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

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Answer 1 · 2016-08-10T17:05:13

#k = -14#

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#