How do you solve for x in #4-6x \le 2(2x+3)#?

1 Answer
Nico Ekkart
Dec 19, 2014

To solve an inequality of the first order (this is one because there isn't a higher exponent of x, than 1), you have to first bring every factor of x to one part, and every number to the other part.

#4-6x <= 2(2x+3)#
#4-6x <= 4x + 6# (apply the distributive property)
#4-6x-6 <= 4x+6-6# (substract #6# from both sides)
#-2-6x <= 4x#
#-2-6x+6x <= 4x+6x# (add #6x# to both sides)
#-2 <= 10x#
#(-2)/10 <= (10x)/10# (divide both sides by ten#""^"(1)"#)
#-1/5 <= x#
This is your answer, you can flip it if that seams easier:
#x >= -1/5#
The set of solutions is all the numbers that are greater than or equal to #-1/5#.

Hope this helped.

(1): If you were to divide by a negative number, you would have to flip the inequality sign. In this case, #<=# would become #>=#

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