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3 years ago

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

1 answer:

3 0

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

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