Chapter 11.1, Problem 51AYU

To determine

To find: The solution of the given system of equations.

Answer to Problem 51AYU

$x=1,y=3,z=-2$

Explanation of Solution

Given:

$x+y-z=6$

$3x-2y+z=-5$

$x+3y-2z=14$

Calculation:

The system of equations is:

$x+y-z=6$

$3x-2y+z=-5$

$x+3y-2z=14$

Adding the first and second equation we get,

$4x-y=1$

Multiplying the second equation by 2 and then adding to the third equation we get,

$2\left(3x-2y+z\right)+\left(x+3y-2z\right)=2\left(-5\right)+14$

$6x-4y+2z+x+3y-2z=-10+14$

$7x-y=4$

Thus we have the following two equations in $x$, $y$:

$7x-y=4$

$4x-y=1$

Subtracting we get,

$\left(7x-y\right)-\left(4x-y\right)=4-1$

$7x-y-4x+y=3$

$3x=3$

$x=\frac{3}{3}$

$x=1$

Substituting $x=1$ in the equation $4x-y=1$ we get,

$4\left(1\right)-y=1$

$4-y=1$

$y=4-1$

$y=3$

Now substituting $x=1$, $y=3$ in the first given equation $x+y-z=6$ we get,

$1+3-z=6$

$4-z=6$

$z=4-6$

$z=-2$

Thus $x=1,y=3,z=-2$ is the solution to the given system of equations.