Question Number 28554 by naka3546 last updated on 27/Jan/18
$$\frac{\left({a}\:−\:{b}\right)}{\left({c}\:−\:{d}\right)}\:\:=\:\:\mathrm{3} \\ $$$$\frac{\left({a}\:−\:{c}\right)}{\left({b}\:−\:{d}\right)}\:\:=\:\:\mathrm{4} \\ $$$$\frac{\left({a}\:−\:{d}\right)}{\left({b}\:−\:{c}\right)}\:\:=\:\:? \\ $$$$ \\ $$
Answered by ajfour last updated on 27/Jan/18
$${a}−{b}=\mathrm{3}\left({c}−{d}\right)\:\:...\left({i}\right) \\ $$$${a}−{c}=\mathrm{4}\left({b}−{d}\right)\:\:\:\:\:...\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$$\Rightarrow\:{b}−{c}=\mathrm{3}\left({b}−{c}\right)+{b}−{d} \\ $$$${or}\:\:\mathrm{2}\left({b}−{c}\right)={d}−{b}\:\:\:\:...\left({I}\right) \\ $$$$\left({i}\right)−\mathrm{3}\left({ii}\right): \\ $$$$−\mathrm{2}{a}−{b}+\mathrm{3}{c}=\mathrm{3}{c}−\mathrm{3}{d}−\mathrm{12}{b}+\mathrm{12}{d} \\ $$$$\Rightarrow\:−\mathrm{2}\left({a}−{d}\right)=−\mathrm{11}\left({b}−{d}\right)\:...\left({II}\right) \\ $$$$\left({II}\right)\boldsymbol{\div}\left({I}\right): \\ $$$$\frac{{a}−{d}}{{b}−{c}}=−\mathrm{11}\:. \\ $$
Answered by mrW2 last updated on 27/Jan/18
$${a}−{b}=\mathrm{3}{c}−\mathrm{3}{d}\:\:\:...\left({i}\right) \\ $$$${a}−{c}=\mathrm{4}{b}−\mathrm{4}{d} \\ $$$${a}−\mathrm{4}{b}={c}−\mathrm{4}{d}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right) \\ $$$$\mathrm{3}{b}=\mathrm{2}{c}+{d} \\ $$$$\Rightarrow{b}=\frac{\mathrm{2}{c}+{d}}{\mathrm{3}} \\ $$$$\Rightarrow{a}={b}+\mathrm{3}{c}−\mathrm{3}{d}=\frac{\mathrm{2}{c}+{d}+\mathrm{9}{c}−\mathrm{9}{d}}{\mathrm{3}}=\frac{\mathrm{11}{c}−\mathrm{8}{d}}{\mathrm{3}} \\ $$$${a}−{d}=\frac{\mathrm{11}{c}−\mathrm{8}{d}−\mathrm{3}{d}}{\mathrm{3}}=\frac{\mathrm{11}\left({c}−{d}\right)}{\mathrm{3}} \\ $$$${b}−{c}=\frac{\mathrm{2}{c}+{d}−\mathrm{3}{c}}{\mathrm{3}}=−\frac{\left({c}−{d}\right)}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{a}−{d}}{{b}−{c}}=\frac{\frac{\mathrm{11}\left({c}−{d}\right)}{\mathrm{3}}}{−\frac{\left({c}−{d}\right)}{\mathrm{3}}}=−\mathrm{11} \\ $$