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