SOLUTION: What is the solution? x(3-x) + 2x(x-1)_<(less than or equal too).. x^2 - 3x + 12

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Question 674291: What is the solution?
x(3-x) + 2x(x-1)_<(less than or equal too).. x^2 - 3x + 12
Answer by jokaehler(26) About Me  (Show Source):
You can put this solution on YOUR website!
x(3-x) + 2x(x-1)_<(less than or equal too).. x^2 - 3x + 12

First, we need to use the distributive property on the left side of the equation. (x * 3) + (x * -x) + (2x * x) + (2x * -1)
Which turns out to be...
+3x+-+x%5E2+%2B+2x%5E2+-+2x+%28is+less+than+or+equal+to%29+x%5E2+-+3x+%2B+12+
Then, we need to get "x" on one side (I like "x" on the left side). Subtract "+x%5E2+" and add "+3x+" to both sides.
We end up with...
4x (is less than or equal to) 12
Divide both sides by "4," and we end up with...
x (is less than or equal to) 3