Algebra - Simplify Complex Fraction?

okay

using the rules of indices 3-2x^-1 can be wrote 3-(2/x)

3x^-1+2x^-1 is equal to (3/x) +(2/x)

so

(3-(2/x)) / (3/x) +(2/x)

you want to write the numerator and denominator as a single fraction

((3x-2)/x) / ((3+2)/x)

this is a fraction divided by a fraction. to divide a fraction by a fraction turn one of them upside down and then multiply

((3x-2)/x) * (x/(3+2))

(3x-2)x / (3+2)x

divide the top and bottom of the fraction by x

(3x-2) / (3+2)

(3x-2)/5

0.6x - 0.4

Convert the complex fraction into division:

(3 - 2x^-1) ÷ (3x^-1 + 2x^-1)

To make a negative power positive, take the reciprocal:

(3 - 2/x) ÷ (3/x + 2/x)

Combine the like terms:

(3 - 2/x) ÷ (5/x)

Make the denominators common in the first fraction:

(3x/x - 2/x) ÷ (5/x)

Combine the fractions:

(3x-2/x) ÷ (5/x)

Normally to divide fractions, you would flip the second fraction and multiply but in this instance you don't have to because the denominators are the same and cancel each other out. So just divide the numerators from each other:

(3x - 2) / 5

Multiply the numerator and denominator by x

(recall that x*x^-1 = 1):

(3*x - 2*1)/(3*1 + 2*1) = (3x - 2)/5

Multiply the fraction by 1 in the form of x/x, because this will make the x^-1 factors become 1:

x(3 - 2x^-1)

_____________ =

x(3x^-1 + 2x^-1)

(3x - 2)/(3 + 2) =

(3/5)x - (2/5)

Complicated fractions we are able to divide a difficult number (a + bi) by way of a further elaborate number (c + di) ≠ 0 in two methods. The primary means has already been implied: to transform each problematic numbers into exponential kind, from which their quotient is simply derived. The second means is to express the division as a fraction, then to multiply each numerator and denominator with the aid of the complex conjugate of the denominator. The new denominator is an actual quantity. ___________ The excerpt above is from the supply beneath...