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

Question

Avery Alexander · Accepted Answer

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

#x !=0 and y!=0# #((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.

Kennedy Adams · Answer

undefined To simplify the expression ((3x)/y - x/1) / ((2y)/1 - y/x), we can start by finding a common denominator for each fraction.

For the first fraction, the denominator is y, and for the second fraction, the denominator is x. To find a common denominator, we multiply the first fraction by x/x and the second fraction by y/y.

This gives us ((3x)/y * (x/x) - (x/1) * (y/y)) / ((2y)/1 * (y/y) - (y/x) * (x/x)).

Simplifying further, we get (3x^2 - xy) / (2y^2 - xy).

To find the restrictions, we need to identify any values of x and y that would make the denominator equal to zero.

In this case, the restrictions occur when 2y^2 - xy = 0.

To solve this equation, we can factor out a common term: y(2y - x) = 0.

This equation is satisfied when either y = 0 or 2y - x = 0.

Therefore, the restrictions for the given expression are y = 0 and 2y - x = 0.