# Rational inequality?

• Dec 12th 2010, 02:32 PM
homeylova223
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. (Cool)
• Dec 12th 2010, 02:36 PM
Plato
Quote:

Originally Posted by homeylova223
Can anyone help with this rational inequality 7/(x+1)>7

You need to be sure that $x+1>0~\&~x+1<1$.
• Dec 12th 2010, 02:38 PM
Quote:

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. (Cool)

$\displaystyle\frac{7}{x+1}>7$

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

$\Rightarrow\ -1
• Dec 12th 2010, 02:43 PM
snowtea
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
• Dec 12th 2010, 02:58 PM
Soroban
[soze=3]Hello, homeylova223![/size]

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

• Dec 12th 2010, 02:59 PM
Soroban
Hello, homeylova223!

Quote:

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

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

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

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

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

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

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

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

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

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

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

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