How do you solve #-6abs(5-2y) = -9#?

1 Answer
Aug 23, 2015

#y = 7/4" "# or #" "y = 13/4#

Explanation:

Start by isolating the modulus on one side of the equation. This can be done by dividing both sides by #-6#

#(color(red)(cancel(color(black)(-6))) * |5 - 2y|)/color(red)(cancel(color(black)(-6))) = ((-9))/((-6)#

#|5 - 2y| = 3/2#

Since you're dealing with the absolute value of an expression, you must take into account the fact that the expression can be negative or positive.

  • #5-2y>0 implies |5-2y| = 5 - 2y#

Your equation becomes

#5 - 2y = 3/2#

#2y = 5 - 3/2#

#y = (10 - 3)/2 * 1/2 = color(green)(7/4)#

  • #5 - 2y <0 implies |5 - 2y| = -(5-2y)#

This time, the equation becomes

#-(5 - 2y) = 3/2#

#-5 + 2y = 3/2#

#2y = (3 + 10)/2#

#y = 13/2 * 1/2 = color(green)(13/4)#

Your original equation has two possible solutions,

#y = 7/4" "# or #" "y = 13/4#