Question Number 152596 by liberty last updated on 30/Aug/21
Answered by som(math1967) last updated on 30/Aug/21
![let ((3x+y)/(4y+3))=((4y+1)/9)=((11)/(3x+y))=k [∵a=b=c] 3x+y=k(4y+3) 4y+1=9k 11=k(3x+y) ∴3x+5y+12=k(3x+5y+12) ∴k=1[ ∵ x,y ∈N ∴3x+5y+12≠0] ∴4y+1=9 ∴y=2 11=3x+y x=3 x=3 ,y=2 ans](https://www.tinkutara.com/question/Q152598.png)
$$\boldsymbol{{let}}\:\frac{\mathrm{3}\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{4}\boldsymbol{{y}}+\mathrm{3}}=\frac{\mathrm{4}\boldsymbol{{y}}+\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{11}}{\mathrm{3}\boldsymbol{{x}}+\boldsymbol{{y}}}=\boldsymbol{{k}} \\ $$$$\left[\because\boldsymbol{{a}}=\boldsymbol{{b}}=\boldsymbol{{c}}\right] \\ $$$$\mathrm{3}\boldsymbol{{x}}+\boldsymbol{{y}}=\boldsymbol{{k}}\left(\mathrm{4}\boldsymbol{{y}}+\mathrm{3}\right) \\ $$$$\mathrm{4}\boldsymbol{{y}}+\mathrm{1}=\mathrm{9}\boldsymbol{{k}} \\ $$$$\mathrm{11}=\boldsymbol{{k}}\left(\mathrm{3}\boldsymbol{{x}}+\boldsymbol{{y}}\right) \\ $$$$\therefore\mathrm{3}\boldsymbol{{x}}+\mathrm{5}\boldsymbol{{y}}+\mathrm{12}=\boldsymbol{{k}}\left(\mathrm{3}\boldsymbol{{x}}+\mathrm{5}\boldsymbol{{y}}+\mathrm{12}\right) \\ $$$$\therefore\boldsymbol{{k}}=\mathrm{1}\left[\:\:\because\:\boldsymbol{{x}},\boldsymbol{{y}}\:\in\mathbb{N}\:\:\therefore\mathrm{3}\boldsymbol{{x}}+\mathrm{5}\boldsymbol{{y}}+\mathrm{12}\neq\mathrm{0}\right] \\ $$$$\therefore\mathrm{4}\boldsymbol{{y}}+\mathrm{1}=\mathrm{9}\:\therefore\boldsymbol{{y}}=\mathrm{2} \\ $$$$\mathrm{11}=\mathrm{3}\boldsymbol{{x}}+\boldsymbol{{y}} \\ $$$$\boldsymbol{{x}}=\mathrm{3} \\ $$$$\boldsymbol{{x}}=\mathrm{3}\:,\boldsymbol{{y}}=\mathrm{2}\:\boldsymbol{{ans}} \\ $$
Answered by Rasheed.Sindhi last updated on 30/Aug/21
$$\: \\ $$$$\begin{array}{|c|}{\frac{{a}}{{b}}=\frac{{c}}{{d}}=\frac{{e}}{{f}}=\frac{{a}+{c}+{e}}{{b}+{d}+{f}}}\\\hline\end{array} \\ $$$$\frac{\mathrm{3}{x}+{y}}{\mathrm{4}{y}+\mathrm{3}}=\frac{\mathrm{4}{y}+\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{11}}{\mathrm{3}{x}+{y}}=\frac{\mathrm{3}{x}+\mathrm{5}{y}+\mathrm{12}}{\mathrm{3}{x}+\mathrm{4}{y}+\mathrm{12}}=\mathrm{1} \\ $$$$ \\ $$$$\begin{cases}{\mathrm{4}{y}+\mathrm{1}=\mathrm{9}\Rightarrow{y}=\mathrm{2}}\\{\mathrm{3}{x}+{y}=\mathrm{11}\Rightarrow\mathrm{3}{x}+\mathrm{2}=\mathrm{11}\Rightarrow{x}=\mathrm{3}}\end{cases} \\ $$
Commented by liberty last updated on 13/Sep/21
$${great} \\ $$