i need help simplifying
Do you know the answer? I think I have it, but I'm not sure if the calculations are completely correct.
no i dont, i have simplified the 10x^4 to equal 2x^2 and the 15x^2 to equal 3. thats as far as i got
I have \[2x^3+3x^2 \over 2(4x-3)\]
that is one of the options my question has. can you explain how you got that answer?
Yes. Let me write it down for you.
ok
\[{6x+9 \over 15x^2} \div {16x-12 \over 10x^4}\] you can cancel the denominators as far as possible, since cross multiplying will make it so that you can. ----->\[{6x+9\over 3} \div {16x-12 \over 2x^2}\]
Do you get this so far?
After this, you can now factor the 6x-9 so that you can cancel out the 3. \[{3(2x+3) \over 3} \div {16x-12 \over 2x^2}\]
\[(2x+3) \div {4(4x+3) \over 2x^2}\]
Now, you can cancel out the 2. \[(2x+3) \div {2(4x+3) \over x^2}\]
Can you carry on from here, or should I show you the last steps?
dont you have to flip the 16x-12/2x^2 so that the 2x^2 is the numerator?
That's if you want to multiply straight instead of doing it at the end... You'd have to change the divide sign to the times sign. eg -----> \[(2x+3) \times {x^2 \over 2(4x+3)}\]
So, yes. You could do it like this, and now times the fractions. :) It would give you the same answer.
would it be easier to simplify everything then multiply?
For me, yes. And it can take less time, but if you're not used to it, then it might be easier to first multiply.
So, do you understand it now?
yeah a little more. can you show me how you got the final answer?
Sure.
Now, that it's flipped, you can easily just multiply the fractions -----> \[2x^2(x+3) \over 2(4x-3)\]------->\[2x^3 +3x^2 \over 2(4x-3)\]----> Which is the final answer.
thank you!
Um hold on. I made a little mistake in explaning.
It should be \[x^2 \]not \[2x^2\] and it should be (2x+3) not (x+3)
I just made that mistake in the last part. The other steps were correct :)
i like ur pic. @order
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