Chapter 11.1, Problem 48AYU

To determine

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

Answer to Problem 48AYU

Solution:

Consistent, infinitely many solutions.

Explanation of Solution

Given:

  $2x-3y-z=0$

  $3x+2y+2z=2$

  $x+5y+3z=2$

Calculation:

The system of equations is:

  $2x-3y-z=0$

  $3x+2y+2z=2$

  $x+5y+3z=2$

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

  $2(2x-3y-z)+(3x+2y+2z)=0+2$

  $4x-6y-2z+3x+2y+2z=2$

  $7x-4y=2$

  $4y=7x-2$$y=\frac{1}{4}(7x-2)$

Adding all the three given equations we get,

  $6x+4y+4z=4$

Substituting $4y=7x-2$ in the above expression we get,

  $6x+(7x-2)+4z=4$

  $6x+7x-2+4z=4$

  $13x-2+4z=4$

  $4z=4+2-13x$

  $4z=6-13x$

  $z=\frac{1}{4}(6-13x)$

Thus the solution set to the given system of equations is:

  $\{\left(x,\frac{1}{4}(7x-2),\frac{1}{4}(6-13x)\right):x$ is real $\}$ since $x$ can be any real number the number of solutions to this system of equations is infinite.

Hence the given system of equations is consistent and has infinite many solutions.