+1124 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..
+118680 Hi Juriemagic,
You can multiply by 6/6 but multiplying just by six makes it 6 times bigger
What you can do is this
\(\frac{6}{6}*\left[{x^2 \over 6}-{3x \over 2}-{x \over 3}+3\right]\\ =\frac{1}{6}*\left\{\frac{6}{1}*\left[{x^2 \over 6}-{3x \over 2}-{x \over 3}+3\right]\right\}\\ =\frac{1}{6}(x^2-9x-2x+18)\)
+1124 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?
