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

yes!

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?

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

+4

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

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

It looks much more clean this way :P

does the c cancel in 23c and 6c?

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

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

Yw :)

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

super helpful, thank you so so much!

Yw!