SOLUTION: Can someone help me solve these equations: (w+3)^-1/3=1/3 and (x+2)^3/2=-1 I have no clue where to start.

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Question 208136This question is from textbook Elementary and intermediate algebra
: Can someone help me solve these equations:
(w+3)^-1/3=1/3 and
(x+2)^3/2=-1
I have no clue where to start.



 This question is from textbook Elementary and intermediate algebra

Found 2 solutions by Alan3354, ankor@dixie-net.com:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
(w+3)^-1/3=1/3
Raise both sides to the 3rd power
(w+3)^-1 = 1/27
Invert
w+3 = 27
w = 24
--------
(x+2)^3/2=-1
Square both sides
(x+2)^3 = 1
x+2 = 1
x = -1 (ignoring the complex number solutions)

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
%28w%2B3%29%5E%28-1%2F3%29+=+1%2F3
:
Reciprocal gets rid of the negative exponent
1%2F%28%28w%2B3%29%5E%281%2F3%29%29+=+1%2F3
:
Multiply both side by 3
3%2F%28%28w%2B3%29%5E%281%2F3%29%29+=+1
:
Cube both sides:
27%2F%28%28w%2B3%29%29+=+1
:
27 = w + 3
:
w = 24
:
:
Check solution on calc: enter %2824%2B3%29%5E%28-1%2F3%29 = .333
;
;
;
%28x%2B2%29%5E%283%2F2%29 = -1
:
Square both sides
%28x%2B2%29%5E3 = +1
:
Find the cube root of both sides:
x + 2 = 1
:
x = 1 -2
x = -1
The problem here is if you substitute -1 for x in the original equation,
you get +1 not -1, so there is no solution here