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Originally Posted by Eraser147 Click image to enlarge. Hi Eraser147! As a first step, can you replace each occurrence of x by (b/3)?
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
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 $\displaystyle \frac{\frac{b}{3}+8}{\frac{b^2}{9}+2}$
^ Thankyou.
I see my problem now. I didn't multiply both sides the denominator of both the equation's numerator and demoninator.
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?
$\displaystyle \frac{\frac{b}{3}+8}{\frac{b^2}{9}+2}$ multiply 3 both top and bottom, and 9 both top and bottom.
Lastly, I simplify from $\displaystyle {\frac{9b+216}{\frac{3b^2}{1}+54}$ which equates to $\displaystyle {\frac{3b+72}{\frac{b^2}{1}+18}$
Last edited by Eraser147; Oct 6th 2012 at 02:16 PM.
Originally Posted by Eraser147 Lastly, I simplify from $\displaystyle {\frac{9b+216}{\frac{3b^2}{1}+54}$ which equates to $\displaystyle {\frac{3b+72}{\frac{b^2}{1}+18}$ It should be $\displaystyle \frac{\frac{b}{3}+8}{\frac{b^2}{9}+2}=\frac{3b+72} {b^2+18}$
You've got it!!
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