Consider the following linear system $$ \begin{cases} Y_{1}+Y_2 & =&2\alpha_1+\gamma_1+\gamma_2\\ Y_{3}+Y_4&=&2\alpha_2+2\gamma_1\\ Y_{5}+Y_6&=&2\alpha_3+\gamma_2+\gamma_3\\ Y_{7}+Y_8&=&2\alpha_4+2\gamma_3\\ Y_1+Y_3+Y_4&=&\alpha_1+2\alpha_2+3\gamma_1\\ Y_2+Y_6&=&\alpha_1+\alpha_3+2\gamma_2\\ Y_5+Y_7+Y_8&=&\alpha_3+2\alpha_4+3\gamma_3\\ \end{cases} $$ with unknowns $x\equiv (\alpha_1,\alpha_2,\alpha_3,\alpha_4, \gamma_1,\gamma_2, \gamma_3)$.
Question: Could you help me to show that this system has a unique solution with respect to $x$ if and only if one component among $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4, \gamma_1,\gamma_2, \gamma_3\}$ is set equal to a known value?
My attempt: I have tried (without success, could you help if you think this is the right path?) to show that one equation among those listed above is a linear combination of the others. Once that is shown, we are left with a linear system in $6$ equations and $7$ unknowns. How do we go from there?