Find x and y for (y/2)-x=2 and 6x-(3y/2)=3 using gaussian elimination
Find x and y for (y/2)-x=2 and 6x-(3y/2)=3 using gaussian elimination
$\displaystyle \frac{y}{2} - x = 2$
$\displaystyle 6x - \frac{3y}{2} = 3$.
$\displaystyle 3R_1 \to R_1$
$\displaystyle -3x + \frac{3y}{2} = 6$
$\displaystyle 6x - \frac{3y}{2} = 3$
$\displaystyle R_2 + R_1 \to R_2$
$\displaystyle -3x + \frac{3y}{2} = 6$
$\displaystyle 3x = 9$.
Solve for $\displaystyle x$:
$\displaystyle x = 3$.
Back substitute and solve for $\displaystyle y$:
$\displaystyle -3(3) + \frac{3y}{2} = 6$
$\displaystyle -9 + \frac{3y}{2} = 6$
$\displaystyle \frac{3y}{2} = 15$
$\displaystyle 3y = 30$
$\displaystyle y = 10$.
So the solution is $\displaystyle (x, y) = (3, 10)$.
Then you should have said that.
Rewrite the system as
$\displaystyle \left[\begin{matrix}-1 & \frac{1}{2} \\ 6 & -\frac{3}{2}\end{matrix}\right]\left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}2 \\ 3\end{matrix}\right]$.
Then follow the exact same process I said above.
Hello, brumby_3!
Solve by Gaussian elimination: . $\displaystyle \begin{array}{cccc}\dfrac{y}{2}-x &=& 2 & [1] \\ \\[-3mm] 6x-\dfrac{3y}{2} &=& 3 & [2]\end{array}$
We have: .$\displaystyle \left|\begin{array}{cc|c} \text{-}1 & \frac{1}{2} & 2 \\ \\[-4mm] 6 & \text{-}\frac{3}{2} & 3 \end{array}\right| $
. $\displaystyle \begin{array}{c} \\ R_2+6R_1\end{array}\left|\begin{array}{cc|c}
\text{-}1 & \frac{1}{2} & 2 \\ \\[-4mm] 0 & \frac{3}{2} & 15 \end{array}\right|$
. . . . $\displaystyle \begin{array}{c} \\ \frac{2}{3}R_2\end{array} \left|\begin{array}{cc|c}
\text{-}1 & \frac{1}{2} & 2 \\ 0 & 1 & 10 \end{array}\right| $
$\displaystyle \begin{array}{c}R_1-\frac{1}{2}R_2 \\ \\ \end{array}
\left|\begin{array}{cc|c} \text{-}1 & 0 & \text{-}3 \\ 0 & 1 & 10 \end{array}\right|$
. . .$\displaystyle \begin{array}{c}\text{-}1\!\cdot\! R_1 \\ \\ \end{array}
\left| \begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & 10 \end{array}\right| $
Therefore: .$\displaystyle x \:=\:3,\;y\:=\:10$