Chapter 11.1, Problem 53AYU

To determine

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

Answer to Problem 53AYU

$x=-3,y=\frac{1}{2},z=\text{ 1}$

Explanation of Solution

Given:

$x+\text{ 2}y-z=-3$

$2x-\text{ 4}y+z=-7$

$-2x+\text{ 2}y-\text{ 3}z=\text{ 4}$

Calculation:

The system of equations is:

$x+\text{ 2}y-z=-3$

$2x-\text{ 4}y+z=-7$

$-2x+\text{ 2}y-\text{ 3}z=\text{ 4}$

Adding the first and second equations we get,

$3x-2y=-10$

$2y=3x+10$

Multiplying the second equation by 3 and then adding to the third equation we get,

$3\left(2x-4y+z\right)+\left(-2x+2y-3z\right)=3\left(-7\right)+4$

$6x-12y+3z-2x+2y-3z=-21+4$

$4x-10y=-17$

$4x-5\left(2y\right)=-17$

Substituting $2y=3x+10$ in the above equation we get,

$4x-5\left(3x+10\right)=-17$

$4x-15x-50=-17$

$-11x=-17+50$

$-11x=33$

$x=\frac{33}{-11}$

$x=-3$

Substituting $x=-3$ in $2y=3x+10$ we get,

$2y=3\left(-3\right)+10$

$2y=-9+10$

$2y=1$

$y=\frac{1}{2}$

Substituting $x=-3$ and $y=\frac{1}{2}$ in the first given equation $x+\text{ 2}y-z=-3$ we get,

$-3+2\left(\frac{1}{2}\right)-z=-3$

$-3+1-z=-3$

$z=-2+3$

$z=1$

Thus $x=-3,y=\frac{1}{2},z=\text{ 1}$ is the solution for the given system of equations.