How do you simplify #(5p-10)/(2-p)#?

Answer 1

#-5#

Consider #5p-10#
Factor out the 5 giving: #5(p-2)#
Notice that the #p-2# is the other way round to that in the denominator. We can 'play' with this by 'adjusting' signs

You can manipulate things mathematically in any way you want as long as you include a function that returns everything to its initial values.

Lets multiply #5(p-2)# by #(-1)# giving:#-5(p-2)# But this changes the values so lets make adjustments so that when we multiply by #(-5)# we end up with the original #5p-10#
Write #5(p-2)# as #-5(2-p)#

It now agrees with the denominator, allowing us to cancel.

#(-5(2-p))/(2-p) color(white)("ddd")=color(white)("ddd") (-5)xx(2-p)/(2-p)color(white)("ddd") =color(white)("ddd") -5#
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Answer 2

To simplify the expression (5p-10)/(2-p), you can start by factoring out a negative sign from the numerator to get (-1)(10-5p)/(2-p). Then, you can rearrange the terms in the numerator to get (-1)(-5p+10)/(2-p). Finally, you can simplify further by canceling out the negative signs to get (5p-10)/(p-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.

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.

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.

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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