Solving Radical Equations using LCM/LCD

[swampy]

Problem is: Have problem 2/x+2 = 1/x - 2 this is what have done so far

we are suppose to find the LCD of unlike denominators what I came up with is 4x^2 as the LCD is my equivalent expression then 2/4x^2] = 1/4x^2 now do I take the 2/4x^2] multiply it by an expression equivalent to 1. what is the expression equivalent to 1?

The answer is 5 -- but I don't know how to get it. What am I forgetting or not getting... the teacher doesn't take the time to explain and I stayed after to ask and he did not have time ... I am so lost on this

Solving Rational Equations --- this problem is under the Solving Rational Equations -- but to resolve you have to find the LCM -- but can't figure how to get the LCM via the LCD.

Ok.. we have this... problem 2/(x+1) = 1/(x-2 ) ok -- problem was wrong but if we take what Loren said to do then we have 2/x + 2 = 1/x - 2 and the LCD is X is this correct?

 

[Loren]

Re: LCM unlike Denominators

Your first job is to learn how to use parenthesis to clarify what is meant.

2/x+2 = 1/x - 2 means \(\displaystyle \frac{2}{x}+2=\frac{1}{x}-2\).

If that is what you mean, then the first step is to multiply both sides of the equation by the lcd, x.

 

[Mrspi]

Problem is: Have problem 2/x+2 = 1/x - 2 this is what have done so far

we are suppose to find the LCD of unlike denominators what I came up with is 4x^2 as the LCD is my equivalent expression then 2/4x^2] = 1/4x^2 now do I take the 2/4x^2] multiply it by an expression equivalent to 1. what is the expression equivalent to 1?

The answer is 5 -- but I don't know how to get it. What am I forgetting or not getting... the teacher doesn't take the time to explain and I stayed after to ask and he did not have time ... I am so lost on this

Solving Rational Equations --- this problem is under the Solving Rational Equations -- but to resolve you have to find the LCM -- but can't figure how to get the LCM via the LCD.

Ok.. we have this... problem 2/(x+1) = 1/(x-2 ) ok -- problem was wrong but if we take what Loren said to do then we have 2/x + 2 = 1/x - 2 and the LCD is X is this correct?

I'll bet any money that your problem is really this:

   2            1
------    =  ------
 x + 1        x - 2

The LCM of the two denominators (and thus the LCD for the fractions) is (x + 1)(x - 2).

If you multiply both sides of the equation by (x + 1)(x - 2), the denominators of the fractions will DIVIDE OUT, leaving you with an equation that has no fractions:

   2                            1
------ * (x + 1)(x - 2)    =  ------  *(x + 1)(x - 2)
 x + 1                         x - 2

2(x - 2) = 1(x + 1)

Ok...now you have this equation:

2(x - 2) = 1(x + 1)

Hopefully, you can solve that for x. Be sure to check your solutions...you should realize that x cannot be either -1 or 2, because that will make the denominator of one of your original fractions 0, and you can't have 0 as the denominator of a fraction.

 

[Deleted member 4993]

Ok.. we have this... problem 2/(x+1) = 1/(x-2 ) ok -- problem was wrong but if we take what Loren said to do then we have 2/x + 2 = 1/x - 2 and the LCD is X is this correct?

No .. if the problem is:

\(\displaystyle \frac{2}{x+1} \, = \, \frac{1}{x-2}\)

Then your denominators are (x+1) and (x-2)

you do not have any factor common between those (they are relatively prime - like 9 and 16)

then the LCM of those two expressons are

\(\displaystyle (x-2)\cdot(x+1)\)

multiply both sides of the equation with the expression above:

\(\displaystyle \frac{2}{x+1} \cdot (x-2)\cdot(x+1) \, = \, \frac{1}{x-2}\cdot (x-2)\cdot(x+1)\)

eliminate common factors from the numerators and the denominators - to get:

\(\displaystyle {2}\cdot (x-2) \, = \, {1}\cdot (x+1)\)

Now simplify and solve for 'x'

 

[Denis]

Problem is: Have problem 2/x+2 = 1/x - 2
...blah blah blah....
The answer is 5 -- but I don't know how to get it.

Seems it's agreed equation should be 2/(x+2) = 1/(x - 2)

WHERE did you see/get 5 as answer? It is NOT 5; answer is 6.
Do a criss-cross multiplication:
2(x - 2) = 1(x + 2)
2x + 4 = x + 2
x = 6

You need to be CLEARER; you typed this:
"we are suppose to find the LCD of unlike denominators what I came up with is 4x^2 as the LCD is my equivalent expression then 2/4x^2] = 1/4x^2 now do I take the 2/4x^2] multiply it by an expression equivalent to 1. what is the expression equivalent to 1?"

That has no punctuation (so hard to follow): show it to your English Lit teacher!
You say you came up with "4x^2 as the LCD": but did not show HOW.
The rest is as cluttered as the Folies Bergeres dressing room!