I would like some help doing these two problems. If you could, please explain the problems, that would be even better.

I need these simplified, but just don't see how it can be done.

1/(1+x/y)

x+1/x^2+2x-8 - x/4x-8

Question

I would like some help doing these two problems. If you could, please explain the problems, that would be even better.

I need these simplified, but just don't see how it can be done.

1/(1+x/y)

x+1/x^2+2x-8 - x/4x-8

anonymous · Answer

Calmat?

anonymous · Answer

To simplify the first one, think of it as two fractions, one on top of the other.  The bottom term is 1+x/y.  If we get the common denominator as y, 
$$1+\frac{ x }{ y }=\frac{ y+x }{ y }$$   Now we have to divide 1 by this denominator.  To do so just requires us to multiply 1 by the reciprocal of the fraction.  So,

$$1\div \frac{ y+x }{ y }=1\times \frac{ y }{ y+x }=\frac{ y }{ y+x }$$.

The second one requires a little more work.

$$\frac{ x+1 }{ x ^{2}+2x-8 }-\frac{ x }{ 4x-8 }=\frac{ x+1 }{ (x+4)(x-2) }-\frac{ x }{ 4(x-2)}$$
Now we have to find a common denominator.   Both denominators have a common factor of (x-2), so that is their greatest common factor.  Multiply it by the factors which they don't have in common.  This leads us to an LCD of : 4(x-2)(x+4).  So, we must convert each denominator to this LCD.

This leads us to the following:

$$\frac{ 4(x+1) }{4(x+4)(x-2)  }-\frac{ x(x+4) }{4(x-2)(x+4)  }$$

Now, multiply each numerator out.  This leads us to:
$$\frac{ 4x+4-x ^{2}-4x }{4(x+4)(x-2)}$$

Combine like terms in the numerator. 
$$\frac{ -x ^{2} +4}{ 4(x+4)(x-2) }$$ 

Finally, factor out a  -1 from the numerator and factor the result as the difference of squares.
$$\frac{ -1(x-2)(x+2) }{ 4(x+4)(x-2) }$$ 

From here, the common factor x-2 may be reduced.  Leaving your final answer as:

$$\frac{ -(x+2) }{ 4(x+4) }$$

anonymous · Answer

Sorry, took me a while to type that second one out.

anonymous · Answer

No problem, it is perfectly all right. Your explanations are superb! Thank you so very much for taking the time to help mt out!

anonymous · Answer

no problem.  You are most welcome.

anonymous · Answer

These are spelled out so easily that even I can understand them, ha. The only class I ever really have an issue with is Algebra 2. This helps me out more than you know!

anonymous · Answer

Good, I am glad it helps you.

anonymous · Answer

Thanks again!