Question

Help please. Can someone please explain how you calculate the restricted value(s) of x for the equation

Answer 1

Here is the problem. Also what those value(s) represent?

Answer 2

Make the denominators zero. Done.

Answer 3

When you solve an equation for x, you want to find values for x such that those values satisfy the equation (i.e., the equality holds) but you also want to play the "game" by the rules, that is, you don't want anything illegal or you'll find stuff that are untrue. In this particular case, you cannot divide by 0, so you want values of x such that all those denominators aren't equal to zero. Mathematically, \( x + 2 = 0\) and \( 2x = 0 \) aren't valid values for x, hence they are restricted. Can you solve both equations for x?

Answer 4

Ok, thank you. So would I use the cross multiplication or LCD method? And why? Would it be the LCD method?

Answer 5

To solve the equation, that is? If all you need is to find the restricted values, all you have to do is set the denominators to zero and voila. If you need to solve it too, I would multiply by \({x+2} \) on both sides. I think that's called cross multiplication methdo.

Answer 6

LCD is the only method that there is. That other one should not ever be mentioned.

My views. I welcome others.

Answer 7

Ok I think you are right about the cross multiplication method, now that you explained it. So let me make sure I understand this: So to calculate the restricted value of x for the problem I would set the denominators to zero? What do those values represent? But then to take it further and solve the problem, I need to multiply by x + 2 on both sides? So is that all of the steps to solve the equation using the cross multiplication method? And what would the solution be?

Answer 8

That method that shall not be named has NO application with three terms. Just let it die a miserable death. Once you have determined restrictions, you have committed to the denominators not being zero. As they are NOT zero, you can multiply both sides of the equation by the non-zero expression.

Answer 9

Being a bit pedantic, those values represent discontinuities in the solution interval of the equation. In English, it basically means that those values don't follow the rules of the game and you can't divide by zero. To avoid such problems, you find values of x which are forbidden beforehand, so you don't accidently proof something which is untrue. I will post an example of that afterwards. Yeah, I would multiply both sides by x + 2. To be honest, both LCD and cross-multiplication seem fairly easy on this problem, but I never liked multiplying polynomials. This is what will happen: \[ (2+x)(\frac{3}{2+x} + \frac{1}{2x}) = (2+x)(\frac{4}{2+x})\]Distribute it over, and: \[ 3 + \frac{2+x}{2x} = 4 \] Can you solve it from here?

Answer 10

I think so

Answer 11

The simplest division by zero that I can think of is: \[ 0 * 1 = 0 * 2 = 0 \] Then: \[\frac{0}{0} * 1 = \frac{0}{0} * 2 = \frac{0}{0}\] But any number divided by itself is one, so:
\[ 1*1 = 1*2 = 1 \implies 1 = 2\]This is a rather artificial example, but you can see how these problems happen (and happen fairly often) so you always have to take care. Also, kudos for your English. As I'm a non-native, sometimes I make some mistakes but it's always refreshing to see someone writing properly :)

Answer 12

Thank you so much, I appreciate it.

Answer 13

No problem at all. Good luck!

Answer 14

Thanks again.

Answer 15

@bmp My behavior on this issue is not pedantic at all. It is based on the preponderance of errors I have observed over the years. CM is over-used and under-comprehended and it looks SO EASY to the beginning student that they NEVER get past it. I do think it should be written from math texts and never mentioned again - Like "FOIL".

Answer 16

I understand, thanks.

Answer 17

@tkhunny Oh, I didn't mean you were being pedantic, I understand you and I certainly agree. I said that I was being pedantic beforehand (because I gave a somewhat complicated description of what a discontinuity is before giving a more intuitive one). Sorry if you understood like that, it wasn't my intention at all.

Answer 18

@bmp No worries. I think we're still getting along, ok.