Matrix Equation, Solving for Variables. I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question:
$
\begin{align}
\begin{bmatrix}
a-b & b+c\\
3d+c & 2a-4d
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 & 1\\
7 & 6
\end{bmatrix}
\end{align}
$
What I have done so far is:
$
\begin{align}
\begin{bmatrix}
a-b-8 & b+c-1\\
3d+c-7 & 2a-4d-6
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}
\end{align}
$

And the solving for variables,
$$
a-b-8 = 0
$$
$$
a-b = 8
$$
$$
a = \frac{8}{-b}
$$

$$
b+c-1=0
$$
$$
b+c=1
$$
$$
b=\frac{1}{c}
$$

$$
3d+c-7=0
$$
$$
3d+c=7
$$
$$
3d=\frac{7}{c}
$$
$$
d=\frac{7}{3c}
$$

$$
2a-4d-6=0
$$
$$
2a-4d=6
$$
$$
\frac{16}{-2b}-\frac{28}{12c}=6
$$

Am I going about this correctly? Or am I just doing this completely incorrect?
 A: Error 1: $a-b=8$ is not the same as $a = \frac{8}{-b}$ but $a = 8 + b$
Error 2: $b+c=1$ is not the same as $b = \frac{1}{c}$ but $b = 1-c$
Error 3: $3d+c=7$ ... same as errors 1 and 2
Error 4: $2a-4d=6$ Incorrect substitution because of previous errors. Fix the previous errors and then subsititue to fix this error
A: How familiar are you with basic linear algebra? As @response has posted, you have numerous algebra mistakes. However, as for the method used to solve the problem, I think it's worth noting that you can easily transform this system to one more familiar looking and then use standard technique of Gaussian elimination. That is, rewrite the system as follows:
$
\begin{align}
\begin{bmatrix}
1 & -1 & 0 &  0\\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 3 \\
2 & 0 & 0 & -4 \\
\end{bmatrix}
\end{align}
$
$
\begin{align}
\begin{bmatrix}
a \\
b \\
c \\
d \\
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 \\
1\\
7 \\
6
\end{bmatrix}
\end{align}
$
And then solve.
A: The best way is follow the solution given by @Alex Wertheim. Let me show you a tedious substitution method to solve this. From $a_{11}$ we have $$b=a-8$$ (dividing by $a$ or $b$ is invalid as we may have zeros). Now from $a_{12}$ entry we have $$c=-b+1=-a+9,$$ and using $a_{21}$ $$d=\dfrac{-c+7}{3}=\dfrac{a-2}{3}.$$ Finally from the last entry $a_{22}$ we have $$a-2d=a-\dfrac23(a-2)=3.$$ Find $a$ from this simple linear equation and substitute back to find $b, c$ and $d.$ 
