Quadratic Formula??

(1) -8x - 12y = 36  ->   8x + 12y = -36

(2)  5y -   6x = 41  ->  -6x +   5y =  41

$$\boxed{\begin{array}{rcrcr}
8x & + & 12y &=& -36 \\
-6x & + & 5y &=& 41
\end{array}}$$

\boxed{\begin{array}{rcrcr}
 8x & + & 12y &=& -36 \\
-6x & + & 5y &=&  41 
\end{array}}

 $$\\x=
\frac{\begin{vmatrix}
-36 & 12 \\
41 & 5
\end{vmatrix}}{\begin{vmatrix}
8& 12 \\
-6 & 5
\end{vmatrix}}
=\frac{-36*5-41*12}{8*5-(-6)*12}
=\frac{-180-492}{40+72}
=\frac{-672}{112}=-6\\
\boxed{x=-6}
\\y=
\frac{\begin{vmatrix}
8 & -36 \\
-6 & 41
\end{vmatrix}}{\begin{vmatrix}
8& 12 \\
-6 & 5
\end{vmatrix}}
=\frac{8*41-(-6)(-36)}{8*5-(-6)*12}
=\frac{328-216}{40+72}
=\frac{112}{112}=1\\
\boxed{y=1}$$

 \\x=
\frac{\begin{vmatrix}
  -36 & 12 \\
  41 & 5  
\end{vmatrix}}{\begin{vmatrix} 
  8& 12 \\
  -6 & 5  
\end{vmatrix}}
=\frac{-36*5-41*12}{8*5-(-6)*12}
=\frac{-180-492}{40+72}
=\frac{-672}{112}=-6\\
\boxed{x=-6}
\\y=
\frac{\begin{vmatrix}
  8 & -36 \\
  -6 & 41  
\end{vmatrix}}{\begin{vmatrix}
  8& 12 \\
  -6 & 5  
\end{vmatrix}}
=\frac{8*41-(-6)(-36)}{8*5-(-6)*12}
=\frac{328-216}{40+72}
=\frac{112}{112}=1\\
\boxed{y=1}

−8x − 12y = 36
-6x +  5y  = 41         We can use the elimination method to solve this

We'll eliminate "x"   by multiplying the top equation by -6 on both sides and by multiplying the bottom equation by 8 on both sides   .....this gives us

48x + 72y = -216

-48x + 40y = 328       Now....add the equations together

112y = 112                So...it's clear that y = 1

To find "x,"  substitute 1 for y in any of the equations....I'll use -6x + 5y = 41

-6x + 5(1) = 41

-6x + 5 = 41        subtract 5 ftom both sides

-6x = 36             divide by -6 on both sides

x = -6

So....x = -6 and y = 1......you should verify that these "work" in the other equations...I think you will find that they do......