#|3x|<=|2x-5|#
There are four solutions;
#1st#
#-3x <= -(2x - 5)#
#2nd#
#3x <= 2x - 5#
#3rd#
#3x <=-(2x - 5)#
#4th#
#-3x <= 2x - 5#
From the #1st#
#-3x <= -(2x - 5)#
#-3x <= -2x + 5#
Add #2x# to both sides;
#-3x + 2x <= -2x + 5 + 2x#
#-x <= 5#
Multiply through by Minus #(-)#
#-(-x) <= -(5)#
Note: When you divide or multiplty an inequality sign by a negative value, the sign changes..
#x >= -5#
From the #2nd#
#3x <= 2x - 5#
Subtracting both sides by #2x#;
#3x - 2x <= 2x + 5 - 2x#
#x <= 5#
From the #3rd#
#3x <=-(2x - 5)#
Removing the bracket;
#3x <= -2x + 5#
Add #2x# to both sides;
#3x + 2x <= -2x + 5 + 2x#
#5x <= 5#
Dividing both sides by the coefficient of #x#;
#(5x)/5 <= 5/5#
#(cancel5x)/cancel5 <= cancel5/cancel5#
#x <=5#
From the #4th#
#-3x <= 2x - 5#
Subtracting #2x# from both sides;
#-3x - 2x <= 2x - 5 - 2x#
#-5x <= -5#
Dividing both sides by the coefficient of #x#;
#(-5x)/(-5) <= (-5)/(-5)#
#(cancel(-5)x)/cancel(-5) <= cancel(-5)/cancel(-5)#
#x >= 1#
Hence the possible ranges are;
#x >= -5#
#x <=5#
#x <=5#
#x >= 1#
Therefore;
#1<= x <= 5#
