Question

Solve and show work for the 8 problems:
(7, 8, 10, 11, 13, 14, 19, & 20) of the following link: http://www.kutasoftware.com/FreeWorksheets/Alg2Worksheets/Solving%20Rational%20Equations.pdf It has the answers too, but I haven't looked at them yet. Just look at the questions.
A medal will be rewarded for help.

Answer 1

#1 & 2 especially

Answer 2

# 1


\[\large \frac{1}{6k^2} = \frac{1}{3k^2} - \frac{1}{k}\]


\[\large \frac{1}{6k^2} = \frac{1}{3k^2} - \frac{3k}{3k^2}\]


\[\large \frac{1}{6k^2} = \frac{1-3k}{k^2}\]


\[\large 3*k^2 = 6k^2(1-3k)\]


\[\large 3*k^2 = 6k^2-18k^3\]


\[\large 0 = 6k^2-18k^3 - 3k^2\]


\[\large 6k^2-18k^3 - 3k^2 = 0\]


\[\large -18k^3 + 3k^2 = 0\]


\[\large -3k^2(6k - 1) = 0\]


I'll let you finish. Make sure you check all your potential answers.

Answer 3

@jim_thompson5910 Do #2 and I think I'll be good

Answer 4

can i at least get a please? man the rudeness on this site just astounds me sometimes...

Answer 5

I meant please, I have 2 - 3 questions people have and It's hard to type a response for each one, so I am sorry.
#1: I had multiplied each side with k^2 to cancel some numbers out what would I get with that.

Answer 6

@bahrom7893

Answer 7

multiply everything out by 2n^2:

2+2n=1
2n=1-2
n=-1/2

Answer 8

Nevermind @carson889 Could you do #7, please!? :D

Answer 9

@UnkleRhaukus Click on link and help me on number 1 the quicker the way to do it

Answer 10

I multiplied both sides by 6k^2, but after that I am stuck

Answer 11

that's the the greatest denominator between them all

Answer 12

\[\frac1{6k^2}=\frac1{3k^2}-\frac 1k\\
1=\frac{6k^2}{3k^2}-\frac {6k^2}k\\1=2-6k\\k=\]

Answer 13

k = 1/6

Answer 14

right