Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 161744 by HongKing last updated on 21/Dec/21

$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$$\frac{\mathrm{x}}{\mathrm{y}}+\frac{5}{\mathrm{x}}+\frac{\mathrm{y}-5}{5}=\frac{\mathrm{y}+\mathrm{x}}{\mathrm{y}+5}+\frac{5+\mathrm{y}}{5+\mathrm{x}}$$

Commented by Rasheed.Sindhi last updated on 22/Dec/21

$$Notuniquesolution.Because$$$$only\mathit{o}\mathit{n}\mathit{e}equationisgivenin\mathit{t}\mathit{w}\mathit{o}$$$$variables.$$

Answered by Rasheed.Sindhi last updated on 22/Dec/21

$$\mathrm{y}=5:$$$$\frac{\mathrm{x}}{5}+\frac{5}{\mathrm{x}}=\frac{5+\mathrm{x}}{10}+\frac{10}{5+\mathrm{x}}$$$$\frac{{\mathrm{x}}^2+5^2}{5\mathrm{x}}=\frac{{(5+\mathrm{x})}^2+{10}^2}{10(5+\mathrm{x})}$$$$\frac{{\mathrm{x}}^2+25}{\mathrm{x}}=\frac{25+10\mathrm{x}+{\mathrm{x}}^2+100}{2(5+\mathrm{x})}$$$$({\mathrm{x}}^2+25)(2\mathrm{x}+10)={\mathrm{x}}^3+{10\mathrm{x}}^2+125\mathrm{x}$$$${2\mathrm{x}}^3+{10\mathrm{x}}^2+50\mathrm{x}+250-{\mathrm{x}}^3-{10\mathrm{x}}^2-125\mathrm{x}$$$${\mathrm{x}}^3-75\mathrm{x}+250=0$$$$\mathrm{x}=5,-10$$$$(5,5),(-10,5)aretwoofmany$$$$solutions.$$

Commented by HongKing last updated on 25/Dec/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathrm{Sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com