How do you simplify #(x/9 - 1/x) / (1+ 3/x) #?

1 Answer
Jul 13, 2016

Answer:

#(x/9 - 1/x)/(1/1+3/x) = (x-3)/9#

Explanation:

For this particular problem, we'd have to take note of some algebraic properties such as

#a/b - c/d = (ad - bc) / (bd)#

In this case, we follow this property to start simplifying:

#(x/9 - 1/x)/(1/1+3/x) = ((x^2-9)/(9x)) /((x+3)/(x) )#

We can simplify further by multiplying top and bottom (numerator and denominator) by the reciprocal of #(x+3)/x#, namely #x/(x+3)#.

#(x/9 - 1/x)/(1/1+3/x) = (((x^2-9)/(9x)) * (x/(x+3)))/((cancel((x+3)/(x))) * (cancel(x/(x+3)))) = (cancel(x)(x^2-9))/(9cancel(x)(x+3)#

We've now simplified our expression, although we're not done yet.
Note that we can still factor the top (numerator), that is, #x^2 - 9#.
In order to factor this expression, we proceed as follows:

#x^2 - 9 = 0#

#x^2 = 9#

Solving for #x# gives us

#x = ± sqrt(9)#

#x = ± 3#

Since the solutions are #x = 3# and #x = -3#, we now have our factors and can rewrite the numerator in the following way:

#(cancel(x)(x^2-9))/(9cancel(x)(x+3)) = ((x-3)cancel((x+3)))/(cancel((x+3))*9) = (x-3)/9#