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I have this system of linear equations $$ \begin{cases} x_a+x_b=1 \\ ax_a+bx_b=x \end{cases} $$ I want to write this as $(a-x)x_a+(b-x)x_b=0$.

I tried to add eq. 2 to eq. 1: \begin{align} x_a+ax_a+x_b+bx_b&=1+x \\ x_a(1+a)+x_b(b+1)&=1+x \end{align} Stuck! If I instead subtracts eq. 2 from eq. 1 I get: \begin{align} x_a-ax_a+x_b-bx_b&=1-x \\ x_a(1-a)+x_b(1-b) &=1-x \end{align} Stuck! What have I missed?

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  • $\begingroup$ Multiply equation $1$ by $x$ first. Then subtract equation $2$. $\endgroup$ – Michael Burr Feb 13 '17 at 17:39
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Given any solution $(x_a,x_b)$ of the system

$$\cases{x_a+x_b=1\cr ax_a+bx_b=x}\tag{1}$$

we have :

$$(a-x)x_a+(b-x)x_b=ax_a+bx_b-x(x_a+x_b)=x-x\times 1$$

Hence :

$$(a-x)x_a+(b-x)x_b=0\tag{2}$$

But be aware that we only proved that $(1)\implies (2)$. The two conditions are not equivalent !

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  • $\begingroup$ Thanks! How can we prove $(2) \implies (1)$? $\endgroup$ – JDoeDoe Feb 13 '17 at 18:47
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You can simply do it as -

Multiply equation (1) by x and subtract it from equation (2).

$ax_a+bx_b - x(x_a+x_b)=x - x$

$ax_a+bx_b - x.x_a-x.x_b=0$

$ax_a - x.x_a+ bx_b - x.x_b=0$

$x_a(a-x)+ x_b(b - x)=0$

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