# Thread: Dividing both sides of an inequality by one factor

1. ## Dividing both sides of an inequality by one factor

This is a system of inequations for which I want to find the set of answers.

$(x-1)(x+2)<(x+2)(x-3)$
$(x+3)(x+5)>(x+4)(x+3)$

What I CAN'T do here is the following:

We start with the first inequation $(x-1)(x+2)<(x+2)(x-3)$

1) I divide both sides of the inequation by $(x+2)$
and I get:

$\frac{(x-1)(x+2)}{(x+2)}<\frac{(x+2)(x-3)}{(x+2)}$

2) I cancel the $(x+2)$
in top and bottom of both members of the inequality and I get:

$(x-1)<(x-3)$

In other words:

$x-1

Then solving this inequality I get:

3) $0<-2$

And this is plainly wrong: Then my question is WHY, what's wrong with this procedure.

I know the right one so I don't need the correct procedure, but instead what I'm looking for is an explanation about how come this is incorrect (not just saying because the answer is wrong, I know the correct answer) I thought that perhaps has to do with order of operations...

2. ## Re: Dividing both sides of an inequality by one factor

Originally Posted by querti09
This is a system of inequations for which I want to find the set of answers.

$(x-1)(x+2)<(x+2)(x-3)$
$(x+3)(x+5)>(x+4)(x+3)$

What I CAN'T do here is the following:

We start with the first inequation $(x-1)(x+2)<(x+2)(x-3)$

1) I divide both sides of the inequation by $(x+2)$
and I get:

$\frac{(x-1)(x+2)}{(x+2)}<\frac{(x+2)(x-3)}{(x+2)}$

2) I cancel the $(x+2)$
in top and bottom of both members of the inequality and I get:

$(x-1)<(x-3)$

In other words:

$x-1

Then solving this inequality I get:

3) $0<-2$

And this is plainly wrong: Then my question is WHY, what's wrong with this procedure.

I know the right one so I don't need the correct procedure, but instead what I'm looking for is an explanation about how come this is incorrect (not just saying because the answer is wrong, I know the correct answer) I thought that perhaps has to do with order of operations...
When you divide by the factor $(x+2)$ you don't know if the factor is positive or negative.

If it were negative you would need to flip over the inequality. I hope this helps.

3. ## Re: Dividing both sides of an inequality by one factor

Originally Posted by TheEmptySet
When you divide by the factor $(x+2)$ you don't know if the factor is positive or negative.
If it were negative you would need to flip over the inequality. I hope this helps.
I believe that the coefficient must be positive, since making the “invisible” coefficients show up, the original inequality
would be: $+1(x-1)(x+2)<+1(x+2)(x-3)$
If I'm putting well the “invisible” coefficients then I need to divide by a positive 1 coefficient, and the inequality doesn't change the signs:
$\frac{+1(x-1)(x+2)}{+1(x+2)}<\frac{+1(x+2)(x-3)}{+1(x+2)}$
What I thought as a possibility is to think that this is just something that happens even when the manipulation with the quantities is being done well. I think about it as just a path of solving that doesn't solve the problem, a dead end. Something not to do if I want to get the values for $x$ certainly.

4. ## Re: Dividing both sides of an inequality by one factor

Just multiply everything and put all terms on the left. Both inequalities are in the form of f(x)<0 and g(x)>0, where f and g are quadratic functions. Graph of quadratic function is called parabola, and the inequality f(x)<0 can be translated to the question "for which values of x is the parabola of the function f strictly below the abscissa", and similarly g(x)>0 can be translated to the question "for which values of x is the parabola of the function g strictly above the abscissa". So all you have to do is sketch both parabolas (you just have to find the roots of f(x)=0, g(x)=0 and decide on the orientation of the parabola), and then by simple "visual inspection" you can get the answers to both of the questions. Both answers will be intervals and all you need to do is find their intersection.

5. ## Re: Dividing both sides of an inequality by one factor

Originally Posted by querti09
This is a system of inequations for which I want to find the set of answers.

$(x-1)(x+2)<(x+2)(x-3)$
$(x+3)(x+5)>(x+4)(x+3)$

What I CAN'T do here is the following:

We start with the first inequation $(x-1)(x+2)<(x+2)(x-3)$

1) I divide both sides of the inequation by $(x+2)$
and I get:

$\frac{(x-1)(x+2)}{(x+2)}<\frac{(x+2)(x-3)}{(x+2)}$

2) I cancel the $(x+2)$
in top and bottom of both members of the inequality and I get:

$(x-1)<(x-3)$

In other words:

$x-1

Then solving this inequality I get:

3) $0<-2$

And this is plainly wrong: Then my question is WHY, what's wrong with this procedure.

I know the right one so I don't need the correct procedure, but instead what I'm looking for is an explanation about how come this is incorrect (not just saying because the answer is wrong, I know the correct answer) I thought that perhaps has to do with order of operations...
You can do that if x+ 2> 0 which is, of course, the same as saying x> -2. As long as that is true, then x+ 1< x- 3 which, subtracting x from both sides gives -1< -3. Because that is NOT true, we know that x CANNOT be larger than -2. As long as x< -2, x+2< 0 and dividing both sides by x+2 is dividing by a negative number and so the inequality is reversed: x- 1> x- 3. That reduces to -1> -3 which is TRUE for all x. So the original inequality is true for all x> -2.x^3

You could also have done this by going ahead and doing the indicated multiplication: $(x- 1)(x+2)= x^3+ x- 2$ and $(x+ 2)(x- 3)= x^2- x- 6$. The inequality is just $x^2+ x- 2< x^2- x- 6$. Now you can just add and subtract from each side to get $2x< -4$ so, because 2 is obviously positive, we can divide by 2 to get x< -2.

6. ## Re: Dividing both sides of an inequality by one factor

What I still don't get it Mathoman is:The $x^{2}$ terms, both disappear when I move the second member to the left side of the inequality in order to graph. This makes me really wonder, should this be called a quadratic inequality or a linear inequality? I am confused about this, since when I sketch the inequality I have $2x+4<0$ and $2x+4$ is a polynomial of first grade. What does it make us to label an inequality quadratic or not? The original set of inequalities are quadratic, but later when I made them equal to 0 the sketch I see is a line, not a parabola.
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