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Is there a value of x for the following equation which will make it less than x ?

The question is which is more $$ \frac{3x+1}{x+1}$$ if $$x\not=-1$$ or simply 2 ?

According to the book there is not enough information to solve this problem, but i think the expression is greater than 2. Is there any value of x which makes it less than 2 ?

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Which equation? – Américo Tavares Jul 10 '12 at 23:05
$x=0 {}{}{}{}{}$ – anon Jul 10 '12 at 23:05
Do you mean expression? – Américo Tavares Jul 10 '12 at 23:07
up vote 4 down vote accepted

We know $x\neq -1$, so either $x+1>0$ or $x+1<0$. Let's consider these cases separately (since we're working with inequalities).

If $x+1>0$, then $\frac{3x+1}{x+1}<2$ implies that $3x+1<2(x+1)=2x+2$, so $x<1$. In fact, any $-1<x<1$ will do the trick.

If $x+1<0$, then $\frac{3x+1}{x+1}<2$ implies that $3x+1>2x+2$ (direction of inequality switched since we multiplied by a negative number), so $x>1$, but this is impossible, since $x+1<0$.

Thus, $\frac{3x+1}{x+1}<2$ if and only if $-1<x<1$.

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