Question

How do I solve this problem...?

3/(x^2+3x)+(x+2)/(x+3)<1/x

Answer 1

\[\frac{ 3 }{ x^2+3x}+\frac{ x+2 }{ x+3 }<\frac{ 1 }{ x }\]

Answer 2

I would add -1/x to both sides

Answer 3

I would get common denomonators

Answer 4

So after adding -1/x what would the other side look like...?

Answer 5

\[ \frac{ 3 }{ x^2+3x}+\frac{ x+2 }{ x+3 }- \frac{ 1 }{ x }<0 \]
now do what cc3 suggested. find a common denominator for the 3 fractions
it looks like x(x+3)

Answer 6

So I'd divide it by x+3...? I mean that was my gut feel but then what would happen to the 1/x?

Answer 7

no,
multiply the 2nd fraction by x/x
multiply the last equation by (x+3)/(x+3)

now all three fractions will have the same denominator.
you can combine the tops

Answer 8

So that would leave me with \[\frac{ 3 }{ x^2+3x }+\frac{ 2 }{ x^2+3 }-\frac{ x+3 }{ x^2+3 }<0\] ...?

Answer 9

the bottoms should be all the same...

For example, notice for the last fraction
\[ \frac{1}{x}\cdot \frac{(x+3)}{(x+3)} = \frac{(x+3)}{x(x+3)}= \frac{(x+3)}{x^2+3x}\]

Answer 10

re do the middle fraction

Answer 11

OHhhhh so it would actually be
\[\frac{ 3 }{ x^2+3 }+\frac{ 2 }{ x^2+3 }+\frac{ x+3 }{ x^2+3 }<0\]

Answer 12

no, the minus sign on the third fraction should still be there.
the bottom is x^2 + 3x or x(x+3) (which is how I would write it)

Answer 13

and the middle fraction is still not correct.

Answer 14

I'm sorry, could you walk me through this ;3;...? I'm not really good at this kind of thing...

Answer 15

\[ \frac{ 3 }{ x^2+3x}+\frac{ x+2 }{ x+3 }- \frac{ 1 }{ x }<0 \]

multiply the second fraction by \( \frac{x}{x} \)
what do you get?
multiply the third fraction by \( \frac{(x+3)}{(x+3)} \)

Answer 16

\[\frac{ 3 }{ x^2+3x }+\frac{ x^2+2 }{ x^2+3 }-\frac{ x+3 }{ x^2+3 }\]?

Answer 17

I guess you forgot how the distributive rule works.

you should leave the bottom x(x+3)
but if you expand it, you multiply x times *everything inside the parens*
you would get x*x + 3*x or x^2 +3x (NOT x^2+3)

up top in the second fraction it is x*(x+2)
that expands into x*x + 2*x or x^2+2x (NOT x^2 + 2)

Answer 18

\[\frac{ 3 }{ x^2+3x }+\frac{ x^2+2x }{ x^2+3 }-\frac{ x+3 }{ x^2+3x }\] ?

Answer 19

\[\frac{ 3 }{ x ^{2}+3 }+\frac{ x+2 }{ x+3 }-\frac{ 1 }{x }<0\]
\[or \frac{ 3+x \left( x+2 \right)-1\left( x+3 \right) }{ x \left( x+3 \right) }<0\]
\[or~\frac{ 3+x ^{2}+2x-x-3 }{x \left( x+3 \right) }<0\]
\[or~\frac{ x \left( x+1 \right) }{ x \left( x+3 \right) }<0\]
\[or~\frac{ x+1 }{ x+3 }<0\]
either x+1>0,x>-1 and x+3<0,x<-3
or x+1<0,x<-1,x+3>0,x>-3
hence solution is (-3,-1)