Gilbert Strang, Introduction to Linear Algebra 5th Edition, Worked Example 3.1 B p.129 In this example $V_1$ is described as all the linear combinations of $v_1 = (1,1,0,0)$ and $v_2 = (1,1,1,0)$ and $v_3 = (1,1,1,1)$.
You are then asked to describe it as a combination of .... and the soutions of ...
The first part is obvious.
The solution for the second part is $V_1 = $ all solutions of $v_1 - v_2 = 0$
I would have thought is should be  $x_1*v_1 + x_2*v_2 + x_3*v_3$ for any vector x?.
Can someone please explain how he came to the solution given?
 A: Your idea is correct but you need to expand on it as follows: We want to know which vectors $\mathbf{b}=\begin{bmatrix}b_1\\b_2\\b_3\\b_4\end{bmatrix}$ can be expressed as a linear combination of the three given vectors?
So we want to solve the following system
$$\left[\begin{array}{lll|l}1&1&1&b_1\\1&1&1&b_2\\0&1&1&b_3\\0&0&1&b_4\end{array}\right].$$
When you row reduce then you get
$$\left[\begin{array}{lll|l}1&1&1&b_1\\0&1&1&b_3\\0&0&1&b_4\\0&0&0&b_2-b_1\end{array}\right].$$
For this to be consistent, we need $\color{magenta}{b_1-b_2=0}$. Thus
$$V_1=\left\{\begin{bmatrix}b_1\\b_2\\b_3\\b_4\end{bmatrix} \, \Bigg| \, b_1-b_2=0.\right\}$$
Note: If you look at the given vectors $\mathbf{v}_i$, they all satisfy this condition that their first two coordinates are same, so any linear combination of these will have the same property. By the method shown above, we have concluded that this is the only defining property we need to describe this space $V_1$.
Perhaps your usage of $v_1, v_2$ for vectors as opposed to $\mathbf{v}_1, \mathbf{v}_2$ is what might have lead to the confusion. Generally bold faced letters are used to denote vectors and italicized letters to denote the coordinates.
