## anonymous one year ago (5c-4d)/(2c)-(8c-7d)/(3c)+4 write as a single fraction

1. anonymous

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

2. anonymous

yes!

3. anonymous

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

4. anonymous

sorry am i cross multiplying?

5. anonymous

The $\pm$ implies it works for either sign

6. anonymous

Not quite

7. anonymous

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

8. anonymous

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

9. anonymous

So yes you did it right!

10. anonymous

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

11. anonymous

oh okay, but wouldnt the denominator be 6c?

12. anonymous

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

13. anonymous

Close but you have $6c^2$for the denominator

14. anonymous

opps haha forgot the ^2

15. anonymous

:)

16. anonymous

Keep going, you're on the right track

17. anonymous

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

18. anonymous

would I now just simplify it?

19. anonymous

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

20. anonymous

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

21. anonymous

+4

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

so it'll be 24c^2

26. anonymous

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

27. anonymous

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

28. 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.

29. anonymous

It looks much more clean this way :P

30. anonymous

does the c cancel in 23c and 6c?

31. anonymous

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

32. anonymous

$\frac{ c }{ c^2 } = \frac{ 1 }{ c }$

33. anonymous

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

34. anonymous

Yw :)

35. 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 $\frac{ 1 }{ 3 } + \frac{ 4 }{ 5 }$

36. anonymous

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

37. anonymous

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

38. anonymous

super helpful, thank you so so much!

39. anonymous

Yw!