Chapter 11.1, Problem 48AYU

To determine

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

Answer to Problem 48AYU

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,

$2\left(2x-3y-z\right)+\left(3x+2y+2z\right)=0+2$

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

$7x-4y=2$

$4y=7x-2$

$y=\frac{1}{4}\left(7x-2\right)$

Adding all the three given equations we get,

$6x+4y+4z=4$

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

$6x+\left(7x-2\right)+4z=4$

$6x+7x-2+4z=4$

$13x-2+4z=4$

$4z=4+2-13x$

$4z=6-13x$

$z=\frac{1}{4}\left(6-13x\right)$

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

$\left\{\left(x,\frac{1}{4}\left(7x-2\right),\frac{1}{4}\left(6-13x\right)\right):x\text{ is real}\right\}$.

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.