How do you solve #1/8+2/t=17/(8t)#?

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Answer 1 · 2015-10-16T10:47:23

#t=1#.

First of all, write the left member as

#1/8+2/t = (t+2*8)/(8t) = (t+16)/(8t)#

(note that #t# must not be zero, otherwise we would have #0# at the denominator).

So we have

#(t+16)/(8t) = 17/(8t)#

since the denominators are equal, the equation holds if and only if the numerators are equal. This means

#t+16=17#. Solving for #t#, we get #t=17-16=1#.

CHECK: for #t=1#, the equation becomes

#1/8+2=17/8#, which is indeed true.