Question

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

Answer 1

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

Answer 2

yes!

Answer 3

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

Answer 4

sorry am i cross multiplying?

Answer 5

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

Answer 6

Not quite

Answer 7

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

Answer 8

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

Answer 9

So yes you did it right!

Answer 10

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

Answer 11

oh okay, but wouldnt the denominator be 6c?

Answer 12

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

Answer 13

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

Answer 14

opps haha forgot the ^2

Answer 15

:)

Answer 16

Keep going, you're on the right track

Answer 17

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

Answer 18

would I now just simplify it?

Answer 19

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

Answer 20

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

Answer 21

+4

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

so it'll be 24c^2

Answer 26

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

Answer 27

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

Answer 28

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.

Answer 29

It looks much more clean this way :P

Answer 30

does the c cancel in 23c and 6c?

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

Yw :)

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

super helpful, thank you so so much!

Answer 39

Yw!