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With $$x_1=a-ax_2-2x_3$$ can we eliminate $$x_1$$: $$x_2(3-a)-2x_3=2-a$$$$x_2(2-a^2)+4ax_3=-a^2$$ And then using $$x_3=\frac{x_2(3-a)-(2-a)}{2}$$ to eliminate $$x_3$$. Then you will get $$x_2(2-a^2)+2a(3-a)x_2-2a(2-a)=-a^2$$
With $$x_1=a-ax_2-2x_3$$ can we eliminate $$x_1$$: $$x_2(3-a)-2x_3=2-a$$$$x_2(2-a^2)+4ax_3=-a^2$$ And then using $$x_3=\frac{x_2(3-a)-(2-a)}{2}$$ to eliminate $$x_3$$.
With $$x_1=a-ax_2-2x_3$$ can we eliminate $$x_1$$: $$x_2(3-a)-2x_3=2-a$$$$x_2(2-a^2)+4ax_3=-a^2$$ And then using $$x_3=\frac{x_2(3-a)-(2-a)}{2}$$ to eliminate $$x_3$$. Then you will get $$x_2(2-a^2)+2a(3-a)x_2-2a(2-a)=-a^2$$
With $$x_1=a-ax_2-2x_3$$ can we eliminate $$x_1$$: $$x_2(3-a)-2x_3=2-a$$$$x_2(2-a^2)+4ax_3=-a^2$$ And then using $$x_3=\frac{x_2(3-a)-(2-a)}{2}$$ to eliminate $$x_3$$.
