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Solve for x: |x + 2| + 16 = 14 A. x = −32 and x = −4 B. x = −4 and x = 0 C. x = 0 and x = 28 D. No solution
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@HectorH
i think its D
No solution
ok i was right thnx :)
We subtract 16 from both sides to get \[\left| x+2 \right| = -2\] Then thats impossible since absolute value bars keep it from going negative.
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yup :) thnx
@jkl5149 its no solution
@Janu16 Yes.
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