## Simplifying Complex Fraction

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
EdSewersPark
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### Simplifying Complex Fraction

Question: $\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}}$

I'm guessing I have to get rid of the fractions by multiplying by the LCD, but I feel like I am missing a step in even getting that.

stapel_eliz
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EdSewersPark wrote:Question: $\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}}$

I'm guessing I have to get rid of the fractions by multiplying by the LCD...

Yes, that would be the first step. You may want to factor the lower fraction's denominator first, though, so you're clearly aware of what the LCM actually is.

MrAlgebra
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### Re: Simplifying Complex Fraction

EdSewersPark wrote:Question: $\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}}$

I'm guessing I have to get rid of the fractions by multiplying by the LCD, but I feel like I am missing a step in even getting that.

I would be inclined to start like this. Keeping in mind that

$\frac{\frac{a}{b}}{\frac{c}{d}} = (\frac{a}{b})(\frac{d}{c})$

We have

$\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}}$

$(5+\frac{x+3}{x-1})(\frac{1-x^2}{3x-1})$

Then find the LCD for $(5+\frac{x+3}{x-1})$ and go from there.

EdSewersPark
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### Re: Simplifying Complex Fraction

$\frac{(x-1)}{(x-1)}$ $\frac{(5)}{(1)}$ and $\frac{(x+3)}{(x-1)}$ $\frac{(1)}{(1)}$ which equals $\frac{5x-5}{x-1}\,+\,\frac{x+3}{x-1}$ which equals $\frac{6x-2}{x-1}$

Then I have $(\frac{6x-2}{x-1})(\frac{1-x^2}{3x-1})$

This is where I draw a blank, do I need to change $(1-x^2)$ so that it appears in the denominator? Or do I need to find the LCD for $(x-1)$ and $(3x-1)$?

stapel_eliz
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EdSewersPark wrote:$\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}}$

You have two denominators: x - 1 and 1 - x2 = (1 - x)(1 + x). With a "minus" sign, the first denominator becomes -(1 - x). Take the "minus" sign through the numerator, and you get a subtraction on top.

Use the method illustrated here and multiply, top and bottom, by (1 - x2). This will immediately simplify the fraction considerably!

EdSewersPark
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### Re: Simplifying Complex Fraction

$(\frac{2(3x-1)}{x-1})(\frac{(1-x)(x-1)}{3x-1})$

Can I cancel out both (x-1)'s and (3x-1)'s at this point?

Would 2(1-x) be the wrong answer?

stapel_eliz
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EdSewersPark wrote:$(\frac{2(3x-1)}{x-1})(\frac{(1-x)(x-1)}{3x-1})$

Can I cancel out both (x-1)'s and (3x-1)'s at this point?

Would 2(1-x) be the wrong answer?

Since 1 - x and x - 1 are not the same, no, they cannot be "cancelled". Instead, try making the change explained earlier. Then see about cancelling.

MrAlgebra
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### Re: Simplifying Complex Fraction

EdSewersPark wrote:$(\frac{2(3x-1)}{x-1})(\frac{(1-x)(x-1)}{3x-1})$

Can I cancel out both (x-1)'s and (3x-1)'s at this point?

Would 2(1-x) be the wrong answer?

You're right. It should go:

$(\frac{2(3x-1)}{x-1})(\frac{(1-x)(x-1)}{3x-1})$

$(\frac{2}{x-1})(\frac{(1-x)(x-1)}{1})$ Canceled out 3x - 1.

$(\frac{2}{1})(\frac{(1-x)(1)}{1})$ Canceled out x - 1.

$2(1-x)$ All that's left.

Donut2CoffeeCup
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### Re: Simplifying Complex Fraction

Tidy up the top, ie: make it a single fraction:

$\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}} = \frac{\frac{5x-5+x+3}{x-1}}{\frac{3x-1}{1-x^2}} = \frac{\frac{6x-2}{x-1}}{\frac{3x-1}{1-x^2}}$

Now use $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$ to get

$\frac{(6x-2)}{(x-1)} \times \frac{1-x^2}{3x-1}$

Factorise $(6x-2)=2(3x-1)$

And note (very carefully) that $1-x^2=(1-x)(1+x)$

If you can't see this - reverse the step to check: $(1-x)(1+x)=1-x^2$

So far we have:

$\frac{2(3x-1)}{(x-1)} \times \frac{(1-x)(1+x)}{3x-1}$

To set up the `cancelling' of the $(x-1)$ term write $(x-1) = -(1-x)$, so that

$-\frac{2(3x-1)}{(1-x)} \times \frac{(1-x)(1+x)}{3x-1}$

Cancel the $(1-x)$ terms:

$-\frac{2(3x-1)}{1} \times \frac{(1+x)}{3x-1}$

Cancel the $(3x-1)$ terms so that you get:

$-\frac{2}{1} \times \frac{(1+x)}{1} = -2(1+x)$