[tex]lg(2x-3y)=\frac{1}{2}(lgx+lgy)\\
lg(2x-3y)=\frac{1}{2}lg(xy)\\
lg(2x-3y)=lg((xy)^{\frac{1}{2}})\\
lg(2x-3y)=lg(\sqrt{xy})\rightarrow \sqrt{xy}=2x-3y\\\\
xy=(2x-3y)^2\\
4x^2-13xy+9y^2=0\\\\
\text{Scriem solutiile lui x ale ecuatiei}\\
\Delta = 169y^2-16\cdot9y^2=y^2(169-144)=25y^2\rightarrow \sqrt\Delta=5y\\
x_{1,2}=\frac{13\pm 5y}{8}\\
x_1=\frac{9}{4}y\\
x_2=y[/tex]
Daca x ar fi egal cu y, atunci nu s-ar mai respecta conditia de existenta 2x > 3y ==> Singura solutie este x = (9/4) * y
[tex]\frac{x}{y}= \frac{\frac{9}{4}y}{y} =\frac{9}{4}[/tex]
