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How do you solve for n in #1/2n = 2p #?

Answer 1

#n=4p#

Given:#" "color(brown)(1/2n=2p)#

Need to have #n# on its own. So if we can change #1/2# into 1 then #1xxn=n#. As long as we obey the rule that; "what we do to one side of the = we do the same thing to the other side"; we will be fine.

Multiply both sides by#" "color(blue)(2)#

#color(brown)(1/2n=2p# becomes

#color(brown)(" " color(blue)(2xx)1/2 n" "=" "color(blue)(2xx)2p)#

But #2xx1/2" is the same as "2/2=1" "#so we have:

#1xxn=4p#

#n=4p#

'~~~~~~~~~~~~~~~~~~~~~~~~~~ With a bit of practice this type of question can be done totally in the head. Keep at it and you will get there!

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Answer 2

Isaiah Anthony

To solve for ( n ) in ( \frac{1}{2}n = 2p ), you can multiply both sides of the equation by 2 to isolate ( n ). This yields ( n = 4p ).

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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.