How do you solve #(a-6)/5=7/12#?

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Answer 1 · 2018-08-06T10:20:33

#a=107/12#

Cross-multiply to get

#12a-72=35#

So

#12a=107#

#a=107/12#

Answer 2 · 2018-08-06T14:27:46

#color(green)(a = 107/12 = 8 11/12#

#(a-6)/5 = 7/12#

#a-6 = (5 * 7) / 12 = 35/12#

#a = 6 + 35/12#

#a= (72 + 35) / 12#, taking L C M.

#color(green)(a = 107/12 = 8 11/12#