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If \(a\), \(b\), \(c\), and \(d\) are non-zero numbers such that \(\frac{a}{b} = \frac{c}{d}\) and \(\frac{a}{d} = \frac{b}{c}\), which of the following must be true?
A. \(|a| = |c|\) B. \(|b| = |d|\) C. \(|a| = |d|\) D. \(|b| = |a|\) E. \(|b| = |c|\)
If \(a\), \(b\), \(c\), and \(d\) are non-zero numbers such that \(\frac{a}{b} = \frac{c}{d}\) and \(\frac{a}{d} = \frac{b}{c}\), which of the following must be true?
A. \(|a| = |c|\) B. \(|b| = |d|\) C. \(|a| = |d|\) D. \(|b| = |a|\) E. \(|b| = |c|\)
Given: (i) \(\frac{a}{b} = \frac{c}{d}\) and (ii) \(\frac{a}{d} = \frac{b}{c}\).
Now, from (i) \(a=\frac{bc}{d}\), substitute \(a\) in (ii) \(\frac{bc}{d^2}=\frac{b}{c}\). Cross-multiply and reduce by \(b\): \(c^2=d^2\), which means that \(|c|=|d|\), but this option is not listed among the answer choices so we should try to get the one which is listed.
Again, from (i) \(c=\frac{ad}{b}\), substitute \(c\) in (ii): \(\frac{a}{d}=\frac{b^2}{ad}\). Cross-multiply and reduce by \(d\): \(a^2=b^2\), which means that \(|a|=|b|\).
In order for those equalities to be true you can see that c=b or a=b because those letters appear on either side of the equality. There is no c=d in the answers so choose a=b.
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Hi, i have followed following steps.. let me know if it's correct a/b = c/d ad=bc (1) a/d=b/c ac=bd (2) adding (1) and (2) ad+ac=bc+bd a(d+c)=b(c+d) cancelling (d+c) on both sides a=b or b=a Is this method correct.. as it is mentioned that a,b,c,d are non zero I have divided (c+d) both the sides.. but still I am not getting mod(a) = mod(b) expression. Could you please, if this approach is correct
is cross multiplication allowed? We don't know the sign of any number. Please suggest.
We are concerned about the sign when dealing with inequalities: we should keep the sign if we multiply by a positive value and flip the sign when we multiply by a negative value.
For equations we can multiply/divide by a variable regardless of its sign (providing it's not 0) or cross-multiply for that matter.