I don't get it:(

[Mayapotter00]

Hi guys,
This is an example from my math book explaining rational equations, however, when I tried multiplying the common denominator (which is x-1) all of a sudden we don't multiply it with -2, and I wonder why? Actually I don't understand the example at all, could someone explain the steps for me?
Thank you so much!!!

 

[Mr. Bland]

The example is multiplying both sides of the equation by [MATH](x - 1)[/MATH] in order to eliminate the fractions. Personally, I would add a step to make it more intuitive...

Addition and subtraction on fractions is done more easily when the denominators match. We are given the following:

[MATH]\frac{x - 2}{x - 1} - 2 = -\frac{1}{x - 1}[/MATH]​

That lonely [MATH]2[/MATH] there is conceptually the fraction [MATH]\frac{2}{1}[/MATH], which has a different denominator from the others. We need to get it in the form of [MATH]\frac{?}{x - 1}[/MATH], but how do we do that? As pointless as it might seem, we can multiply it by [MATH]1[/MATH]. Doing so will not change its value, and will not affect any other parts of the equation, so it can be useful.

Observe that a fraction in the form of [MATH]\frac{x}{x} = 1[/MATH], so we can get away with doing this:

[MATH]2 * \frac{x - 1}{x - 1} = \frac{2x - 2}{x - 1}[/MATH]​

This gives us three fractions with the same denominator, making our equation as follows:

[MATH]\frac{x - 2}{x - 1} - \frac{2x - 2}{x - 1} = -\frac{1}{x - 1}[/MATH]​

Now if we multiply both sides by [MATH](x - 1)[/MATH], it's trivial to see how we get to the next step:

[MATH](x - 2) - (2x - 2) = -1[/MATH]
[MATH]x - 2 - 2x + 2 = -1[/MATH]​

And this catches us back up with what was done in the example.

 

[Dr.Peterson]

Hi guys,
This is an example from my math book explaining rational equations, however, when I tried multiplying the common denominator (which is x-1) all of a sudden we don't multiply it with -2, and I wonder why? Actually I don't understand the example at all, could someone explain the steps for me?
Thank you so much!!!

But they did multiply the -2 by (x-1)! They distributed the multiplication to every term, including the -2. In the case of the fractions, the (x-1) cancels with the denominator; in the case of the non-fraction, there is no denominator, so they had to actually carry out the multiplication (distributing the -2 over the terms of x-1). Is that what you are referring to?

 

[lookagain]

@ Mayapotter00 -- Did you post the whole example? If so, then the example
is wrong, because that solution of x = 1 is the only candidate, and it does not
check in the original equation.

The answer is "no solution."

The original equation can be transformed into:

\(\displaystyle \dfrac{x - 2}{x - 1} \ + \ \dfrac{1}{x - 1} \ = \ 2 \ \ \implies\)

\(\displaystyle \dfrac{x - 1}{ x - 1} \ = \ 2 \)

This reduces to 1 = 2 when \(\displaystyle \ x \ne 1, \ \) but this is impossible.

 

[Mayapotter00]

The example is multiplying both sides of the equation by [MATH](x - 1)[/MATH] in order to eliminate the fractions. Personally, I would add a step to make it more intuitive...

Addition and subtraction on fractions is done more easily when the denominators match. We are given the following:

[MATH]\frac{x - 2}{x - 1} - 2 = -\frac{1}{x - 1}[/MATH]​

That lonely [MATH]2[/MATH] there is conceptually the fraction [MATH]\frac{2}{1}[/MATH], which has a different denominator from the others. We need to get it in the form of [MATH]\frac{?}{x - 1}[/MATH], but how do we do that? As pointless as it might seem, we can multiply it by [MATH]1[/MATH]. Doing so will not change its value, and will not affect any other parts of the equation, so it can be useful.

Observe that a fraction in the form of [MATH]\frac{x}{x} = 1[/MATH], so we can get away with doing this:

[MATH]2 * \frac{x - 1}{x - 1} = \frac{2x - 2}{x - 1}[/MATH]​

This gives us three fractions with the same denominator, making our equation as follows:

[MATH]\frac{x - 2}{x - 1} - \frac{2x - 2}{x - 1} = -\frac{1}{x - 1}[/MATH]​

Now if we multiply both sides by [MATH](x - 1)[/MATH], it's trivial to see how we get to the next step:

[MATH](x - 2) - (2x - 2) = -1[/MATH]​

​

[MATH]x - 2 - 2x + 2 = -1[/MATH]​

And this catches us back up with what was done in the example.

Thank you so much!!!

 

[Mayapotter00]

But they did multiply the -2 by (x-1)! They distributed the multiplication to every term, including the -2. In the case of the fractions, the (x-1) cancels with the denominator; in the case of the non-fraction, there is no denominator, so they had to actually carry out the multiplication (distributing the -2 over the terms of x-1). Is that what you are referring to?

Yes, I saw that later on. Thank you so much!!

 

[Mayapotter00]

@ Mayapotter00 -- Did you post the whole example? If so, then the example
is wrong, because that solution of x = 1 is the only candidate, and it does not
check in the original equation.

The answer is "no solution."

The original equation can be transformed into:

\(\displaystyle \dfrac{x - 2}{x - 1} \ + \ \dfrac{1}{x - 1} \ = \ 2 \ \ \implies\)

\(\displaystyle \dfrac{x - 1}{ x - 1} \ = \ 2 \)

This reduces to 1 = 2 when \(\displaystyle \ x \ne 1, \ \) but this is impossible.

Hii! No, I didn't post the whole example. The answer is no solution, as you said. However I just didn't get the steps. Thank you so much for your answer!!!

 

[Mr. Bland]

This reduces to 1 = 2 when \(\displaystyle \ x \ne 1, \ \) but this is impossible.

Perhaps more apparent is that if [MATH]x = 1[/MATH], then the denominator [MATH]\frac{?}{x - 1}[/MATH] becomes zero.