- #1
XJellieBX
- 40
- 0
Problem:
[tex]\frac{3}{|x+1|-1}[/tex]+[tex]\frac{2}{x}[/tex]<1
My Solution:
There are 2 cases:
1) x[tex]\geq0[/tex],[tex]\frac{3}{x+1-1}[/tex]+[tex]\frac{2}{x}[/tex]<1
[tex]\frac{3}{x}[/tex]+[tex]\frac{2}{x}[/tex]<1
And you end up with... x>5.
2) x<0, [tex]\frac{3}{-x-1-1}[/tex]+[tex]\frac{2}{x}[/tex]<1
[tex]\frac{3x-2x-4}{-x^{2}-2x}[/tex]<1
... a few reductions later...
[tex]x^{2}[/tex]+3x-4>0
And the solution set for this case is, -4>x>1
My Question:
Is there something I'm missing or something else I need to do?
[tex]\frac{3}{|x+1|-1}[/tex]+[tex]\frac{2}{x}[/tex]<1
My Solution:
There are 2 cases:
1) x[tex]\geq0[/tex],[tex]\frac{3}{x+1-1}[/tex]+[tex]\frac{2}{x}[/tex]<1
[tex]\frac{3}{x}[/tex]+[tex]\frac{2}{x}[/tex]<1
And you end up with... x>5.
2) x<0, [tex]\frac{3}{-x-1-1}[/tex]+[tex]\frac{2}{x}[/tex]<1
[tex]\frac{3x-2x-4}{-x^{2}-2x}[/tex]<1
... a few reductions later...
[tex]x^{2}[/tex]+3x-4>0
And the solution set for this case is, -4>x>1
My Question:
Is there something I'm missing or something else I need to do?