How do you graph and solve #|3y+2|=|2y-5|#?

2 Answers

Answer:

#y in { 3/5, -7 }#

Explanation:

or #3y +2 = 2y - 5# (i) or #3y + 2 = -2y + 5# (ii)

(i) #Rightarrow y = -7#

(ii) #Rightarrow 5y = 3#

Jan 2, 2016

Answer:

#y=-7,3/5#

Explanation:

Find the points when the term inside the absolute value switches sign.

#abs(3y+2)#

#3y+2=0#
#y=-2/3#
#3y+2<0# when #y<-2/3#, #>0# when #y> -2/3#

#abs(2y-5)#

#2y-5=0#
#y=5/2#
#2y-5<0# when #y<5/2#, #>0# when #y>5/2#

From this, we have three distinct ranges of numbers: #(-oo,-2/3),(-2/3,5/2)#, and #(5/2,+oo)#.

#color(blue)((-oo,-2/3)#

In this set, both of the terms inside the absolute value functions will be negative. Take the negative versions of each of the absolute value expressions.

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

Solve. The answer is only valid if #y<-2/3#.

#3y+2=2y-5#
#y=-7#

This is a valid answer.

#color(blue)((-2/3,5/2)#

Here, the #3y+2# term will be positive but #2y-5# will be negative. Take the opposite version of only the #2y-5# term and solve.

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

This is also a valid answer, since #-2/3<3/5<5/2#.

#color(blue)((5/2,+oo)#

From the first set, we know this will result in an answer of #-7#. Though it is invalid for this range, it was valid for #(-oo,-2/3)#.

Thus, #y=-7,3/5#.

graph{abs(3x+2)-abs(2x-5) [-19.8, 20.75, -8.48, 11.79]}