Do We Need Boundaries for Fraction Equations?

[Callmelucky]

Homework Statement

Do we always need to exclude solutions that would give 0 in the denominator of a fraction?

Relevant Equations

1/x

When working with fractions and when we have a fraction or equation with fractions like this one for example $\frac{x}{x-1}+\frac{x}{x+1}=\frac{9}{4}$ do we always need to set boundaries? Like, do we always need to write that x can't be a number that would give the denominator 0? In this particular case x can't be 1 and -1. Because that would equal 0 and we can't have 0 in the denominator.

I know it's kind of a stupid question and the answer is kind of obvious but I am not sure if that should be stated in the problem by the author or if I am supposed to do that.

Thank you.

 

[FactChecker]

Homework Statement:: Do we always need to exclude solutions that would give 0 in the denominator of a fraction?
Relevant Equations:: 1/x

When working with fractions and when we have a fraction or equation with fractions like this one for example $\frac{x}{x-1}+\frac{x}{x+1}=\frac{9}{4}$ do we always need to set boundaries? Like, do we always need to write that x can't be a number that would give the denominator 0? In this particular case x can't be 1 and -1. Because that would equal 0 and we can't have 0 in the denominator.

I know it's kind of a stupid question and the answer is kind of obvious but I am not sure if that should be stated in the problem by the author or if I am supposed to do that.

Thank you.

You should not call it a solution point. You should keep track of those invalid points and make sure that your final answer does not include them as valid.

ADDED: Sometimes intermediate steps are not valid for certain x-values. Although they don't work in the intermediate calculation, those x-values should be checked in the original problem. They might work.

 

[Callmelucky]

You should not call it a solution point. You should keep track of those invalid points and make sure that your final answer does not include them as valid.

Thank you.

 

[Mark44]

When working with fractions and when we have a fraction or equation with fractions like this one for example $\frac{x}{x-1}+\frac{x}{x+1}=\frac{9}{4}$ do we always need to set boundaries?

It's fairly clear in the equation above that x can be neither 1 nor -1.
If you multiply both sides of this equation by $4(x - 1)(x + 1)$ you get $4x(x + 1) + 4x(x - 1) = 9$. At this point you should* state $x \ne 1$ and $x \ne -1$, since there are no apparent restrictions on the possible solutions for x.
(* - by "should" I mean, if you don't do this, you are opening yourself up to trouble.)

Here's a different problem (admittedly very contrived) that shows why it's important to list restrictions:
$\frac{x^2 -1}{x - 1} + \frac 1 {x - 1} = \frac 1{x - 1}$

If we blithely multiply both sides by x - 1, we get $x^2 - 1 + 1 = 1$, which simplifies to $x^2 = 1$, with solutions $x = \pm 1$. One of these is a solution to the original equation and the other is not.

 

[Callmelucky]

I know, I was just confused because in some problems it's stated by the author that x can't be a number that would make a denominator 0. So I didn't know if I was supposed to write what x can't be or not.

 

[Callmelucky]

I have another question, about something I haven't seen before.

Here is the screenshot from photomath app(pic below).

It obviously makes sense, and the solutions do match with ones in my textbook. But I didn't expect that, I have only divided both sides with (a+b-x) and got x=c as the only solution.
Is there some kind of rule or something that would explain this.?

Thank you.

 

[PeroK]

But I didn't expect that, I have only divided both sides with (a+b-x) and got x=c as the only solution.
Is there some kind of rule or something that would explain this.?

You can only cancel $a+b - x$ if this expression is not zero. Whenever you cancel anything like this, you automatically have to write "or $a + b - x = 0$.

This rule can be expressed as:
$$ac = bc \ \Rightarrow \ a = b \ \text{or} \ c = 0$$

 

[Callmelucky]

You can only cancel $a+b - x$ if this expression is not zero. Whenever you cancel anything like this, you automatically have to write "or $a + b - x = 0$.

This rule can be expressed as:
$$ac = bc \ \Rightarrow \ a = b \ \text{or} \ c = 0$$

I see, thank you.

 

[Callmelucky]

When the equation is set by the author at the beginning, does that automatically mean that there are no 0s in the denominator?
For example, I have an equation like this $\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$ and I want to add stuff on the right side and in order to do that I need the smallest common denominator, which is abx, but if any of them is 0 I can't do that.

Thank you

 

[PeroK]

When the equation is set by the author at the beginning, does that automatically mean that there are no 0s in the denominator?
For example, I have an equation like this $\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$ and I want to add stuff on the right side and in order to do that I need the smallest common denominator, which is abx, but if any of them is 0 I can't do that.

Thank you

If any of these is zero, then one of your fractions is undefined. The original equation implies that none of the denominators is zero.

 

[Callmelucky]

If any of these is zero, then one of your fractions is undefined. The original equation implies that none of the denominators is zero.

Thank you.
There is one more thing, if I multiply both sides with (a+b+x)(abx) do I need to write that a+b+x$\neq$0 and a$\neq$0, b$\neq$0, x$\neq$0. Because on the left side a can be 0 so can b and x. In that case that would not apply to all a's and b's.
Thank you.

 

[FactChecker]

You should note those things to the best of your ability. It can be difficult to keep track of all the exceptions. It is always good to check your final answers in the original equations. That way, any division by zero will become apparent.

 

[PeroK]

Thank you.
There is one more thing, if I multiply both sides with (a+b+x)(abx) do I need to write that a+b+x$\neq$0 and a$\neq$0, b$\neq$0, x$\neq$0. Because on the left side a can be 0 so can b and x. In that case that would not apply to all a's and b's.
Thank you.

Technically, that is implied by your original equation and could be stated explicitly at the beginning.

If at any stage you divide by a quantity, then you introduce an additional constraint that that quantity is non zero.

 

[Callmelucky]

Technically, that is implied by your original equation and could be stated explicitly at the beginning.

If at any stage you divide by a quantity, then you introduce an additional constraint that that quantity is non zero.

Thank you. But what if it happens that one of the denominators becomes 0 somewhere in the procedure? That fraction is invalid, but if it happens, does that mean that we just leave that fraction out of the equation and keep going with the procedure or something else?

 

[PeroK]

Thank you. But what if it happens that one of the denominators becomes 0 somewhere in the procedure? That fraction is invalid, but if it happens, does that mean that we just leave that fraction out of the equation and keep going with the procedure or something else?

You'll need to provide an example of what you mean by that.

 

[Callmelucky]

You'll need to provide an example of what you mean by that.

for example if equation $\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$ is step in solving some other equation and it happens for a to be 0. The denominator on the left side will not be 0 and the rest of the stuff on the right side is valid. Is that then $\frac{1}{b+x}=\frac{1}{b}+\frac{1}{x}$
It's not from the problem, it's just curiosity, I'm not even sure if that can happen.

 

[PeroK]

If you know $a =0$, then you can replace $a$ by $0$.

 

[Callmelucky]

If you know $a =0$, then you can replace $a$ by $0$.

I know that but what happens with fraction 1/0, and with equation after plugging that in? My calculator says math error. Does that mean that the entire equation is invalid?

For example $\frac{1}{b+x}=\frac{1}{0}+\frac{1}{2}$. My calculator gives error on $\frac{1}{5}+\frac{1}{0}+\frac{1}{2}$

 

[PeroK]

I know that but what happens with fraction 1/0, and with equation after plugging that in? My calculator says math error. Does that mean that the entire equation is invalid?

Yes. The entire equation is invalid.

 

[PeroK]

for example if equation $\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$ is step in solving some other equation and it happens for a to be 0. The denominator on the left side will not be 0 and the rest of the stuff on the right side is valid. Is that then $\frac{1}{b+x}=\frac{1}{b}+\frac{1}{x}$
It's not from the problem, it's just curiosity, I'm not even sure if that can happen.

This makes no sense. That first equation is not valid with $a =0$.

 

[Callmelucky]

Yes. The entire equation is invalid.

this is what I wanted to know. Thank you.

 

[MidgetDwarf]

I have another question, about something I haven't seen before.

Here is the screenshot from photomath app(pic below).

It obviously makes sense, and the solutions do match with ones in my textbook. But I didn't expect that, I have only divided both sides with (a+b-x) and got x=c as the only solution.
Is there some kind of rule or something that would explain this.?

Thank you.

They used the zero product property.