Question

Solve the system of equations:x + 3y - z = -4 2x - y + 2z = 13 3x - 2y - z = -9

Answer 1

Answer:

The solution to the system of equations is

$\begin{gathered} x=\frac{179}{13} \\ \\ y=-\frac{279}{39} \\ \\ z=-\frac{48}{13} \end{gathered}$

Explanation:

Giving the system of equations:

$\begin{gathered} x+3y-z=-4\ldots\ldots\ldots\ldots\ldots\ldots..........\ldots\ldots\ldots\ldots.\ldots\text{.}\mathrm{}(1) \\ 2x-y+2z=13\ldots\ldots...\ldots\ldots\ldots\ldots..\ldots..\ldots\ldots\ldots\ldots\ldots.(2) \\ 3x-2y-z=-9\ldots\ldots\ldots.\ldots\ldots\ldots\ldots....\ldots\ldots.\ldots\ldots\ldots\text{.}\mathrm{}(3) \end{gathered}$

To solve this, we need to first of all eliminate one variable from any two of the equations.

Subtracting (2) from twice of (1), we have:

$5y-4z=-21\ldots\ldots\ldots\ldots\ldots.\ldots.\ldots..\ldots..\ldots\ldots.\ldots..\ldots\text{...}\mathrm{}(4)$

Subtracting (3) from 3 times (1), we have

$3y-5z=-3\ldots\ldots...\ldots\ldots..\ldots\ldots\ldots\ldots\ldots.\ldots\ldots\ldots\ldots\ldots..\ldots\ldots(5)$

From (4) and (5), we can solve for y and z.

Subtract 5 times (5) from 3 times (4)

$\begin{gathered} 13z=-48 \\ \\ z=-\frac{48}{13} \end{gathered}$

Using the value of z obtained in (5), we have

$\begin{gathered} 3y-5(-\frac{48}{13})=-3 \\ \\ 3y+\frac{240}{13}=-3 \\ \\ 3y=-3-\frac{240}{13} \\ \\ 3y=-\frac{279}{13} \\ \\ y=-\frac{279}{39} \end{gathered}$

Using the values obtained for y and z in (1), we have

$\begin{gathered} x+3(-\frac{279}{39})-(-\frac{48}{13})=-4 \\ \\ x-\frac{279}{13}+\frac{48}{13}=-4 \\ \\ x-\frac{231}{13}=-4 \\ \\ x=-4+\frac{231}{13} \\ \\ x=\frac{179}{13} \end{gathered}$