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How do you simplify #(17x ^ { 2} v ^ { 3} - 16x v ) \div ( - 2x ^ { 2} v ^ { 3} )#?

Answer 1

Eva Abramson

#-17/2+8/(xv^2)=8x^-1v^-2-17/2#

#(17x^2v^3-16xv) -: (-2x^2v^3)=(17x^2v^3-16xv)/(-2x^2v^3)#

We can undo the fraction into two separate fractions:

#(17x^2v^3)/(-2x^2v^3)-(16xv)/(-2x^2v^3)#

And now we can cancel terms:

#(17cancelx^2cancelv^3)/(-2cancelx^2cancelv^3)-(16cancelxcancelv)/(-2x^cancel2v^cancel3)=-17/2+8/(xv^2)#

We can write this a bit neater this way:

#-17/2+8/(xv^2)=8x^-1v^-2-17/2#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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