How do you solve #(x+2)/6=(x-1)/12#?

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How do you solve #(x+2)/6=(x-1)/12#?

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Answer 1 · 2017-10-08T08:40:50

#x=-5#

Explanation:

By cross multiplication,
#12(x+2)=6(x-1)#
#12x+24=6x-6#
#12x-6x=-24-6#
#6x=-30#
Divide both sides by #color(red)(6)#
#cancel(6x)/cancelcolor(red)(6) = (-30)/color(red)(6)#
#x=-5#

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How do you solve #(x+2)/6=(x-1)/12#?

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