I'm finishing up a problem and don't understand how the answer is what it is... Here goes... -28/3Y = -28 (the step says to divide both sides by -28/3) and the answer should be Y=3 HOW?????
Is this what it looks like?\[\large \frac{-28}{3}y=-28\]The \(\large y\) is not in the denominator of that fraction, correct?
sorry, was answering someones post too... and yes correct it is how you put it up there...
If we had\[\large 4y=4\] We would divide by 4 to isolate the \(\large y\) term, yes? Basically what we did was, we divided by the size of the `coefficient` on \(\large y\). So it's clear that the -28/3's will cancel on the y side right? We need to revisit what happens when we divide by a fraction. (That will tell us what's happneing on the right side of the equation).
\[\large \dfrac{\cancel{\dfrac{-28}{3}}y}{\left(\cancel{\dfrac{-28}{3}}\right)}=\dfrac{-28}{\left(\dfrac{-28}{3}\right)}\] Understand that these will cancel on the left since they're the same value?
yes, exactly....
oh my...... and when I put it in the calculator the way you have it written... it comes out 3
When we divide by a fraction, we can `flip it` and rewrite the operation as multiplication. Here's an example.\[\large \frac{\left(\dfrac{1}{2}\right)}{\left(\dfrac{3}{4}\right)} \qquad = \qquad \frac{1}{2}\cdot\frac{4}{3}\] See how I `flipped`, (took the reciprocal) of the bottom one?
Oh yes, if you feel more comfortable just using a calculator ~ Then make sure you use some brackets.
for some reason, every way I put it in before I would get the answer 1/3...... and that is wrong!!!!
yep, I used the brackets, but I misplaced them!!!! Thanks for your help... medal is on the way if it will let me give it to you!!!! (now I can move on to the resat of my work)!THANKS
\[\large \frac{-28}{\left(\dfrac{-28}{3}\right)} \qquad = \qquad -28\cdot \frac{3}{-28}\] See how I flipped the bottom fraction and rewrote it as multiplication using that rule?
heh np XD
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