How do you simplify and find the restrictions for #((3x)/y -x/1) / ((2y)/1 - y/x)#?

1 Answer
May 17, 2017

Answer:

#= (x^2(3-y))/(y^2(2x-1))#

#x !=0 and y!=0#

Explanation:

#((3x)/y -x/1) / ((2y)/1 - y/x)" " x !=0 and y!=0#

Before we start with any simplifying we can see that the denominators at the top and bottom have an #x# and a #y#
Therefore, they may not be #0#
We might find further restrictions later.

Write the complex fraction as a division of two fractions:

#((3x)/y -x/1) div ((2y)/1 - y/x)# find LCD and subtract as usual,.

#= ((3x-xy))/y div ((2xy-y))/x" "larr# factorise the numerators:

#= (x(3-y))/y div (y(2x-1))/x" "larr# multiply by the reciprocal

#=(x(3-y))/y xx x/(y(2x-1))#

#= (x^2(3-y))/(y^2(2x-1))#

The original restrictions are confirmed.