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..

juriemagic Jun 5, 2023

#1**+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

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

Melody Jun 6, 2023

#2**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