# more help needed

• July 21st 2008, 08:03 AM
Laura901
more help needed
Is this right? I'm dividing complex fractions....

8x^2 divided by 4x
over
x^2 -25 divided by 3x+15

I flipped the other fraction and changed the problem to multiplication

8x^2 ovwer x^2 -25 divided by 3x +15 divided by 4x

I factored x^2 -25 and got x+5 and x-5 and divided 8 by 4 and got two and 4 divided by 4 is 1 or x

and now I have 2 times 3x plus 15 over x...is that correct?
• July 21st 2008, 10:04 AM
Simplicity
Quote:

Originally Posted by Laura901
Is this right? I'm dividing complex fractions....

8x^2 divided by 4x
over
x^2 -25 divided by 3x+15

I flipped the other fraction and changed the problem to multiplication

8x^2 ovwer x^2 -25 divided by 3x +15 divided by 4x

I factored x^2 -25 and got x+5 and x-5 and divided 8 by 4 and got two and 4 divided by 4 is 1 or x

and now I have 2 times 3x plus 15 over x...is that correct?

You are getting confused with constant and variables ( $x$). They cannot cancel each other out. Constants can only cancel out constants and variables can only cancel variables. Let me explain the procedure:

$\frac{8x^2}{4x} \div \frac{x^2-25}{3x+15}$

You correctly said to flip a fraction and multiply, hence:

$\frac{8x^2}{4x} \times \frac{3x+15}{x^2 - 25}$

Simplify the expressions. $\frac{8x^2}{4x}$ can be simplied to $2x$ as the denominator has $4x^1$ and rules of indices says $\frac{x^a}{x^1} = x^{a-1}$ so we have $\frac{8x^2}{4x} = 2x^{2-1} = 2x^1 = 2x$ (Notice, we only let constants work with constant $\left(\frac84=2\right)$ and variables work with variable $\left(\frac{x^2}{x} = x\right)$). For $\frac{3x+15}{x^2 - 25}$, the numerator and the denominator can be factorised to get $\frac{3(x+5)}{(x+5)(x-5)}$. We have $(x+5)$ on the numerator and the denominator so we can cancel it out to get $\frac{3}{(x-5)}$ So we get:

$2x \times \frac{3}{(x-5)}$

For fractions, when we have to multiply, we multiply the numerator (top) together and we multiply the denominator (bottom) togather. So we get:

$\frac{2x}{1} \times \frac{3}{(x-5)} = \frac{(3)(2x)}{(1)(x+5)} = \frac{6x}{x+5}$