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Question Number 173369 by AgniMath last updated on 10/Jul/22

Answered by mr W last updated on 10/Jul/22

$$\frac{{by}+{cz}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }=\frac{{cz}+{ax}}{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{{ax}+{by}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={k},\:{say} \\$$$${ax}+{by}={k}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\:\:\:\:...\left({i}\right) \\$$$${by}+{cz}={k}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\:\:\:\:\:...\left({ii}\right) \\$$$${cz}+{ax}={k}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\:\:\:...\left({iii}\right) \\$$$$\Sigma: \\$$$${ax}+{by}+{cz}={k}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\:\:\:...\left({iv}\right) \\$$$$\left({iv}\right)−\left({ii}\right): \\$$$$\Rightarrow{ax}={ka}^{\mathrm{2}} \:\Rightarrow\frac{{x}}{{a}}={k} \\$$$${similarly} \\$$$$\Rightarrow\frac{{y}}{{b}}={k},\:\frac{{z}}{{c}}={k} \\$$$$\Rightarrow\frac{{a}}{{x}}=\frac{{b}}{{b}}=\frac{{c}}{{z}}={k} \\$$

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