Explanation:
The set of all vectors in real numbers squared (R^2) that satisfy a linear equation like 3x + 4y = 1 does not necessarily form a subspace of R^2. In order for a set to be a subspace, it must satisfy three conditions:
1. The zero vector (0, 0) must be in the set.
2. The set must be closed under vector addition, which means that if you take any two vectors from the set and add them together, the result must also be in the set.
3. The set must be closed under scalar multiplication, which means that if you take any vector from the set and multiply it by any scalar (real number), the result must also be in the set.
In this case, the set of vectors that satisfy 3x + 4y = 1 does not meet the second condition. For example, if you take two vectors (x1, y1) and (x2, y2) from this set and add them together, the result may not satisfy 3x + 4y = 1. Therefore, the set is not closed under vector addition, and it does not form a subspace of R^2.
So, the correct option is:
Option 2: False