Question Number 73896 by liki last updated on 16/Nov/19
Commented by liki last updated on 16/Nov/19
$$…{Anyone}\:{to}\:{assist}\:{me}.. \\ $$
Answered by mr W last updated on 16/Nov/19
$$\mathrm{10}−{x}={y}−\mathrm{10} \\ $$$$\Rightarrow{x}+{y}−\mathrm{20}=\mathrm{0} \\ $$$$\frac{{x}}{\mathrm{1}}=\frac{{y}}{{x}} \\ $$$$\Rightarrow{y}={x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}−\mathrm{20}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{5}\right)\left({x}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{5}\:{or}\:\mathrm{4} \\ $$$$\Rightarrow{y}=\mathrm{25}\:{or}\:\mathrm{16} \\ $$
Commented by liki last updated on 16/Nov/19
$$..{thanks}\:{sir}.. \\ $$
Answered by $@ty@m123 last updated on 16/Nov/19
$$\mathrm{10}−{x}={y}−\mathrm{10} \\ $$$$\Rightarrow{x}+{y}=\mathrm{20}\:…\left(\mathrm{1}\right) \\ $$$$\frac{{x}}{\mathrm{1}}=\frac{{y}}{{x}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} ={y}\:\:…..\left(\mathrm{2}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:\&\left(\mathrm{2}\right) \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{20}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{80}}}{\mathrm{2}}\:=\frac{−\mathrm{1}\pm\mathrm{9}}{\mathrm{2}}\:=\mathrm{4},−\mathrm{5} \\ $$$$\Rightarrow{y}=\mathrm{16},\:\mathrm{25} \\ $$
Commented by liki last updated on 16/Nov/19
$$….{Thanks}\:{alot}\:{sir}.. \\ $$