<s>[INST] Solving For Variables In Simultaneous Equations I'm doing some work in linear algebra and these came up and I realized I don't know how to solve them as they have quadratics in them. I'm sure I've done this before but if someone could give me a crash course on how to find the values of the variables, it would be most appreciated. For the following questions, let $a,b,c,d \in \mathbb{R}$
1)
$\begin{cases}
a^2+b^2=a\\
(a+d)b=b\\
b^2+d^2=d
\end{cases}$
2)
$\begin{cases}
a^2+bc=a\\
(a+d)b=b\\
(a+d)c=c\\
bc+d^2=d
\end{cases}$
 [/INST] I tried to solve 1) in your question.
$$\begin{cases}
a^2+b^2=a ----(1)\\
(a+d)b=b  ----(2)\\
b^2+d^2=d ----(3)
\end{cases}$$
when $b=0$, the equations (1) and (3) boil down to $a^2=a$ and $d^2=d$. In this case, we have the solutions below
$$\{a=0,b=0,d=0\},\\
\{a=0,b=0,d=1\},\\
\{a=1,b=0,d=0\},\\
\{a=1,b=0,d=1 \}.$$

If $b\ne0$, then equations (1), (2) and (3) becomes
\begin{cases}
a^2-a +b^2=0 ----(4)\\
a+d =1  -------(5)\\
d^2-d +b^2= 0----(6)
\end{cases}
From (4),(5)and (6), we can see $a$ and $d$ are the solutions of the equation $x^2-x+b^2=0$, where $b$ can be any real number. So in this case, the solutions could be
$$\left\{a=\frac{1}{2} \left(1-\sqrt{1-4 b^2}\right),d=\frac{1}{2} \left(\sqrt{1-4 b^2}+1\right)\right\},\\
\left\{a=\frac{1}{2} \left(\sqrt{1-4 b^2}+1\right),d=\frac{1}{2} \left(1-\sqrt{1-4 b^2}\right)\right\}.
$$
Hint: your question 2) can be solved in a similar way.
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