# Math Help - Difficulty with Inequality

1. ## Difficulty with Inequality

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)?

2. ## Re: Difficulty with Inequality

Originally Posted by Amakeb
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)?
Yes, if $z> -x$ then $z^2-x^2-1>-1~.$

What if $z<-x~?$

3. ## Re: Difficulty with Inequality

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!

4. ## Re: Difficulty with Inequality

Originally Posted by Amakeb
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)?
No you cannot multiply by z+x whether z+x is positive or negative it leads to contradiction, -1 > -1 or -1<-1

5. ## Re: Difficulty with Inequality

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.

6. ## Re: Difficulty with Inequality

Originally Posted by thevinh
The fatal mistake in solving inequality is to multiply both side to get rid of the denominators. ...
That's correct. In this particular question there is no need for a chart. All is needed is to find a relationship between z and x. whether (z + x) is positive or negative the inequality holds for z > x.

7. ## Re: Difficulty with Inequality

Originally Posted by votan
No you cannot multiply by z+x whether z+x is positive or negative it leads to contradiction, -1 > -1 or -1<-1
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.

9. ## Re: Difficulty with Inequality

Hello, Amakeb!

$\text{I was having difficulty with a question: }\:\frac{z^2 - x^2 - 1}{z+x} \:>\: \frac{-1}{z+x}$

$\text{Could I just multiply both sides by }(z+x)\,?$ . No

First, note that $z+x \,\ne\,0.$

Then take thevinh's suggestion: move all terms to the left.

. . $\frac{z^2-x^2-1}{z+x} + \frac{1}{z+x} \:>\:0 \quad\Rightarrow\quad \frac{z^2-x^2-1 + 1}{z+x} \:>\:0$

. . $\frac{z^2-x^2}{z+x} \:>\:0 \quad\Rightarrow\quad \frac{(z-x)(z+x)}{z+x} \:>\:0$

Since $z+x\,\ne\,0$, we can cancel: . $z-x \:>\:0$

Therefore, the inequality holds if $z \,>\,x.$