Finding the value of $a$ for which this $2\times2$ system does not have a unique solution The system of linear equations below has a unique solution for all but one value of $a$ : 
$\begin{align} 4 x - 4 y = 8 \\ 34 x + ay = 68 \\ \end{align}$
What is this exceptional value for $a$?
Not even sure where to begin on this. Thanks in advance!
 A: A = -34
Having A as -34 makes your 2nd equation
34x - 34y = 68
or
x - y = 2
Your 1st equation is
4x - 4y = 8
or
x - y = 2
This makes these equations same.. hence they will overlap and will have infinite solutions
A: Hint: $\text{det}A = 0$ with $A = \begin{bmatrix} 4 & -4\\ 34 & a \\ \end{bmatrix}$,
and generally speaking, if we can express a system of $n$ linear equations in $n$ variables in the form $AX = B$, then the system will not have a unique solution if $A$ is not invertible, and this means $\text{det}A = 0$.
A: By gaussian elimination we have
$$\begin{pmatrix}4 & -4 & |& 8 \\ 34 & a & | & 68\end{pmatrix}\sim\begin{pmatrix}4 & -4 & | & 8 \\ 0 & a+34 & | & 0\end{pmatrix}$$
If $a+34\neq 0$, then the system has only one solution. For $a+34=0$ there are infinite many solutions.
A: Rewrite the first equation to:
$$4x-4y=8\quad\Longrightarrow\quad x = 2+y.$$
Substitute this into the second equation:
$$34(2+y)+ay=68\quad\Longrightarrow\quad (34+a)y=0.$$
If $(34+a)\neq0$, then $y=0$ and $x=2$ is the unique solution. However, if $(34+a)=0$, $y$ can be anything, so that there is an infinite number of solutions. The latter happens when $=-34$.
A: As you might know, linear equation is the equation of locus for a line. The solution of two linear equations is essentially the point of intersection of the lines. For there to be no solutions, the lines must never intersect and be parallel. Mathematically this means that:$$\frac{4}{34}=\frac{-4}{a}$$$$a=-34$$
