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- Mar 20th 2011, 06:43 PMPhreshSolve the inequality
- Mar 20th 2011, 06:51 PMProve It
You have $\displaystyle \displaystyle \frac{3}{a + 1} \geq 3$.

First, note that $\displaystyle \displaystyle a + 1 \neq 0 \implies a \neq -1$.

In order to solve for $\displaystyle \displaystyle a$, you need to consider two cases, since multiplying or dividing by a negative number reverses the inequality sign...

Case 1: $\displaystyle \displaystyle a + 1 < 0 \implies a < -1$.

Then $\displaystyle \displaystyle \frac{3}{a + 1} \geq 3$

$\displaystyle \displaystyle 3 \leq 3(a + 1)$

$\displaystyle \displaystyle 1 \leq a + 1$

$\displaystyle \displaystyle 0 \leq a$.

So in this case, you are told $\displaystyle \displaystyle a < -1$ and $\displaystyle \displaystyle a \geq 0$. Since there are not any values of $\displaystyle \displaystyle a$ that satisfy both conditions, there are not any solutions in this case.

Case 2: $\displaystyle \displaystyle a + 1 > 0 \implies a > -1$.

$\displaystyle \displaystyle \frac{3}{a + 1} \geq 3$

$\displaystyle \displaystyle 3 \geq 3(a + 1)$

$\displaystyle \displaystyle 1 \geq a + 1$

$\displaystyle \displaystyle 0 \geq a$.

So in this case, the conditions are $\displaystyle \displaystyle a >-1$ and $\displaystyle \displaystyle a \leq 0$.

Therefore, the solution in this case is $\displaystyle \displaystyle -1 < a \leq 0$.

So the solution of your inequality is $\displaystyle \displaystyle -1 < a \leq 0$. - Mar 20th 2011, 06:53 PMpickslides
Wow, that is a big post, what have you tried?

- Mar 21st 2011, 04:31 AMHallsofIvy