# Simultaneous Equation

• December 1st 2009, 01:25 AM
s2951
Simultaneous Equation
x-3y=14 -(1)
4x+2y=0 -(2)

What is x & y?
thanks.
• December 1st 2009, 01:36 AM
Stroodle
Which part are you having trouble with?

From the first equation you can see that $x=14+3y$
So plug $14+3y$ into the second equation in the place of $x$ then solve that for $y$. Once you have the value of $y$ plug that back into either equation to get the value of $x$.
Let us know if you have trouble with this.
• December 1st 2009, 01:38 AM
mrmohamed
Quote:

Originally Posted by s2951
x-3y=14 -(1)
4x+2y=0 -(2)

What is x & y?
thanks.

hi
from (2) y = -2 x
put y = -2x in (1)
x + 6x = 14 7x = 14 x = 2 , y = -4
• December 1st 2009, 01:46 AM
s2951
Quote:

Originally Posted by mrmohamed
hi
from (2) y = -2 x
put y = -2x in (1)
x + 6x = 14 7x = 14 x = 2 , y = -4

x-3y=14 (1)
4x+2y=0 (2)

(1)
x=14+3y

(1) into (2)
4(14+3y)+2y=0
56+12y+2y=0
56+14y=0
14y=-56
y=-4

(2)
4x+2(-4)=0
4x+-8=0
4x=8
x=2

Got it, so i have to use Substitution Method. Can i use Elimination Method ?
• December 1st 2009, 01:54 AM
Bacterius
Wouldn't be useful here. Elimination method is only useful when the fact of adding or substracting together both equations gets rid of an unknown (like $-2x$ in the first one and $2x$ in the second). Here you wouldn't get anything useful [I never ever use the elimination method, oddly enough I prefer using substitution in all problems ...] but if you are asked to do the Elimination method ...
• December 1st 2009, 02:53 AM
Raoh
$\left\{\begin{matrix}
x-3y=14\\
4x+2y=0
\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}
x-3y=14\\
2x+y=0
\end{matrix}\right.$

Gauss Elimination,
$\Rightarrow \left\{\begin{matrix}
x-3y=14\\
14y=-56
\end{matrix}\right.$

Therefore,the solution is $[x=2,y=-4]$.