How do you solve #abs(t+1)=4t +3#?

2 Answers
Mar 2, 2018

Answer:

Appliying definition of absolute value. See details

Explanation:

We define absolute value of a number as

#absx=x# if #x>=0# and

#absx=-x# if #x<=0#

With this in mind, lets aplly to our equation in t

#abs(t+1)=t+1# if #t+1>=0#, it say: #t>= -1#

#abs(t+1)=-(t+1)# if #t+1<0#, it say: #t< -1#

In the first case (#t>= -1#) :

#t+1=4t+3#

#1-3=4t-t#; thus #t=-2/3# This value is valid because is bigger than -1

In the second case (#t< -1#) :

#-(t+1)=4t+3#

#-t-1=4t+3#

#t=-4/5# this value is invalid because our initial restriction #t<-1# is not verified by #t=-4/5#

Mar 2, 2018

Answer:

#t=-2/3#

Explanation:

#"the value inside the absolute value bars can be "#
#"positive or negative"#

#"thus there are 2 possible solutions"#

#t+1=4t+3larrcolor(blue)"positive inside bars"#

#"subtract "(t+1)" from both sides"#

#rArr0=3t+2#

#rArr3t=-2rArrt=-2/3larrcolor(red)"possible solution"#

#-t-1=4t+3larrcolor(blue)"negative inside bars"#

#rArr5t=-4rArrt=-4/5larrcolor(red)"possible solution "#

#color(blue)"As a check"#

#|-2/3+1|=|1/3|=1/3" and "-8/3+9/3=1/3#

#"both sides are equal hence "x=-2/3" is a solution"#

#|-4/5+1|=1/5" and "-16/5+15/5=-1/5#

#1/5!=-1/5" hence "t=-4/5" is not a solution"#