Hi all! Really happy to find this forum. I was having difficulty with a question.
(z^2 - x^2 - 1)/(z+x) > -1/(z+x)
Could I just multiply both sides by (z+x)?
Thanks for your response! I actually don't even need to solve the equation. I just need to know if I can divide both sides by (z+x). To answer your question, here is what I am thinking:
Dividing both sides by (z+x) isn't allowed. Given these are variables, we don't know if their sum is a negative or positive value. If positive, z^2 - x^2 - 1 > -1. However, if the value is negative, we'd have to flip the sign, such that z^2 - x^2 - 1 < -1. Because we can't know which is correct, we cannot divide by (z+x).
Am I correct? Is there anything else I should consider?
Thanks a lot!
The fatal mistake in solving inequality is to multiply both side to get rid of the denominators. The best way to solve it is to bring everything to one side of the inequality then make common denominator and solve the inequality (top and bottom) separately. Lastly, do a sign chart to obtain the solution.
How do you get that? If z+ x is positive, multiplying both sides by it gives z^2- x^2- 1> -1 or z^2> x^2. Given that z+ x is positive, z> -x so that z^2> x^2 is automatically satisfied. This inequality is true for all x, z such that z+x> 0 or z> -x.
If z+ x is negative, multiplying both sides by it gives z^2- x^2- 1< -1 so that z^2- x^2< 0 or z^2< x^2. Given that z+ x is negative, z< -x so that z^2< x^2 is automatically satisfied. This inequality is true for all x, z such that z+ x< 0 or z< -x.
Rather than giving a contradiction, this inequality is solved for all (x, y) except, of course, for z= -x.
Hello, Amakeb!
$\displaystyle \text{I was having difficulty with a question: }\:\frac{z^2 - x^2 - 1}{z+x} \:>\: \frac{-1}{z+x}$
$\displaystyle \text{Could I just multiply both sides by }(z+x)\,?$ . No
First, note that $\displaystyle z+x \,\ne\,0.$
Then take thevinh's suggestion: move all terms to the left.
. . $\displaystyle \frac{z^2-x^2-1}{z+x} + \frac{1}{z+x} \:>\:0 \quad\Rightarrow\quad \frac{z^2-x^2-1 + 1}{z+x} \:>\:0$
. . $\displaystyle \frac{z^2-x^2}{z+x} \:>\:0 \quad\Rightarrow\quad \frac{(z-x)(z+x)}{z+x} \:>\:0 $
Since $\displaystyle z+x\,\ne\,0$, we can cancel: .$\displaystyle z-x \:>\:0 $
Therefore, the inequality holds if $\displaystyle z \,>\,x.$