color(red)(t = 3)t=3 is a solution.
color(red)(t = 1)t=1 is an extraneous solution.
SOLVE
|2t-3| = t|2t−3|=t
We need to write two different equations without the absolute value symbols and solve for tt.
These equations are:
(1): (2t-3) = t(2t−3)=t
(2): -(2t-3) = t−(2t−3)=t
Solve Equation 1:
2t-3 = t2t−3=t
Subtract tt from each side.
t-3 = 0t−3=0
Add 33 to each side.
t = 3t=3
Solve Equation 2:
−(2t-3) = t−(2t−3)=t
Remove parentheses.
-2t+3= t−2t+3=t
Add 2t2t to each side.
3 = 3t3=3t
Divide each side by 33.
t = 1t=1
The solutions are t = 1t=1 and t = 3t=3.
CHECK FOR EXTRANEOUS SOLUTIONS:
If t = 1t=1,
|2t-3|=t|2t−3|=t
|2(1)-3|= 3|2(1)−3|=3
|2-3| = 5|2−3|=5
|-1| =5|−1|=5
1=51=5
This is impossible, so t=1t=1 is an extraneous solution.
If t = 3t=3,
|2t-3| = t|2t−3|=t
|2(3) - 3| = 3|2(3)−3|=3
|6 - 3| = 3|6−3|=3
|3| = 3|3|=3
3 = 33=3
t=3t=3 is a solution.
