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)
$\begin{cases} a^2+bc=a\\ (a+d)b=b\\ (a+d)c=c\\ bc+d^2=d \end{cases}$
On
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.
1)
$\begin{cases} a^2+b^2=a\\ (a+d)b=b\\ b^2+d^2=d \end{cases}$
in equation $2$ divide both side by b $\begin{cases} a^2+b^2=a\\ a+d=1\\ b^2+d^2=d \end{cases}$
from eqn $2$ and $3$
$$a=1-d$$ $$b^2=d-d^2$$
put these value in eqn $1$
$$(1-d)^2+d-d^2=1-d\implies d=0$$ it will give $a=1,b=0$
2) $\begin{cases} a^2+bc=a\\ (a+d)b=b\\ (a+d)c=c\\ bc+d^2=d \end{cases}$
in eqn $2,3$ divide both side by b and c
$ a^2+bc=a\\ a+d=1\\ a+d=1\\ bc+d^2=d $
$ a^2+bc=a\\ a+d=1\\ bc+d^2=d $
from eqn $1$ and $2$ $$bc=a-a^2$$ $$d=1-a$$
put these value in $3$ $$a-a^2+(1-a)^2=1-a\implies a=0$$ that will give $d=1$ and either of b and c will 0. $$$$