How do you solve #3(x+3) <= 5x-3#?

1 Answer
Jim G.
Mar 17, 2017

#x>=6#

Explanation:

The first step is to distribute the bracket on the left side of the inequality.

#rArr3x+9<= 5x-3#

Collect terms in x on the left side and numeric values on the right side.

subtract 5x from both sides.

#3x-5x+9<=cancel(5x)cancel(-5x)-3#

#rArr-2x+9<= -3#

subtract 9 from both sides.

#-2xcancel(+9)cancel(-9)<= -3-9#

#rArr-2x<= -12#

To solve for x, divide both sides by - 2

#color(blue)"NOTE"# when we multiply/divide an inequality by a negative value we must #color(red)"reverse the symbol"#

#(cancel(-2) x)/cancel(-2)>=(-12)/(-2)larrcolor(red)" reverse symbol"#

#rArrx>=6" is the solution"#

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