How do you solve #12/t+t-8=0#?

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Answer 1 · 2017-12-01T18:16:38

#t=6 or t=2#

You have equation with a fraction. You can get rid of the denominator by multiplying each term by #t#

#(color(blue)(txx)12)/t + color(blue)(txx)t - color(blue)(txx)(8)=0 xxt#

#12+t^2 -8t =0" "larr# re-arrange the terms

#t^2 -8t +12 =0" "larr# factorise

#(t-6)(t-2)=0#

Set each factor equal to #0#

#t-6=0" "rarr t =6#

#t-2=0" "rarr t=2#