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Hi friends,

I understand if I really seem very daft for asking this question, but I'm going to ask anyway...if we have this:

\({x^2 \over 6}-{3x \over 2}-{x \over 3}+3\)

 

we get the LCD, which is 6

so we have:

\({x^2-9x-2x+18} \over 6\)

 

finally we have:

\({x^2-11x+18} \over 6\)

 

WHY CAN'T WE JUST DO THIS:

we have:

\({x^2 \over 6}-{3x \over 2}-{x \over 3}+3\)

just multiply with 6 straight through, and get:

\({x^2-11x+18}\)

 

NO LCD...

 

Sorry if this is really a stupid question..

 Jun 5, 2023
 #1
avatar+118607 
+1

Hi Juriemagic,

 

You can multiply by 6/6  but multiplying just by six makes it 6 times bigger

What you can do is this

 

\(\frac{6}{6}*\left[{x^2 \over 6}-{3x \over 2}-{x \over 3}+3\right]\\ =\frac{1}{6}*\left\{\frac{6}{1}*\left[{x^2 \over 6}-{3x \over 2}-{x \over 3}+3\right]\right\}\\ =\frac{1}{6}(x^2-9x-2x+18)\)

 Jun 6, 2023
 #2
avatar+1124 
0

Hi Melody,

 

okay, but then that answer you got is just a different format to indicate your are still deviding by 6, so it really is exactly the same thing as having an LCD, not so?...So, when can I just multiply straight through using the highest denominator, to end up not having an LCD or a fraction in front of an expression?

juriemagic  Jun 7, 2023
 #3
avatar+118607 
0

You can multiply through by the lowest common denominator ,but then you have divide the answer by the same number. 

You did not do that.

 

Yes

 

\(\frac{1}{6}(x^2-9x-2x+18) = \frac{x^2-11x+18}{6}\)

Melody  Jun 9, 2023

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