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)
$\begin{cases}
a^2+bc=a\\
(a+d)b=b\\
(a+d)c=c\\
bc+d^2=d
\end{cases}$

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.
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