Question

absolute value! check my work please!
|2t-3| = 3t-2

i got
case 1 t=-1 extraneous
case 2 t= 1 extraneous

so my answer is no solution

Answer 1

@myininaya

Answer 2

@zepdrix @dan815 @jim_thompson5910

Answer 3

t = -1 and t = 1 are correct.
If we try t = -1 we get LHS = |-2-3| = 5 and RHS = -3-2 = -5. (extraneous)
If we try t = 1 we get LHS = |2-3| = 1 and RHS = 3-2 = 1. t = 1 is a solution.

Answer 4

okay so only case 2 works?
and case 1 is extraneous?

Answer 5

Yes.

Answer 6

okay thank you @aum

Answer 7

You are welcome.

Answer 8

visually you can compare the graphs of y = |2x-3| and y = 3x-2

wherever they intersect is where your solution will be

turns out they only intersect at one point: (1,1)

x = 1 ---> t = 1 is your only solution

see attached for the graph

Answer 9

@MikeZack123
When you check those solutions, make sure both sides are the same (if they are both the same it is a solution)
If both sides are not the same when pluggin (testing your number) in your number, then it is extraneous (or not a solution)

Answer 10

yes thank you @jim_thompson5910 and @myininaya :) much appreciated