If \(\frac{3}{4x + 3} + \frac{8}{9y + 8} + \frac{2305}{11z + 2305} = 7\), what is the value of \(\frac{4x}{4x + 3} + \frac{9y}{9y + 8} + \frac{11z}{11z + 2305}\)?
–4
Each required fraction can be paired with a given one: \(\frac{4x}{4x + 3} + \frac{3}{4x + 3} = \frac{4x + 3}{4x + 3} = 1\).
In the same way, \(\frac{9y}{9y + 8} = 1 - \frac{8}{9y + 8}\) and \(\frac{11z}{11z + 2305} = 1 - \frac{2305}{11z + 2305}\).
Adding the three: required value = \(3 - \left(\frac{3}{4x + 3} + \frac{8}{9y + 8} + \frac{2305}{11z + 2305}\right) = 3 - 7 = -4\).
Check: the sum of all six fractions is (1) + (1) + (1) = 3, and 7 + (–4) = 3, so the given sum and the required sum together give exactly 3.
Hence, –4.
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