+0  
 
0
562
2
avatar

Consider the system

−8x − 12y = 36
5y − 6x = 41


Find x.

Guest Jun 3, 2014

Best Answer 

 #2
avatar+20025 
+5

(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}

heureka  Jun 4, 2014
 #1
avatar+90001 
+5

−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......

CPhill  Jun 3, 2014
 #2
avatar+20025 
+5
Best Answer

(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}

heureka  Jun 4, 2014

13 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.