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Fraction algebra [Solved!]

My question

The following question was asked on a sample test:

`3/((x+2b)(x-b))` ` + 2/((3b-x)(x+2b))` ` +1/((x-3b)(b-x))`

We tried answering two ways to find the LCD: first, by negating the denominator only; second, by negating both the numerator and the denominator.

We do not know which is the right solution, i.e. which method is right and it is VERY CONFUSING.

So with fractions like this, do we negate only the denominator to get a suitable LCD, or do we negate both the numerator AND the denominator?
Thank you so much.

Relevant page

5. Equivalent Fractions

What I've done so far

The first way: when the third denominator was multiplied by -1 to get the LCD, the denominator was negated but the numerator was not.

`= 3 / [(x+2b)(x-b)]` ` + 2 / [(3b-x)(x+2b)]` ` + 1 / [(3b-x)(x-b)] `

`= [3(3b-x) + 2(x-b) + (1)(x+2b)] / [(x+2b)(x-b)(3b-x)]`

`= [(9b - 3x +2x - 2b + x + 2b)] / [(x+2b)(x-b)(3b-x)]`

`= (9b) / [(x+2b)(x-b)(3b-x)]`

The second way: when the third denominator was multiplied by -1, the numerator and the denominator were negated to get:

`3 / [(x+2b)(x-b)]` ` + 2 / [(3b-x)(x+2b)]` ` + (-1) / [(3b-x)(x-b)]`

`= [3(3b-x) + 2(x-b) + (-1)(x+2b)] / [(x+2b)(x-b)(3b-x)]`

`= [ (9b - 3x + 2x - 2b - x - 2b) ] / [(x+2b)(x-b)(3b-x)]`

` = [ 5b - 2x ] / [(x+2b)(x-b)(3b-x)]`

X

The following question was asked on a sample test:
 
`3/((x+2b)(x-b))` `  + 2/((3b-x)(x+2b))` ` +1/((x-3b)(b-x))`
 
We tried answering two ways to find the LCD: first, by negating the denominator only; second, by negating both the numerator and the denominator.
 
We do not know which is the right solution, i.e. which method is right and it is VERY CONFUSING.

So with fractions like this, do we negate only the denominator to get  a suitable LCD, or do we negate both the numerator AND the denominator?
Thank you so much.
Relevant page

<a href="/factoring-fractions/5-equivalent-fractions.php">5. Equivalent Fractions</a>

What I've done so far

The first way:  when the third denominator was multiplied by -1 to get the LCD, the denominator was negated but the numerator was not.
 
`= 3 / [(x+2b)(x-b)]` `  + 2 / [(3b-x)(x+2b)]` ` +  1 / [(3b-x)(x-b)] `                                       

`= [3(3b-x) + 2(x-b) + (1)(x+2b)]  /  [(x+2b)(x-b)(3b-x)]`                          
 
`= [(9b - 3x +2x - 2b + x + 2b)] / [(x+2b)(x-b)(3b-x)]`

`= (9b) / [(x+2b)(x-b)(3b-x)]`
 
The second way:  when the third denominator was multiplied by -1,  the numerator and the denominator were negated to get:

`3 / [(x+2b)(x-b)]` `  + 2 / [(3b-x)(x+2b)]` ` +  (-1) / [(3b-x)(x-b)]`

`= [3(3b-x) + 2(x-b)  +  (-1)(x+2b)]  /  [(x+2b)(x-b)(3b-x)]`

`= [ (9b - 3x + 2x - 2b - x - 2b) ] / [(x+2b)(x-b)(3b-x)]`

` = [ 5b - 2x ] / [(x+2b)(x-b)(3b-x)]`

Re: Fraction algebra

Please use the math input system. It's easy, and others can read your question much more easily.

Your first answer is correct, but some of your explanation is not quite right (which is maybe where the confusion has arisen).

For this problem we need to remember that in general,

`-(x-y) = y-x`.

That's what you are using on the 3rd fraction, but you are actually doing it twice.

The first use was:

`-(x-3b) = 3b-x`

The second use was:

`-(b-x) = x-b`

So you are actually multiplying by `-1` twice. The nett effect of that is `-1xx-1=1`, so you have (correctly) not changed the value of that 3rd fraction.

In other words, the 3rd fraction was modified as follows:

`1/((3b-x)(x-b)) ` `= 1/((-1xx(3b-x))(-1xx(x-b)))` `= 1/((x-3b)(b-x))`

I usually prefer to multiply top and bottom of a fraction by the same number. So I would have written:

`1/((3b-x)(x-b)) ` `= (-1)/(-1xx(3b-x)) xx (-1)/(-1xx(x-b))` `= (+1)/((x-3b)(b-x))`

It gives us the same result, of course.

Hope it helps.

X

Please use the math input system. It's easy, and others can read your question much more easily.

Your first answer is correct, but some of your explanation is not quite right (which is maybe where the confusion has arisen).

For this problem we need to remember that in general, 

`-(x-y) = y-x`.

That's what you are using on the 3rd fraction, but you are actually doing it twice.

The first use was:

`-(x-3b) = 3b-x`

The second use was:

`-(b-x) = x-b`

So you are actually multiplying by `-1` twice. The nett effect of that is `-1xx-1=1`, so you have (correctly) not changed the value of that 3rd fraction.

In other words, the 3rd fraction was modified as follows:

`1/((3b-x)(x-b)) ` `= 1/((-1xx(3b-x))(-1xx(x-b)))` `= 1/((x-3b)(b-x))`

I usually prefer to multiply top and bottom of a fraction by the same number. So I would have written:

`1/((3b-x)(x-b)) ` `= (-1)/(-1xx(3b-x)) xx (-1)/(-1xx(x-b))` `= (+1)/((x-3b)(b-x))`

It gives us the same result, of course.

Hope it helps.

Re: Fraction algebra

It helps! Thank you.

X

It helps!  Thank you.

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