Co-ordinate Geometry: Equation of the line Find the equation of the line that passes through the point of intersection of the lines
$3 + 2 − 1 = 0$ and $2 −  + 7 = 0$ and is perpendicular to the line $4 + 4 = 7$.  
Solve for x:
$3 + 2 − 1 = 0$
$2 −  + 7 = 0$ (multiply by $2$)  
$3 + \not{2y} − 1 = 0$
$4 − \not{2} + 14 = 0$ 
$7x + 13 = 0$
$x =\frac{-13}{7}$.
Sub $x =\frac{-13}{7}$  into $3 + 2 − 1 = 0$
To get $3\left(\frac{-13}{7}\right) + 2 − 1 = 0$
Which reduces to $y=\frac{23}{7}$.
Line is perpendicular to $4 + 4 = 7$ so $m$ must be $1$.
Equation of line formula: $y-y_1=m(x-x_1)$.
$y-\frac{23}{7}=1(x-\frac{-13}{7})$
$y-1x-\frac{36}{7}=0$ $($Multiply by $7$$)$
$7y-7x-36=0$.  
Been at this for about 2 hours now so I'd appreciate any help or guidance, thank you.

The point of intersection is indeed $\left(-\dfrac{13}7,\dfrac{23}7\right)$. A line perpendicular to $4x+4y=7$ has its linear coefficients proportional to $(1,-1)$ (because $4\cdot1-4\cdot1=0$), hence
$$\left(x+\dfrac{13}7\right)-\left(y-\dfrac{23}7\right)=0$$
or
$$7x-7y+36=0.$$
