Chapter 11.1, Problem 43AYU

To determine

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

Answer to Problem 43AYU

Solution:

  $x=2,y=-1,z=1$

Explanation of Solution

Given:

  $x-2y+3z=7$

  $2x+y+z=4$

  $-3x+2y-2z=-10$

Calculation:

The system of equations is:

  $x-2y+3z=7$

  $2x+y+z=4$

  $-3x+2y-2z=-10$

From the first equation we get,

  $x=7+2y-3z$

Substituting $x=7+2y-3z$ in the second equation we get,

  $2(7+2y-3z)+y+z=4$

  $14+4y-6z+y+z=4$

  $5y-5z=-10$

  $y-z=-2$

  $y=-2+z$

Substituting $y=-2+z$ in the expression $x=7+2y-3z$ weget,

  $x=7+2(-2+z)-3z$

  $x=7-4+2z-3z$

  $x=3-z$

Substituting $x=3-z$ and $y=-2+z$ in the third equation we get,

  $-3(3-z)+2(-2+z)-2z=-10$

  $-9+3z-4+2z-2z=-10$

  $3z=-10+9+4$

  $3z=3$

  $z=\frac{3}{2}$

  $z=1$

Substituting $z=1$ in the expression for $y=-2+z$ we get,

  $y=-2+1$

  $y=-1$

Substituting $z=1$ in the expression for $x=3-z$ we get,

  $x=3-1$

  $x=2$

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