Question 651624
the original equation is:
9*|1-x| <= -36
divide both sides of this equation by 9 to get:
|1-x| <= -36/9 which simplifies to:
|1-x| <= -4
this leads to 2 equations:
(1-x) <= -4
-(1-x) <= -4
so far so good.
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solving for (1-x) <= -4:
remove parentheses to get:
1-x <= -4
subtract 1 from both sides of the equation to get:
-x <= -5
multiply both sides of the equation by -1 to get:
x >= 5
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solving for -(1-x) <= -4:
remove parentheses to get:
-1+x <= -4
add 1 to both sides of the equation to get:
x <= -3
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your 2 solutions are:
x >= 5 or x <= -3
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this appears to agree with what you derived.
placing it in interval notation would indicate the following:
(-infinity,-3] union [5,+infinity)
i believe this is where you went wrong.
x <= -3 means that x goes from minus infinity up to and including -3.
that's where the interval notation of (-infinity,-3] comes in.
note that this is interval notation and not set notation.
x >= 5 means that x goes from 5 to plus infinity.  since 5 is included in the solution set, the interval becomes:
[5,+infinity)
the (-infinity means that the value is greater than minus infinity.
the +infinity) means that the value is less than plus infinity.
the -3] means that the value is less than or equal to -3.
the [5 means that the value is greater than or equal to 5.
not again that this is interval notation, not set noation.
in set notation, the answer would be shown as:
{x | x <= -3 or x >= 5}
or:
{x element of real numbers | x <= -3 or x >= 5}
i'd say you got the answer correct but displayed it wrong.
here's a reference on interval notation and set notation.
<a href = "http://www.regentsprep.org/Regents/math/ALGEBRA/AP1/IntervalNot.htm" target = "_blank">http://www.regentsprep.org/Regents/math/ALGEBRA/AP1/IntervalNot.htm</a>