30.6k views
0 votes
Solve for x: 3|x-3| 2=14

a. No solution
b. x=-1,x=8.3
c. x=0,x=7
d. x=-1,x=7

User Killian
by
8.0k points

1 Answer

4 votes
I am looking on the answers, and there is only one case, when a or b or c or d pass: 3|x-3| + 2 = 14. So I assume, that before two is plus. Then:

3|x-3|+2=14 |minus 2
3|x-3|=12 |divide 3
|x-3|=4

From absolute value definition you've got two ways:

x-3=4 or x-3=-4
x=7 or x=-1

And answer d) passes

User Karansardana
by
8.7k points

No related questions found