Cho hệ pt : (a+1)x-y=3 và ax+y=a . Xác định giá trị của a để hệ ptrinh có nghiệm duy nhất thỏa mãn đk x+y>0

$\begin{array}{l} \left\{ \begin{array}{l} \left( {a + 1} \right)x - y = 3\\ ax + y = a \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} y = a - ax\,\,\,\,\,\,\,\,\,\left( 1 \right)\\ \left( {2a + 1} \right)x = a + 3\,\,\,\,\,\,\left( 2 \right) \end{array} \right.\\ Hpt\,\,co\,\,nghiem\,\,duy\,\,nhat\,\,\, \Leftrightarrow \left( 2 \right)\,\,\,co\,\,nghiem\,\,duy\,\,nhat\\ \Leftrightarrow 2a + 1 \ne 0 \Leftrightarrow a \ne - \frac{1}{2}.\\ \Rightarrow x = \frac{{a + 3}}{{2a + 1}}.\\ \Rightarrow y = a - ax = a - \frac{{{a^2} + 3a}}{{2a + 1}} = \frac{{2{a^2} + a - {a^2} - 3a}}{{2a + 1}} = \frac{{{a^2} - 2a}}{{2a + 1}}.\\ \Rightarrow Hpt\,\,co\,\,nghiem\,\,duy\,\,nhat\,\,\left( {x;\,\,y} \right) = \left( {\frac{{a + 3}}{{2a + 1}};\,\,\frac{{{a^2} - 2a}}{{2a + 1}}} \right).\\ \Rightarrow Theo\,\,de\,\,bai\,\,ta\,\,co:\,\,x + y > 0\\ \Rightarrow \frac{{a + 3}}{{2a + 1}} + \frac{{{a^2} - 2a}}{{2a + 1}} > 0\\ \Leftrightarrow \frac{{{a^2} - 2a + a + 3}}{{2a + 1}} > 0\\ \Leftrightarrow \frac{{{a^2} - a + 3}}{{2a + 1}} > 0\\ \Leftrightarrow 2a + 1 > 0\\ \Leftrightarrow a > - \frac{1}{2}.\\ Vay\,\,a > - \frac{1}{2}\,\,thoa\,\,man\,\,bai\,\,toan. \end{array}$