm having trouble solving this problem.


\(\displaystyle \mbox{7. }\, \dfrac{\dfrac{3}{x\, +\, 2}\, +\, \dfrac{2}{3}}{\dfrac{2x}{x\, +\, 2}\, -\, \dfrac{1}{x}}\)


If I want a common denominator for both top and bottom, does it matter what I multiply by? For instance, If I only look at the top half, should I multiply by 3 on both sides or x + 2?
 

Attachments

  • IMG_0068(new2).jpg
    IMG_0068(new2).jpg
    328.2 KB · Views: 2
Last edited by a moderator:
FritoTaco said:
Hello, I'm having trouble solving this problem.


\(\displaystyle \mbox{7. }\, \dfrac{\dfrac{3}{x\, +\, 2}\, +\, \dfrac{2}{3}}{\dfrac{2x}{x\, +\, 2}\, -\, \dfrac{1}{x}}\)


If I want a common denominator for both top and bottom, does it matter what I multiply by? For instance, If I only look at the top half, should I multiply by 3 on both sides or x + 2?
Click to expand...

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33
 
Last edited by a moderator:
FritoTaco said:
Hello, I'm having trouble solving this problem. If I want a common denominator for both top and bottom, does it matter what I multiply by? For instance, If I only look at the top half, should I multiply by 3 on both sides or x + 2?
Click to expand...

HINT: \(\displaystyle \dfrac{{\frac{3}{{x + 2}} + \frac{2}{3}}}{{\frac{{2x}}{{x + 2}} - \frac{1}{x}}} = \dfrac{{9x + 2x(x + 2)}}{{6{x^2} - 3(x + 2)}}\)
 
Denis said:
Take your time; go step by step.
Let k = x+2. Then you have:

numerator = 3/k + 2/3 = (2k + 9) / (3k)

denominator = 2x/k - 1/x = (2x^2 - k) / (kx)

So invert denominator and expression becomes:
(2k + 9) / (3k) * kx / (2x^2 - k)

Continue...
Click to expand...

In the denominator, why does 2x become squared?
 
FritoTaco said:
Hello, I'm having trouble solving this problem.


\(\displaystyle \mbox{7. }\, \dfrac{\dfrac{3}{x\, +\, 2}\, +\, \dfrac{2}{3}}{\dfrac{2x}{x\, +\, 2}\, -\, \dfrac{1}{x}}\)


If I want a common denominator for both top and bottom, does it matter what I multiply by? For instance, If I only look at the top half, should I multiply by 3 on both sides or x + 2?
Click to expand...
Hi FritoTaco,

Methinks you need a quick lesson in adding/subtracting fractions.

\(\displaystyle \frac{a}{b} +\frac{c}{d} = \frac{ad}{bd}+\frac{bc}{bd}\) ... get a common denominator (in this case, simply multiply the 2 original denominators)

\(\displaystyle = \frac{ad+bc}{bd}\) ... add the numerators over the common denominator

Now let's take the numerator of your problem and deal with that first:

\(\displaystyle \frac{3}{(x+2)}+\frac{2}{3} \) ... common denominator will be \(\displaystyle 3(x+2)\)

\(\displaystyle =\frac{3*3}{3(x+2)}+ \frac{2(x+2)}{3(x+2)}\)

\(\displaystyle =\frac{9+2(x+2)}{3(x+2)}\)

\(\displaystyle =\frac{2x+13}{3(x+2)}\)

Now do the same to the denominator of the original problem.

When you come to do the division remember that \(\displaystyle \frac{a/b}{c/d} = {\frac{a}{b}}\div{\frac{c}{d}}={\frac{a}{b}}*\frac{d}{c}=\frac{ad}{bc}\)
 
Last edited:
FritoTaco said:
In the denominator, why does 2x become squared?
Click to expand...

As you know, when adding or subtracting fractions, you must first find a common denominator. If we proceed with Denis' suggested substitution of k = x + 2, then we have:

\(\displaystyle \dfrac{2x}{k}-\dfrac{1}{x}\)

For these two fractions, what is the lowest common denominator? What did you multiply k by to get that denominator? And when you multiply the denominator of a fraction by a number, what must you also do to the numerator? What happens when you do that? Apply the same process to the 1/x portion. Then what do you get when you subtract and simplify?
 
You must log in or register to reply here.
Top