Find the line $y= x+\lambda d$ for $(n-1)\times n$ linear system of equations 
Consider the equations
$$\sum_{j=1}^n a_{ij}x_j = b_i, i = 1,\cdots, n-1$$
or equivalently, $Ax = b$ with $A\in \mathbb{R}^{(n-1)\times n}, b\in
 \mathbb{R}^{n-1}$ and $x\in \mathbb{R}^n$ corresponding to $n-1$
  linearly independent hyperplanes. The intersection of these
  hyperplanes gives a line in $\mathbb{R}^n$ in the form
$$y = x + \lambda d$$
with $\lambda\in \mathbb{R}$ and $x,d\in\mathbb{R}^n$. Find how to
  chose $x$ and $d$.

I did an example of a system of two equations: 
$$2x+3y+5z = 1$$
$$4x+y+8z=3$$
The intersection turned out to be 
$$20x+42z=14$$ 
It's a line in $\mathbb{R}^3$. But I can't see it in the way the exercise asks be to. I understand that I will always get a line, but how do I convert this line equation in the form the exercise asks?
 A: I will work with the example you have given. First, note that your solution is not correct. You can see this easily, as it doesn't describe a line, but a plane(you have freedom in the $y$-direction).
The way of specifying a line as 
$$x+\lambda d$$
is called parametrization(as all solutions can be obtained using some value for the parameter $\lambda$).

Let's first look at the real solution set, the said line, and how to obtain a parameterized solution right away. Taking your system of equations
$$\begin{pmatrix}2 &3 &5\\4 &1 &8\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\3\end{pmatrix}$$
we may transform it using Gauss-Jordan elimination into
$$\begin{pmatrix}2 &3 &5\\0 &-5 &-2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}$$
We can now use back-substitution to determine the solutions, i.e. setting $z=c$ and determining the values of $x$ and $y$ in correspondence to $c$. You may then expand the solution vector, in the entries depending on $c$, to an equation $x+cy$ for vectors $x,y$ by pulling out the constants using vector addition and extracting $c$ using the laws of scalar multiplication.

Note, that you may convert the plane you've given as a supposed solution into a parametrized equation as well. For this, we just let $y=c,z=c'$ and can then determine $x$ in correspondence to $z$(and $y$ technically). Note, that as it is a plane, we've needed more parameters to describe the corresponding solutions.

In general, for a linear system of equations, for convenience given as $Ax=b$ for a corresponding coefficient matrix $A$ and solution vector $b$, the solution set $S(A,b)$, given a $x\in S(A,b)$, can be specified as $x+S(A,\mathbf0)=\{x+y\mid y\in S(A,\mathbf0)\}$. The solution set $S(A,\mathbf0)$ is commonly called the kernel of $A$. In other words, the solution set of an arbitrary system of equations $Ax=b$ can always be obtained by a linear shift of the solution set to the homogeneous system $Ax=\mathbf0$ by any already found solution. I don't want to get into this too much, but I've already written an answer to a question asking what the solution space of a linear system is. It has slightly more depth so you might want to check this out.
In your case, the solution set $S(A,\mathbf0)$ will always be a line(as the planes are linearly independent and you have $n-1$ of them, check this), i.e. a one-dimensional subspace of the underlying vector space. Now, as above, we can specify the general solution set $S(A,b)$ as $x+S(A,\mathbf0)$ for an arbitrary $x\in S(A,b)$. If $x\neq\mathbf0$(which happens iff $b\neq 0$, check this as well), the set $S(A,b)$ is called an affine line, a linear shift of a linear subspace, as it does not contain the null vector anymore.
As $S(A,\mathbf0)$ is always one dimensional, you may choose a one-element basis $\{d\}$ for $S(A,\mathbf0)$. We may, given this basis, write
$$S(A,\mathbf0)=\{\lambda d\mid\lambda\in\mathbb{R}\}$$
as every element of $S(A,\mathbf0)$ can be obtained as a linear combination of its basis and since it is one-dimensional, this just amounts to scalar multiplication. Thus
$$S(A,b)=x+S(A,\mathbf0)=\{x+\lambda d\mid\lambda\in\mathbb{R}\}$$
i.e. every element of $S(A,b)$ may be written as $x+\lambda d$. Thus, to answer your question more generally, $x$ and $d$ in the specification of your line of solutions always correspond to any already found solution and to a non-null solution to the equation $Ax=\mathbf{0}$ respectively. This solution will automatically be a basis as above, as the space is only $1$-dimensional.

For some further investigations, you may try to generalize these results,  e.g. can you always describe a solution set to a linear system of equations as such a parametric equation, etc.
