Evaluate and simplify f(b/3)

**Eraser147** · October 6th 2012, 01:28 PM

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**ILikeSerena** · October 6th 2012, 01:38 PM

Originally Posted by Eraser147

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Hi Eraser147!

As a first step, can you replace each occurrence of x by (b/3)?

**Eraser147** · October 6th 2012, 01:42 PM

I don't know how to use the codes in the forum but it will look like this: b/3 + 8 over b^2/9 +2

**Plato** · October 6th 2012, 01:50 PM

Originally Posted by Eraser147

I don't know how to use the codes in the forum but it will look like this: b/3 + 8 over b^2/9 +2

You asked for the code

[TEX]\frac{\frac{b}{3}+8}{\frac{b^2}{9}+2}[/TEX] gives $\frac{\frac{b}{3}+8}{\frac{b^2}{9}+2}$

**Eraser147** · October 6th 2012, 01:57 PM

^ Thankyou.

**Eraser147** · October 6th 2012, 02:00 PM

I see my problem now. I didn't multiply both sides the denominator of both the equation's numerator and demoninator.

**ILikeSerena** · October 6th 2012, 02:02 PM

Good!

Originally Posted by Eraser147

I see my problem now. I didn't multiply both sides the denominator of both the equation's numerator and demoninator.

Uhh... so with what number are you multiplying both numerator and denominator?

**Eraser147** · October 6th 2012, 02:09 PM

$\frac{\frac{b}{3}+8}{\frac{b^2}{9}+2}$ multiply 3 both top and bottom, and 9 both top and bottom.

**Eraser147** · October 6th 2012, 02:11 PM

Lastly, I simplify from ${\frac{9b+216}{\frac{3b^2}{1}+54}$ which equates to ${\frac{3b+72}{\frac{b^2}{1}+18}$

**Plato** · October 6th 2012, 02:46 PM

Originally Posted by Eraser147

Lastly, I simplify from ${\frac{9b+216}{\frac{3b^2}{1}+54}$ which equates to ${\frac{3b+72}{\frac{b^2}{1}+18}$

It should be $\frac{\frac{b}{3}+8}{\frac{b^2}{9}+2}=\frac{3b+72} {b^2+18}$

**ILikeSerena** · October 6th 2012, 03:02 PM

You've got it!!

Thread: Evaluate and simplify f(b/3)