How do you solve #-9( x + 2) + 3( x - 1) \leq 15x#?

1 Answer
Jim G.
Oct 6, 2016

#x>=-1#

Explanation:

First step is to distribute the brackets on the left side of the inequation.

#-9x-18+3x-3<=15x#

simplifying left side.

#-6x-21<=15x#

now collect x terms on left and numeric values on right.

subtract 15x from both sides.

#-6x-15x-21<=cancel(15x)cancel(-15x)#

#rArr-21x-21<=0#

add 21 to both sides.

#-21xcancel(-21)cancel(+21)<=0+21#

#rArr-21x<=21#

To solve for x, divide both sides by - 21. Since this is an inequation, however, when we multiply/divide by a negative value we must #color(blue)"reverse the inequality symbol"#

#(cancel(-21) x)/cancel(-21)>=21/(-21)larr" reverse symbol"#

#rArrx>=-1" is the solution"#

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