Решить систему, пожааааалуйста помогите

$\left\{\begin{matrix} \frac{1}{x}&+ &\frac{1}{y} &= &\frac{3}{2} \\\\ \frac{1}{x^2}&+ &\frac{1}{y^2} &= &\frac{5}{4} \end{matrix}\right.\\\\\\ \frac{1}{x}+\frac{1}{y}=\frac{3}{2}\\\\ \frac{1}{x}=\frac{3y-2}{2y}\\\\ x=\frac{2y}{3y-2}\\\\\\ (\frac{3y-2}{2y})^2+\frac{1}{y}=\frac{5}{4}\\\\ \frac{9y^2-12y+4}{4y^2}+\frac{1}{y^2}=\frac{5}{4}\\\\ \frac{9y^2-12y+8}{4y^2}=\frac{5}{4}\ \ |(\cdot4)\\\\ 9y^2-12y+8=5y^2\\\\ 4y^2-12y+8=0 \ \ |(:4)\\\\$

$y^2-3y+2=0\\\\ D=9-8=1; \sqrt{D}=1\\\\ y_{1/2}=\frac{3\pm1}{2}\\\\ y_1=1;\\ y_2=2$

$x_1=\frac{2\cdot1}{3\cdot1-2}=2\\\\ x_2=\frac{2\cdot2}{3\cdot2-2}=1$

Ответ: $(1;2)(2;1)$