Chapter 11.1, Problem 50AYU

To determine

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

Answer to Problem 50AYU

Solution:

Inconsistent, No solutions.

Explanation of Solution

Given:

  $3x-2y+2z=6$

  $7x-3y+2z=-1$

  $2x-3y+4z=0$

Calculation:

The system of equations is:

  $3x-2y+2z=6$

  $7x-3y+2z=-1$

  $2x-3y+4z=0$

Subtracting the first equation from the second equation we get,

  $(7x-3y+2z)-(3x-2y+2z)=-1-6$

  $7x-3y+2z-3x+2y-2z=-7$

  $4x-y=-7$

Multiplying the first equation by 2 and then subtracting from the third equation we get,

  $(2x-3y+4z)-2(3x-2y+2z)=0-2(6)$

  $2x-3y+4z-6x+4y-4z=-12$

  $-4x+y=-12$

  $4x-y=12$

Thus we have the following two equations:

  $4x-y=-7$

  $4x-y=12$

Both the above equations have the same $\text{LHS}(4x-y),$ but different RHS -7 and 12 .

This is not possible.

Hence the given system of equations is inconsistent and has no solutions.