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How do you evaluate #abs(x-y)+y-1# when x=-3, y=-6?

Answer 1

Zoe Adams

#|x-y|+y-1=-4#, when #x=-3# and #y=-6#

Value of #|x-y|+y-1#, when #x=-3# and #y=-6# can be obtained by substituting these values.

#|x-y|+y-1#

= #|(-3)-(-6)|+(-6)-1#

= #|-3+6|+(-6)-1#

= #|3|-6-1#

= #3-6-1=3-7=-4#

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Answer 2

Roman Alvarez

To evaluate (|x - y| + y - 1) when (x = -3) and (y = -6), substitute these values into the expression:

( |(-3) - (-6)| + (-6) - 1 )

This simplifies to:

( |(-3) + 6| + (-6) - 1 )

( |3| + (-6) - 1 )

Since the absolute value of (3) is (3), this further simplifies to:

( 3 + (-6) - 1 )

Now, add the numbers:

( 3 - 6 - 1 )

( -3 - 1 )

( -4 )

So, ( |x - y| + y - 1 ) equals (-4) when (x = -3) and (y = -6).

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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.