Question

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

Answer 1

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`

Answer 2

`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"`