How do you evaluate #x^2/y + x^2/z# for #x=6, y=-4,# and #z=-2#?

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Answer 1 · 2016-07-19T18:04:25

The answer will be #-27#.

Simply plug in the values and simplify:

#(6^2)/-4 + (6^2)/-2#

#=36/-4 + 36/-2#

#=-9 - 18#

#=-27#

Answer 2 · 2016-07-19T18:08:09

#-27#.

#x^2/y+x^2/z=x^2(1/y+1/z)=(x^2(y+z))/(yz)=(6^2(-4-2))/((-4)(-2))=(36(-6))/8=-27#.