Question

(k^2-14)/(8k^2+3k) divided by (k-2)/(16k+6)
simplify.

My work:
(k^2-14)/(8k^2+3k) divided by (k-2)/(16k+6)
(k^2-14)/(8k^2+3k) times (16k+6)/(k-2)
I crossed out like terms and got:
(k+2)/(k(k+1)) * 2k+2

Am I correct or no??

Answer 1

@satellite73 Please help?

Answer 2

@timo86m please help??

Answer 3

((k^2-14)/(8k^2+3k)) /( (k-2)/(16k+6)) that?

Answer 4

\[\frac{ k^2-14 }{ 8k^2+3k }\times \frac{ 16k + 6 }{ k-2 } \]

now lets factor what we can

\[\frac{ k^2-14 }{ k(8k+3) }\times \frac{ 2(8k + 3) }{ k-2 } \]

now we can cancel out

\[\frac{ k^2-14 }{ k }\times \frac{ 2 }{ k-2 } \]

Answer 5

and it seems you did the first thing right of multiply by reciprocal.

Answer 6

unless there is a typo and the k^2 -14 is really k^2 -4

all I can see to do now is multiply across

Answer 7

(2 k^2-28)/(k^2-2 k)

Answer 8

i'd go with jim

Answer 9

makes sense?

Answer 10

Not really. On the first part, how do you get just k in the denominator??

Answer 11

Oh wait, nevermind

Answer 12

Oh and it was a typo.It is 4 not 14. Dx

Answer 13

(k^2-4)/(8k^2+3k) divided by (k-2)/(16k+6)
Thats what it SHOULD be.
@Jim766

Answer 14

\[\frac{ k^2-4 }{ k}\times \frac{ 2 }{ k-2 }\] now the k^2 -4 factors because it is a difference of 2 squares

\[\frac{ (k +2)(k - 2) }{ k}\times \frac{ 2 }{ k-2 }\] now the k-2 cancels out

Answer 15

\[\frac{ (k +2) }{ k}\times \frac{ 2 }{ 1 }\] now mult straight across

\[\frac{ 2k +4 }{ k}\] and that is what you have left

Answer 16

when you do these types of problems, if it is division, change it to multiplication like you did. once that is done look for ways to factor the different parts. Very often when everything is factored, you can see what has to be cancelled out.
What is left is you answer.

questions?

Answer 17

Nope, makes sense. Thank you

Answer 18

yw