How do you simplify and find the restrictions for #(20+40x)/(20x)#?

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Answer 1 · 2017-04-10T13:21:38

See the entire explanation below:

Factor a #20# out of each term in the numerator;

#((20 xx 1) + (20 xx 2x))/(20x) = (20(1 + 2x))/(20x)#

Now, cancel the common term in the numerator and denominator:

#(color(red)(cancel(color(black)(20)))(1 + 2x))/(color(red)(cancel(color(black)(20)))x) = (1 + 2x)/x#

Because you cannot divide by #0# the restriction is #x != 0#

Answer 2 · 2017-04-10T13:52:33

Same thing as the other solution it just looks different.

#=1/x+2#

I am presenting this solution solely to demonstrate that one situation may take on several forms but still have the same inherent value.

Firstly #x!=0# as the equation becomes 'undefined' at that point.
Basically division by 0 is a definite no no!

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Write as: #20/(20x)+(40x)/(20x)#

#=1/x+2#