[a-b b+c] = [8 1]
[3d+c 2a-4d] [7 6]
I know that since the matrices are equal:
a-b=8
b+c=1
3d+c=7
2a-4d=6
but not sure where to go from here
You have made a good start.
From the first equation $\displaystyle a = 8+b$
From the second $\displaystyle c = 1-b$
From the third (and second) $\displaystyle d = \frac{7-c}{3} \implies \frac{7-(1-b)}{3} = \frac{6+b}{3}$
Now looking at the 4th equation $\displaystyle 2a-4d=6$ we have new information for $\displaystyle a$ and $\displaystyle d$ so lets use it in the 4th.
$\displaystyle
2a-4d=6\implies 2(8+b)-4\left(\frac{6+b}{3}\right)=6
$
Now expand and solve for $\displaystyle b$
$\displaystyle
A\begin{bmatrix}a\\b\\c\\d\end{bmatrix}= \begin{bmatrix}a-b\\b+c\\3d+c\\2a-4d\end{bmatrix}=
a\begin{bmatrix}1\\0\\0\\2\end{bmatrix}+
b\begin{bmatrix}-1\\1\\0\\0\end{bmatrix}+
c\begin{bmatrix}0\\1\\1\\0\end{bmatrix}+
d\begin{bmatrix}0\\0\\3\\-4\end{bmatrix} \\
$
$\displaystyle
=\begin{bmatrix}
1 &-1 &0 &0 \\
0 &1 &1 &0 \\
0 &0 &1 &3 \\
2 &0 &0 &-4 \\
\end{bmatrix}
\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=
\begin{bmatrix}8\\1\\7\\6\end{bmatrix}~,~
\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=
A^{-1}\begin{bmatrix}8\\1\\7\\6\end{bmatrix}=
\begin{bmatrix}5\\-3\\4\\1\end{bmatrix}
$