If \frac{a}{b}=\frac{c}{d} Why does \frac{a+c}{b+d}=\frac{a}{b}=\frac{c}{d}?

feminizmiki 2022-01-06 Answered
If \(\displaystyle{\frac{{{a}}}{{{b}}}}={\frac{{{c}}}{{{d}}}}\) Why does \(\displaystyle{\frac{{{a}+{c}}}{{{b}+{d}}}}={\frac{{{a}}}{{{b}}}}={\frac{{{c}}}{{{d}}}}?\)

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sonorous9n
Answered 2022-01-07 Author has 4874 answers
Sketch: If you have \(\displaystyle{\frac{{{p}}}{{{q}}}}\) and \(\displaystyle{\frac{{\lambda{p}}}{{\lambda{q}}}}\), then
\(\displaystyle{\frac{{{p}+\lambda{p}}}{{{q}+\lambda{q}}}}={\frac{{{\left({1}+\lambda\right)}{p}}}{{{\left({1}+\lambda\right)}{q}}}}={\frac{{{p}}}{{{q}}}}\)
provided \(\displaystyle{1}+\lambda\ne{0}\)
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Donald Cheek
Answered 2022-01-08 Author has 314 answers
Consider \(\displaystyle{\frac{{{a}}}{{{b}}}}={\frac{{{k}{a}}}{{{k}{b}}}}\), Then
\(\displaystyle{\frac{{{a}+{k}{a}}}{{{b}+{k}{b}}}}={\frac{{{\left({k}+{1}\right)}{a}}}{{{\left({k}+{1}\right)}{b}}}}={\frac{{{a}}}{{{b}}}}\)
which is exactly what you noticed, but with \(\displaystyle{a}={1},{b}={2},{k}={2}\)
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Vasquez
Answered 2022-01-11 Author has 8850 answers

We know ad=bc, so ab+bc=ab+ad. If you factor out this equation you get b(a+c)=a(b+d) and then you get \(\frac{a}{b}=\frac{a+c}{b+d}\) . Similarly, you can prove the other

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