Find x and y for (y/2)-x=2 and 6x-(3y/2)=3 using gaussian elimination

• March 26th 2010, 10:53 PM
brumby_3
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
• March 26th 2010, 11:02 PM
Prove It
Quote:

Originally Posted by brumby_3
As above!

$\frac{y}{2} - x = 2$

$6x - \frac{3y}{2} = 3$.

$3R_1 \to R_1$

$-3x + \frac{3y}{2} = 6$

$6x - \frac{3y}{2} = 3$

$R_2 + R_1 \to R_2$

$-3x + \frac{3y}{2} = 6$

$3x = 9$.

Solve for $x$:

$x = 3$.

Back substitute and solve for $y$:

$-3(3) + \frac{3y}{2} = 6$

$-9 + \frac{3y}{2} = 6$

$\frac{3y}{2} = 15$

$3y = 30$

$y = 10$.

So the solution is $(x, y) = (3, 10)$.
• March 26th 2010, 11:08 PM
brumby_3
Hi Prove it,
I believe that I'm supposed to show the answer by using the gaussian elimination by using matrixes?
• March 26th 2010, 11:35 PM
Prove It
Quote:

Originally Posted by brumby_3
Hi Prove it,
I believe that I'm supposed to show the answer by using the gaussian elimination by using matrixes?

Then you should have said that.

Rewrite the system as

$\left[\begin{matrix}-1 & \frac{1}{2} \\ 6 & -\frac{3}{2}\end{matrix}\right]\left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}2 \\ 3\end{matrix}\right]$.

Then follow the exact same process I said above.
• March 27th 2010, 08:56 AM
Soroban
Hello, brumby_3!

Quote:

Solve by Gaussian elimination: . $\begin{array}{cccc}\dfrac{y}{2}-x &=& 2 & [1] \\ \\[-3mm] 6x-\dfrac{3y}{2} &=& 3 & [2]\end{array}$

We have: . $\left|\begin{array}{cc|c} \text{-}1 & \frac{1}{2} & 2 \\ \\[-4mm] 6 & \text{-}\frac{3}{2} & 3 \end{array}\right|$

. $\begin{array}{c} \\ R_2+6R_1\end{array}\left|\begin{array}{cc|c}
\text{-}1 & \frac{1}{2} & 2 \\ \\[-4mm] 0 & \frac{3}{2} & 15 \end{array}\right|$

. . . . $\begin{array}{c} \\ \frac{2}{3}R_2\end{array} \left|\begin{array}{cc|c}
\text{-}1 & \frac{1}{2} & 2 \\ 0 & 1 & 10 \end{array}\right|$

$\begin{array}{c}R_1-\frac{1}{2}R_2 \\ \\ \end{array}
\left|\begin{array}{cc|c} \text{-}1 & 0 & \text{-}3 \\ 0 & 1 & 10 \end{array}\right|$

. . . $\begin{array}{c}\text{-}1\!\cdot\! R_1 \\ \\ \end{array}
\left| \begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & 10 \end{array}\right|$

Therefore: . $x \:=\:3,\;y\:=\:10$