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

- anonymous

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

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- anonymous

\[\frac{ (5c-4d) }{ 2c }-\frac{ (8c-7d) }{ 3c }+4\] is this your expression?

- anonymous

yes!

- anonymous

You may use the following \[\frac{ a }{ b } \pm \frac{ c }{ d } = \frac{ ad \pm bc }{ bd }\] try it out

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## More answers

- anonymous

sorry am i cross multiplying?

- anonymous

The \[\pm \] implies it works for either sign

- anonymous

Not quite

- anonymous

so without much simplifying, its 5c-4d(3c)±2c(8c-7d)/2c(3c)?

- anonymous

It's just like adding/ subtracting regular fractions :)

- anonymous

So yes you did it right!

- anonymous

except you don't have to write p/m I was just stating it would work for either sign

- anonymous

oh okay, but wouldnt the denominator be 6c?

- anonymous

\[\frac{ 3c(5c-4d) -2c(8c-7d)}{ (2c)(3c) }+4\]

- anonymous

Close but you have
\[6c^2\]for the denominator

- anonymous

opps haha forgot the ^2

- anonymous

:)

- anonymous

Keep going, you're on the right track

- anonymous

so I got 15c^2-12cd-16c^2-14cd/6c^2

- anonymous

would I now just simplify it?

- anonymous

Yes keep simplifying and watch the signs \[\frac{ 3c(5c-4d) -2c(8c-7d)}{ (2c)(3c) }+4 \implies \frac{ 15c^2-12cd-16c^2 \color{red}+14cd }{ 6c^2 }+4\]

- anonymous

so I got -1c^2+2cd/6c^2

- anonymous

+4

- anonymous

\[\frac{ -c^2+2cd }{ 6c^2 }+4\] looks good! You can do the same thing with the 4 now, \[4 = \frac{ 4 }{ 1 }\]

- anonymous

and so i multiply 4 by 6c^2 to get a common denominator?

- anonymous

The denominator will stay as 6c^2 you just have to multiply 4 by 6c^2 in the numerator

- anonymous

so it'll be 24c^2

- anonymous

\[\frac{ -c^2+2cd }{ 6c^2 }+4 \implies \frac{ -c^2+2cd+4(6c^2) }{ 6c^2 } \implies \frac{ -c^2+2cd+24c^2 }{ 6c^2 }\]

- anonymous

for final answer I got 23c^2+2cd/6c^2

- anonymous

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.

- anonymous

It looks much more clean this way :P

- anonymous

does the c cancel in 23c and 6c?

- anonymous

Yeah that c got cancelled out so we're left with just 1 c, so here \[\frac{ 23c^2+2cd }{ 6c^2 } = \frac{ \cancel c(23c+2d) }{ 6c^{\cancel 2} } = \frac{ 23c+2d }{ 6c }\]

- anonymous

\[\frac{ c }{ c^2 } = \frac{ 1 }{ c }\]

- anonymous

ohhh.. that makes it clear, haha my gosh thank you so much!

- anonymous

Yw :)

- anonymous

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 \[\frac{ 1 }{ 3 } + \frac{ 4 }{ 5 }\]

- anonymous

it was very helpful, much easier to understand compared to what my teacher taught me :)

- anonymous

Yeah, sometimes teachers do that...these days they complicate a lot of things which come from simple things as I showed you above ^.^

- anonymous

super helpful, thank you so so much!

- anonymous

Yw!

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