Help with 3+(3/ (x-1)) / 3-(3/x)

[l.k.mullins]

I need some help with this.

3+(3/ x-1) / 3-(3/x)

Answer is X^2 / x-1^

Can someone show me how to get to this point

I originally had a typo.

Mark can you show me an example, I am trying to help my girlfriend and it has been a while since I have done this. I don not under stand why the final answer has both the numerator and denominator squared

 

[mmm4444bot]

[3+(3/ (x-1))] / [3-(3/x)]

You need these grouping symbols in red; otherwise, the answer you posted is wrong.

Answer is x^2/(x-1)^2

Hello LK:

Add the two terms in the numerator of the compound fraction. In other words, get a common denominator and add the 3 to the 3/(x - 1). This gives you a single fraction on top.

Subtract the two terms in the denominator of the compound fraction using a common denominator, too: 3 - 3/x. Then you have a single fraction on the bottom.

Now, use the rule that dividing by a fraction is the same as multiplying by its reciprocal. In other words, multiply the fraction on top by the reciprocal of the fraction on the bottom.

If none of this makes any sense, then please let me know, and I will do an example problem for you.

Otherwise, please show whatever work you've been able to do so far, and we'll see where you're stuck.

Cheers,

~ Mark

 

[Deleted member 4993]

I need some help with this.

3+(3/ x-1) / 3-(3/x)

Answer is X^2 / x-1^

Can you simplify (without using calculator and leave the solution in fractions):

\(\displaystyle \frac{3 \, + \, \frac{3}{8 \, - \, 1}}{3 \, - \, \frac{3}{8}}\)

You'll follow the same method for your algebraic expression

Can someone show me how to get to this point

I originally had a typo.

Mark can you show me an example, I am trying to help my girlfriend and it has been a while since I have done this. I don not under stand why the final answer has both the numerator and denominator squared

 

[mmm4444bot]

... I don not under stand why the final answer has both the numerator and denominator squared

Hello LK:

There are squares in both the numerator and denominator because that's how it turns out.

Are you saying you don't understand why this is because you came up with a different result? If so, then we like to see your work.

Here's an example of simplifying a compound fraction.

\(\displaystyle \frac{\frac{1}{7} - \frac{2}{7x}}{\frac{x - 7}{7} + 1}\)

On top, we want to combine the two terms, but we need a common denominator of 7x first.

\(\displaystyle \frac{1}{7} \cdot \frac{x}{x} = \frac{x}{7x}\)

Now we can combine the two terms on top.

\(\displaystyle \frac{x}{7x} - \frac{2}{7x} = \frac{x - 2}{7x}\)

On the bottom, we want to combine the two terms, but we need a common denominator of 7 first, so we rewrite 1 as 7/7.

\(\displaystyle \frac{x - 7}{7} + \frac{7}{7} = \frac{x - 7 + 7}{7} = \frac{x}{7}\)

After these combinations, the compound fraction has a single fraction on top and a single fraction on the bottom.

\(\displaystyle \frac{\frac{x - 2}{7x}}{\frac{x}{7}}\)

Dividing the top by the fraction x/7 is the same as multiplying it by the reciprocal 7/x.

\(\displaystyle \frac{x - 2}{7x} \cdot \frac{7}{x} = \frac{x - 2}{x^2}\)

The compound fraction that you posted simplifies the same way. Find common denominators to combine the terms on top into a single fraction; do the same on the bottom; simplify the results; multiply the top by the reciprocal of the bottom; and simplify that result.

Cheers,

~ Mark