How do you solve #-9( x + 2) + 3( x - 1) \leq 15x#?
1 Answer
Oct 6, 2016
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"#