Math question. how to get from one step to the next?

Question

how does this (x+1/x+2) + (1/(x+1/x+2)) algebraically become this (x+1/x+2) + (x+2/x+1)???

Answer 1

(x+1)............1

------....+.....------

(x+2)............(x+1)

...................--------

....................(x+2)

Anytime you are having 1 as the numerator of complex fractions, just get the last expression which is (x+2) in this case, and put it on top. So.

(x+1)........(x+2)

-------..+ .--------

(x+2).........(x+1)

Answer 2

because the second part is 1 divided by (x+1/x+2)), when dividing by a fraction, you invert and multiply so it's 1 times x+2/x+1 or just x+2/x+1

Answer 3

[1/(x+1/x+2)] = (x+2/x+1) when you...

divide 1 by the (x+1/x+2), which also is the same as multiplying 1 by the reciprocal of (x+1/x+2)

the reciprocal of (x+1/x+2) is (x+2/x+1)

Answer 4

numerator / denominator is the same as

numerator x 1 / denominator

So, if you take the second part

1 / (x + 1 / x + 2) = 1 times (x + 2) / (x + 1)

= (x + 2) / (x + 1)

Hope this helps you!  Please email anytime for further questions.

Answer 5

an important rule here is that anything in the denominator is said to be that number to the power of -1. anything to a negative power must swap sides of the 'vinculum line' (the line of division). in the second bracket you have 1/(x+1/X+2) which is like saying 1/((x+1)*(x+2)^-1), or, simplified and applying the above rule, 1*(x+2)/(x+1).

basically, you have to remember that if a number (X) is divided by a number, which we will call Y, which is divided by another number, Z, it will become XZ/Y. so double division = single division with bottom number on top.