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Question Number 105327 by Anindita last updated on 27/Jul/20

$$\mathrm{If}\:\mathrm{x}\:=\:\frac{\mathrm{11}}{\mathrm{8}}\:\mathrm{and}\:\mathrm{y}\:=\:\frac{\mathrm{66}}{\mathrm{9}}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is} \\$$$$\mathrm{the}\:\mathrm{answer}\:\mathrm{of}\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}−\mathrm{y}}\:?\: \\$$

Commented by Aziztisffola last updated on 27/Jul/20

$$\mathrm{You}\:\mathrm{can}\:\mathrm{try}\:\mathrm{to}\:\mathrm{do}\:\mathrm{it},\:\mathrm{we}\:\mathrm{are}\:\mathrm{here}\:\mathrm{to} \\$$$$\mathrm{correct}\:\mathrm{and}\:\mathrm{to}\:\mathrm{help}\:\mathrm{you}. \\$$

Commented by nimnim last updated on 28/Jul/20

$$\:\:\:\:\frac{{x}}{{y}}=\frac{\mathrm{11}/\mathrm{8}}{\mathrm{66}/\mathrm{9}}=\frac{\mathrm{3}}{\mathrm{4}} \\$$$$\:\:\:\:{by}\:{componendo}\:\&\:{dividendo} \\$$$$\:\:\:\:\frac{{x}+{y}}{{x}−{y}}=\frac{\mathrm{3}+\mathrm{4}}{\mathrm{3}−\mathrm{4}}=−\mathrm{7} \\$$

Commented by Rasheed.Sindhi last updated on 29/Jul/20

$$\mathcal{N}{ice}\:\mathcal{W}{ay}! \\$$

Answered by 1549442205PVT last updated on 28/Jul/20

$$\mathrm{y}=\mathrm{kx}\Rightarrow\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}−\mathrm{y}}=\frac{\left(\mathrm{k}+\mathrm{1}\right)\mathrm{x}}{\left(\mathrm{1}−\mathrm{k}\right)\mathrm{x}}=\frac{\mathrm{k}+\mathrm{1}}{\mathrm{1}−\mathrm{k}}.\mathrm{Since} \\$$$$\mathrm{k}=\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{66}}{\mathrm{9}}:\frac{\mathrm{11}}{\mathrm{8}}=\frac{\mathrm{48}}{\mathrm{9}},\mathrm{so} \\$$$$\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}−\mathrm{y}}=\frac{\frac{\mathrm{48}}{\mathrm{9}}+\mathrm{1}}{\mathrm{1}−\frac{\mathrm{48}}{\mathrm{9}}}=\frac{\mathrm{57}}{−\mathrm{39}}=\frac{−\mathrm{19}}{\mathrm{13}} \\$$

Commented by \$@y@m last updated on 28/Jul/20

$${No}\:{need}\:{to}\:{insert}\:{new}\:{variable}\:{k}. \\$$$${Just}\:{dividing}\:{N}^{{r}} \:\&{D}^{{r}} \:{by}\:{x}\:{is}\:{enough}. \\$$

Answered by 1549442205PVT last updated on 28/Jul/20

$$\mathrm{999}\frac{\mathrm{494}}{\mathrm{495}}−\mathrm{99}\frac{\mathrm{98}}{\mathrm{99}}=\left(\mathrm{999}+\frac{\mathrm{494}}{\mathrm{495}}\right)−\left(\mathrm{99}+\frac{\mathrm{98}}{\mathrm{99}}\right) \\$$$$=\left(\mathrm{999}−\mathrm{99}\right)+\left(\frac{\mathrm{494}}{\mathrm{495}}−\frac{\mathrm{98}}{\mathrm{99}}\right)=\mathrm{900}+\frac{\mathrm{494}−\mathrm{490}}{\mathrm{495}} \\$$$$=\mathrm{900}\frac{\mathrm{4}}{\mathrm{495}} \\$$