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Solve K =1/3(b – 2a) for b
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b=1/(3K)+2a
how yu gt that
\[K=(1/(3(b-2a)))\]\[b-2a=1/3K\]\[b=1/3K+2a\]
it is (1+2ak)/3k
Although I can't tell if 1/3 is multiplying (b-2a) or not.
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xtremely sorry ......ans is (1+6ak)/3k
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