1. ## Rational inequality?

Can anyone help with this rational inequality
7/(x+1)>7
I think its zero but apparently the real answer is -1<x>0.

2. Originally Posted by homeylova223
Can anyone help with this rational inequality 7/(x+1)>7
You need to be sure that $\displaystyle x+1>0~\&~x+1<1$.

3. Originally Posted by homeylova223
Can anyone help with this rational inequality
7/(x+1)>7
I think its zero but apparently the real answer is -1<x>0.
$\displaystyle \displaystyle\frac{7}{x+1}>7$

if $\displaystyle (x+1)<1$ and positive

$\displaystyle \Rightarrow\ -1<x<0$

4. The first thing we all want to do is multiply both sides of the inequality by (x+1), but we need to be careful because the inequality changes if we multiply by a negative value, so 2 cases:

Case 1) x+1 < 0 or x < -1
Then 7 < 7(x + 1) and 7 < 7x + 7 and we get x > 0
so taking all the conditions, we have x < -1 and x > 0, but this is not possible, so case 1 gives no solutions

Case 2) x + 1 > 0 or x > -1
Then 7 > 7(x+1) and 7 > 7x + 7 and we get x < 0
so taking all the conditions we have x > -1 and x < 0, this is a range and can be written as -1 < x < 0

5. [soze=3]Hello, homeylova223![/size]

$\displaystyle \dfrac{7}{x+1}\:>\:7$

$\displaystyle \text{I think it's zero}$ . WHAT is zero?

We have: .$\displaystyle \dfrac{7}{x+1} \:>\:7$

Divide by 7: .$\displaystyle \dfrac{1}{x+1} \:> \:1$

Then we have: .$\displaystyle \displaystyle \frac{1}{x+1} - 1 \:>\:0 \quad\Rightarrow\quad \frac{-x}{x+1} \:>\:0$

Multiply by -1: .$\displaystyle \dfrac{x}{x+1} \:<\:0$

If a fraction is negative: the numerator and denominator have opposite signs.

$\displaystyle [1]\;\begin{array}{cccccccc} x \:>\:0 \\ x+1\:<\:0 & \Rightarrow & x \:<\:-1 \end{array}$

. . . This says: .$\displaystyle x > 0$ and $\displaystyle x < -1$ . . . impossible

$\displaystyle [2]\;\begin{array}{ccccccc}x \:<\:0 \\ x+1\:>\:0 & \Rightarrow & x \:>\:-1 \end{array}$

. . . This says: .$\displaystyle x < 0$ and $\displaystyle x > -1$

This is possible: .$\displaystyle -1 \:<\:x\:<\:0$

6. Hello, homeylova223!

$\displaystyle \dfrac{7}{x+1}\:>\:7$

$\displaystyle \text{I think it's zero}$ . WHAT is zero?

We have: .$\displaystyle \dfrac{7}{x+1} \:>\:7$

Divide by 7: .$\displaystyle \dfrac{1}{x+1} \:> \:1$

Then we have: .$\displaystyle \displaystyle \frac{1}{x+1} - 1 \:>\:0 \quad\Rightarrow\quad \frac{-x}{x+1} \:>\:0$

Multiply by -1: .$\displaystyle \dfrac{x}{x+1} \:<\:0$

If a fraction is negative: the numerator and denominator have opposite signs.

$\displaystyle [1]\;\begin{array}{cccccccc} x \:>\:0 \\ x+1\:<\:0 & \Rightarrow & x \:<\:-1 \end{array}$

. . . This says: .$\displaystyle x > 0$ and $\displaystyle x < -1$ . . . impossible

$\displaystyle [2]\;\begin{array}{ccccccc}x \:<\:0 \\ x+1\:>\:0 & \Rightarrow & x \:>\:-1 \end{array}$

. . . This says: .$\displaystyle x < 0$ and $\displaystyle x > -1$

This is possible: .$\displaystyle -1 \:<\:x\:<\:0$