The graphs of $x+y=10$ and $x-y=2$ are straight lines in the $x,y$ plane. By it's interesecton, we mean the point $P=(x,y)$ where the equations $x+y=10$ and $x-y=2$ are simultaneously satisfied.
What we want in this case is that $x+y-10=0$ and that $x-y-2=0$. This means, we want that
$$x+y-10=x-y-2$$
$$y-10=-y-2$$
$$2y=8$$
$$y=4$$
By substituting $y=4$ in any of our original equations we get that $x=6$. In general, the system $$S:\begin{cases} ax+by+c=0 \cr dx+ey+f=0 \end{cases}$$
has a unique solution $P=(X,Y)$ if and only if $$\frac{a}{b} - \frac{d}{e} \ne 0$$ which simply states that the straight lines aren't parallel: The slope of the first line is $-a/b$ and that if the other is $-d/e$ so we want
$$\eqalign{ & - \frac{a}{b} - \left( { - \frac{d}{e}} \right) \ne 0 \cr & - \frac{a}{b} + \frac{d}{e} \ne 0 \cr & \frac{a}{b} - \frac{d}{e} \ne 0 \cr} $$
You'll understand this from an algebraic point of view, rather than from a geometric one when you read about determinants. If $$\frac{a}{b} - \frac{d}{e}=0$$
Then $$ \left(\begin{matrix} a & b \\ d & e \end{matrix}\right)$$
has no inverse, so the matrix equation
$$ \left(\begin{matrix} a & b \\ d & e \end{matrix}\right)\left(\begin{matrix} x \\ y \end{matrix}\right)= \left(\begin{matrix} c \\ f \end{matrix}\right) $$
has no solution.