Will any value of a free variable satisfy a system of equation? Say I have a reduced row echelon form matrix like this:
$$A=\begin{bmatrix}
1 & 0 & \frac{1}{2} & 0\\ 
0 & 1 & -\frac{1}{3} & 0\\ 
0 & 0 & 0 & 1
\end{bmatrix}$$
The number of unknowns is more than the number of known equations. So I can expect an infinite number of solutions.
And $Ax=b$ is like this:
$$
\begin{bmatrix}
1 & 0 & \frac{1}{2} & 0\\ 
0 & 1 & -\frac{1}{3} & 0\\ 
0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x_{1}\\ 
x_{2}\\ 
x_{3}\\ 
x_{4}
\end{bmatrix}
=
\begin{bmatrix}
b_{1}\\ 
b_{2}\\ 
b_{3}
\end{bmatrix}
$$
Then I can say that my $x$ is like this with $x_{3}$ being a free variable in the equation:
$$\begin{bmatrix}
x_{1}\\ 
x_{2}\\ 
x_{3}\\ 
x_{4}
\end{bmatrix}
=
x_{3}\begin{bmatrix}
\frac{1}{2}\\ 
\frac{1}{3}\\ 
1\\ 
0
\end{bmatrix}+
\begin{bmatrix}
b_{1}\\ 
b_{2}\\ 
0\\ 
b_{3}
\end{bmatrix}
$$
Now, if I let $\; \begin{bmatrix}
b_{1}\\ 
b_{2}\\ 
b_{3}
\end{bmatrix}=\begin{bmatrix}
5\\ 
2\\ 
7
\end{bmatrix}$, then...
$$\begin{bmatrix}
x_{1}\\ 
x_{2}\\ 
x_{3}\\ 
x_{4}
\end{bmatrix}
=
x_{3}\begin{bmatrix}
\frac{1}{2}\\ 
\frac{1}{3}\\ 
1\\ 
0
\end{bmatrix}+
\begin{bmatrix}
5\\ 
2\\ 
0\\ 
7
\end{bmatrix}
$$
At this stage, I can say that for any value that I put into the variable $x_{3}$, I would get an answer that is equals to $b$, right?
So assume I just randomly throw a value into $x_{3}=2$, then...
$$\begin{bmatrix}
x_{1}\\ 
x_{2}\\ 
x_{3}\\ 
x_{4}
\end{bmatrix}
=
\begin{bmatrix}
1\\ 
\frac{2}{3}\\ 
2\\ 
0
\end{bmatrix}+
\begin{bmatrix}
5\\ 
2\\ 
0\\ 
7
\end{bmatrix}=
\begin{bmatrix}
6\\ 
\frac{8}{3}\\ 
2\\ 
7
\end{bmatrix}
$$
From here, I need to tally if the equation really gets back my intended values of $b$, which is $\begin{bmatrix}
b_{1}\\ 
b_{2}\\ 
b_{3}
\end{bmatrix}=\begin{bmatrix}
5\\ 
2\\ 
7
\end{bmatrix}$. So,
$x_{4} = b_{3} =7$
This is correct.
$b_{2}=x_{2}-\frac{1}{3}x_{3}$
$b_{2}=\frac{8}{3}-\frac{2}{3}=2$
This is also correct.
BUT NOW,
$b_{1}=x_{1}+\frac{1}{2}x_{3}$
$b_{1}=6+1=7$
$b_{1}$ suppose to be 6 but somehow, why am I getting 7 instead? This is weird. Where did I go wrong?
Thanks for any help!

I think you've confused yourself somewhere.
You have a mapping from $\mathbb{R}^4 \to \mathbb{R}^3$. $x_3$ is free $x_4 = b_3$, the only stipulation is that:
$x_1/3 + x_2/2 = 8/3$.
The vector you've given: $[6, 8/3, 2, 7]^T$ does not satisfy the last condition.
Edit: Solve for $[x_1,x_2,x_3,x_4]^T$ given a vector $[b_1,b_2,b_3]^T$:
$x_4=b_3$
$x_1 + \dfrac{x_3}{2} = b_1$
$x_2 - \dfrac{x_3}{3} = b_2$
These all imply that 
$\dfrac{x_1}{3} + \dfrac{x_2}{2} = \dfrac{b_1}{3} + \dfrac{b_2}{2}$
Now you want the solution to the vector $[5,2,7]^T$, so
$\dfrac{x_1}{3} + \dfrac{x_2}{2} = \dfrac{5}{3} + 1 = \dfrac{8}{3}$.
The vector you said mapped to $\vec{b}$ was $[6, 8/3, 2, 7]$.
Your proposed $x_1 = 6$ and $x_2 = \dfrac{8}{3}$ do not satisfy this.
