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Asked by Rob

What is the proof of theorem 2.23, where (b/a)+(d/c)=[(bc)+(ad)]/(ac). Thank you :)
12 years ago

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Answered by Steve
You change nothing when you multiply by 1. So,

b/a + d/c = b/a * c/c + d/c * a/a = bc/ac + ad/ac = (bc+ad)/ac
12 years ago
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What is the proof of theorem 2.23, where (b/a)+(d/c)=[(bc)+(ad)]/(ac). Thank you :)

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