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If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the reciprocal of the product of a, b, and c, then a =

A. (1 + bc)/(b + c) B. (1 – bc)/(b + c) C. (1 + b + c)/(bc) D. (1 – b – c)/(bc) E. (1 – b – c)/(b + c)

Kudos for a correct solution.

(1/a)+(1/b)+(1/c) = 1/(a*b*c)

i.e. (bc+ac+ab)/abc = 1/abc

i.e. bc+ac+ab=1

i.e. ac+ab = 1-bc

i.e. a(b+c)= 1-bc

i.e. a= (1 - bc)/(b+c)

Answer: option B
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First, translate the given information into an equation. Go phrase by phrase. The sum of the reciprocals of a, b, and c is 1/a + 1/b + 1/c. Notice that you first take reciprocals, then you add the reciprocals together.

Now, set that equal to “the reciprocal of the product of a, b, and c,” which is 1/(abc). Notice that we first take the product of a, b, and c (which is abc), and then take the reciprocal of that product.

The equation is this: 1/a + 1/b + 1/c = 1/(abc)

Now rearrange to isolate a on one side. Make a common denominator on the left side (abc), so that you can add the fractions: 1/a + 1/b + 1/c = bc/(abc) + ac/(abc) + ab/(abc) = (bc + ac + ab)/(abc)

Since the right side of the original equation is 1/(abc), which happens to have the same denominator, you can set the numerators equal:

bc + ac + ab = 1

Now solve for a:

ac + ab = 1 – bc a(c + b) = 1 – bc a = (1 – bc)/(c + b)

Theoretically, you can solve this problem by plugging numbers for the variables, but finding three consistent values of a, b, and c (to satisfy the complicated condition) is rather difficult. A pure algebraic approach is faster and more secure in this case.

Re: If abc ≠ 0 and the sum of the reciprocals of a, b, and c equals the re [#permalink]

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18 Oct 2017, 06:33

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