Cannot have less than infinitely many solutions but more than one for a particular $b$ in $Ax=b$ $$Ax=b$$
It is not possible to have more than one but less than infinitely many solutions for a particular $b$; if both $x$ and $y$ are solutions then:
$$z=\alpha x+(1-\alpha)y$$
is also a solution for any real $\alpha$.

Can anyone explain this? How does the right side simplify to $z$? And intuitively, why is the $(1-\alpha)$ needed as it seems arbitrary to me? It's not like an inverse or anything.
 A: If you have one solution to $ Ax = b $, say, $ v $, then adding any solution of $ Ax = 0 $ to that vector produces another solution. Since the solutions of $ Ax = 0 $ form a vector space, the vector space is either trivial (zero dimensional) or has infinitely many elements (assuming that the base field is infinite, such as $ \mathbb{R} $ or $\mathbb{Q}$).
A: Say $W = \{x \in V \mid Ax=b\}$ is the set of all solutions of your equation.
Let $x, y \in W$ be two different solutions for the given equation, i.e. $Ax = b,~ Ay = b,~ x \neq 0$. Now choose an $\alpha \in \mathbb{R}$ and consider the vector $\alpha x + (1-\alpha)y$, let's call it $z$.
Then $Az = \alpha Ax+(1-\alpha)Ay = \alpha b + (1-\alpha) b = b$,
so $z$ is also a solution of the given equation, it follows $z \in W$.
Now let's check that changing $\alpha$ will also change $z$: We can rewrite $z$ as $y + \alpha(x-y)$, and since $x \neq y$, we have $x - y \neq 0.$ Thus changing $\alpha$ will indeed give us a different $z$.
This means that for any real $\alpha$ there is a different solution to the equation, hence there are infinitely many.
We can formally write this as an injective function $f : \mathbb{R} \to W$ with $f(\alpha) = y + \alpha(x-y)$. Since $f$ is injective, $W$ must contain at least as many elements as $\mathbb{R}$, so infinitely many.
A: Important to note that this has an implicit assumption that the underlying fields is infinite. 
Simply, if $x_0$ and $x_1$ are both distinct solutions to $Ax = b$ then:
$$\frac{1}{2}x_0 + \frac{1}{2}x_1$$
Must also be a solution (and distinct) since:
$$A(\frac{1}{2}x_0 + \frac{1}{2}x_1) = \frac{1}{2}Ax_0 + \frac{1}{2}Ax_1 = b$$
From this we can derive infinitely many solutions by always using the linearity of $A$ and $\frac{1}{2}$
A: Suppose that $x_0$ is a solution, i.e., $Ax_0=b.$ 
Now let $x$ be another solution in $S(A,b)$. This gives $A(x-x_0)=Ax-Ax_0=b-b=0.$ Thus $(x-x_0)\in S(A,0).$ We can then write $x=x_0+(x-x_0) \in \{x_0\}+S(A,0).$ So $S(A,b) \subset \{x_0\}+S(A,0).$
Next suppose that $y\in S(A,0).$ Now $A(y+x_0)=Ay+Ax_0=b+0=b.$ So $(y+x_0)\in S(A,b).$ Thus $\{x_0\} + S(A,0) \subset S(A,b).$
Putting these together gives $S(A,b)=\{x_0\}+S(A,0).$
So adding any other solution in $S(A,0)$ gives you another solution.
