How do you solve #9/(k-7)=6/k#?

1 Answer
Aug 10, 2016

Answer:

#k = -14#

Explanation:

If an equation has ONE term on each side of the equal sign and there are fractions, then we may cross-multiply. This gets rid of the fractions.

#color(red)(9)/color(blue)((k-7))=color(blue)(6)/color(red)(k)#

#color(red)(9xx k) = color(blue)(6xx(k-7)#

#9k = 6k -42#

#9k - 6k = -42#

#3k = -42#

#k = -14#