How do you simplify #\frac { \frac { x ^ { 2} + x ^ { 2} } { x y } - 2} { \frac { 4x ^ { 2} - 4x ^ { 2} } { 2x y } }#?

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Answer 1 · 2018-01-30T21:56:44

This ends up as a division by 0, so it is undefined.

Simplify

#\frac { \frac { x ^ { 2} + x ^ { 2} } { x y } - 2} { \frac { 4x ^ { 2} - 4x ^ { 2} } { 2x y } }#

1) For clarity, write this as one fraction divided by the other

#{\frac { x ^ { 2} + x ^ { 2} } { x y } - (2)/(1)}# #-:# #{ \frac { 4x ^ { 2} - 4x ^ { 2} } { 2x y } }#

2) Combine the like terms in the numerators

#{\frac { 2x ^ { 2} } { x y } - (2}/(1)}# #-:# #{ \frac { 0 } { 2x y } }#

3) To divide fractions, multiply by the reciprocal of the divisor

#{\frac { 2x ^ { 2} } { x y } - (2}/(1)}# #xx# #{ \frac { 2xy } { 0 } }#

4) No matter how you simplify the fractions, this problem is still going to end with a division by #0#

#( 2x ^ { 2} - 2xy)/(xy)# #xx# #( 2xy )/ { 0 }#

#( 2x { x-y} )/(xy)# #xx# #( 2xy )/ { 0 }#

#( 2x { x-y} )/cancel(xy)^1 # #xx# #( 2cancel(xy) )/ { 0 }#

#(4x (x-y))/(0)##larr# undefined

#color(white)(...............................)# . . . . . . . . . . . . .

I guess you know what happened the last time someone divided by #0#.

Answer 2 · 2018-01-31T19:21:50

I think the question and answer should have been more like:

#((x^2+y^2)/(xy)-2)/((4x^2-4y^2)/(2xy)) = (x-y)/(2(x+y))#

with exclusions #x!=0#, #y!=0# and #x != y#

I suspect a couple of typos in the question, so here's an answer if the question should have been to simplify:

#((x^2+y^2)/(xy)-2)/((4x^2-4y^2)/(2xy))#

Multiplying both numerator and denominator by #xy#, this becomes:

#(x^2-2xy+y^2)/(2(x^2-y^2)) = (color(red)(cancel(color(black)((x-y))))(x-y))/(2color(red)(cancel(color(black)((x-y))))(x+y)) = (x-y)/(2(x+y))#

with exclusions #x!=0#, #y!=0# and #x != y#