Question 1031982: solve the inequality of |3x+1|<2+|2x+3|
Found 4 solutions by josgarithmetic, MathTherapy, ikleyn, solver91311:
Answer by josgarithmetic(39617)   (Show Source):
Answer by MathTherapy(10552)  (Show Source):
Answer by ikleyn(52786)  (Show Source):
Answer by solver91311(24713)  (Show Source):
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Re-write so that the inequality can be decomposed into a piece-wise function:
This inequality has two critical points;  values that make one of the expressions inside of the absolute value bars equal to zero, namely  and 
Hence there are three intervals to consider when creating a piece-wise function definition: ,\ ) ,\ ) and )
Since  and  when  , we can say:
\ =\ -3x\ -\ 1\ -\ (-2x\ -\ 3)\ -\ 2) if 
So we need to find the interval that satisfies both \ -\ 2\ <\ 0) and
\ -\ 2\ <\ 0)
 .
But the intervals ) and ) are disjoint. Hence, there is no part of the solution set in the interval .
Next, for values in ) ,
\ =\ -3x\ -\ 1\ -\ (2x\ +\ 3)\ -\ 2)
So, set the function less than zero and solve:
\ -\ 2\ <\ 0)
Hence, the function is true on the interval )
Finally, for values in )
\ -\ 2\ <\ 0)
So the inequality holds for values in the interval )
That leaves the value to check which we can do in the original inequality:
\ +\ 1|\ -\ |2\left(-\frac{1}{3}\right)\ +\ 3|\ -\ 2\ <\ 0)
So now we can specify the union of the two valid intervals which include the endpoint, and
)
Is the complete solution set interval.
Compare this result to the portion of the graph of ) that is below the -axis.
John
My calculator said it, I believe it, that settles it
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