(5c-4d)/(2c)-(8c-7d)/(3c)+4 write as a single fraction

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(5c-4d)/(2c)-(8c-7d)/(3c)+4 write as a single fraction

Mathematics
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\[\frac{ (5c-4d) }{ 2c }-\frac{ (8c-7d) }{ 3c }+4\] is this your expression?
yes!
You may use the following \[\frac{ a }{ b } \pm \frac{ c }{ d } = \frac{ ad \pm bc }{ bd }\] try it out

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Other answers:

sorry am i cross multiplying?
The \[\pm \] implies it works for either sign
Not quite
so without much simplifying, its 5c-4d(3c)±2c(8c-7d)/2c(3c)?
It's just like adding/ subtracting regular fractions :)
So yes you did it right!
except you don't have to write p/m I was just stating it would work for either sign
oh okay, but wouldnt the denominator be 6c?
\[\frac{ 3c(5c-4d) -2c(8c-7d)}{ (2c)(3c) }+4\]
Close but you have \[6c^2\]for the denominator
opps haha forgot the ^2
:)
Keep going, you're on the right track
so I got 15c^2-12cd-16c^2-14cd/6c^2
would I now just simplify it?
Yes keep simplifying and watch the signs \[\frac{ 3c(5c-4d) -2c(8c-7d)}{ (2c)(3c) }+4 \implies \frac{ 15c^2-12cd-16c^2 \color{red}+14cd }{ 6c^2 }+4\]
so I got -1c^2+2cd/6c^2
+4
\[\frac{ -c^2+2cd }{ 6c^2 }+4\] looks good! You can do the same thing with the 4 now, \[4 = \frac{ 4 }{ 1 }\]
and so i multiply 4 by 6c^2 to get a common denominator?
The denominator will stay as 6c^2 you just have to multiply 4 by 6c^2 in the numerator
so it'll be 24c^2
\[\frac{ -c^2+2cd }{ 6c^2 }+4 \implies \frac{ -c^2+2cd+4(6c^2) }{ 6c^2 } \implies \frac{ -c^2+2cd+24c^2 }{ 6c^2 }\]
for final answer I got 23c^2+2cd/6c^2
That works but we can simplify it more \[\frac{ 23c^2+2cd }{ 6c^2 } = \frac{ c(23c+2d) }{ 6c^2 } = \frac{ 23c+2d }{ 6c }\] notice we factored out a c.
It looks much more clean this way :P
does the c cancel in 23c and 6c?
Yeah that c got cancelled out so we're left with just 1 c, so here \[\frac{ 23c^2+2cd }{ 6c^2 } = \frac{ \cancel c(23c+2d) }{ 6c^{\cancel 2} } = \frac{ 23c+2d }{ 6c }\]
\[\frac{ c }{ c^2 } = \frac{ 1 }{ c }\]
ohhh.. that makes it clear, haha my gosh thank you so much!
Yw :)
Also try that rule I showed you above with regular fractions and you will see it's the exact same thing as adding or subtracting fractions, just a quick little rule. :) So you can even use it for something like \[\frac{ 1 }{ 3 } + \frac{ 4 }{ 5 }\]
it was very helpful, much easier to understand compared to what my teacher taught me :)
Yeah, sometimes teachers do that...these days they complicate a lot of things which come from simple things as I showed you above ^.^
super helpful, thank you so so much!
Yw!

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