Question Number 207576 by efronzo1 last updated on 19/May/24
Commented by A5T last updated on 19/May/24
$$\mathrm{4} \\ $$
Commented by mr W last updated on 19/May/24
$${when}\:{d}={c}:\:\frac{{a}}{{b}}=\frac{\mathrm{6}{x}}{\mathrm{2}{x}}=\mathrm{3} \\ $$$$\Rightarrow\frac{{a}}{{b}}+\frac{{c}}{{d}}=\mathrm{3}+\mathrm{1}=\mathrm{4} \\ $$
Answered by mr W last updated on 19/May/24
Commented by mr W last updated on 19/May/24
$${general}\:{case}: \\ $$$${let}\:\lambda=\frac{{y}}{{x}} \\ $$$$\frac{{k}}{{d}}=\frac{{nx}−{y}}{{mx}}=\frac{{n}−\lambda}{{m}} \\ $$$$\frac{{c}}{{k}}=\frac{{a}}{{b}}=\frac{{mx}+{nx}}{{mx}+{nx}−{y}}=\frac{{m}+{n}}{{m}+{n}−\lambda} \\ $$$$\frac{{k}}{{d}}×\frac{{c}}{{k}}=\frac{{c}}{{d}}=\left(\frac{{n}−\lambda}{{m}}\right)\frac{{a}}{{b}} \\ $$$$\Rightarrow\frac{{c}}{{d}}+\frac{{a}}{{b}}=\left(\frac{{m}+{n}−\lambda}{{m}}\right)\frac{{a}}{{b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{{m}+{n}−\lambda}{{m}}\right)×\left(\frac{{m}+{n}}{{m}+{n}−\lambda}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{m}+{n}}{{m}}=\frac{\mathrm{2}+\mathrm{6}}{\mathrm{2}}=\mathrm{4}\:\checkmark \\ $$
Commented by Tawa11 last updated on 21/Jun/24
$$\mathrm{weldone}\:\mathrm{sir} \\ $$