How do you solve #9c(c-11)+10(5c-3)=3c(c+5)+c(6c-3)-30#?

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Answer 1 · 2017-04-18T04:56:36

The answer is: undetermined/impossible.

First you want to distribute to all of the parentheses:

#9c^2-99c + 50c -30 = 3c^2 +15c + 6c^2 -3c -30#

Then you want to combine like factors.

#9c^2 -49c - 30 = 9c^2 +12c -30#

Now that you have like factors on both sides, you can work to getting "c" by itself on one side.

#-49c=12c#

^^^However, the equation above is impossible because no matter what number you put (besides 0), you will not get the same answer from both sides.

For example:

#-49(1) = 12 (1)#
-49 is not equal to 12,
so impossible or undetermined.