Chapter 11.1, Problem 54AYU

To determine

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

Answer to Problem 54AYU

Solution:

  $x=3,y=-\frac{8}{3},z=\frac{1}{9}$

Explanation of Solution

Given:

  $x+4y-3z=-8$

  $3x-y+3z=12$

  $x+y+6z=1$

Calculation:

The system of equations is:

  $x+4y-3z=-8$

  $3x-y+3z=12$

  $x+y+6z=1$

Adding the first and second equations we get,

  $4x+3y=4$

  $3y=4-4x$

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

  $2(x+4y-3z)+(x+y+6z)=2(-8)+1$$2x+8y-6z+x+y+6z=-16+1$

  $3x+9y=-15$

  $x+3y=-5$

  $3y=-5-x$

We get two expressions for 3y and both must be equal.

Therefore we get,

  $3y=4-4x=-5-x$

  $4-4x=-5-x$

  $-4x+x=-5-4$

  $-3x=-9$

  $3x=9$

  $x=\frac{9}{3}$

  $x=3$

Substituting $x=3\mathrm{in}3y=4-4x$ we get,

  $3y=4-4(3)$

  $3y=4-12$

  $3y=-8$

  $y=-\frac{8}{3}$

Substituting $x=3,y=-\frac{2}{3}$ in the first given equation $x+4y-3z=-8$ we get

  $3+4\left(-\frac{8}{3}\right)-3z=-8$

  $3-\frac{32}{2}-3z=-8$

  $3z=3-\frac{32}{3}+8$

  $3z=11-\frac{32}{3}$

  $3z=\frac{1}{3}$

  $z=\frac{1}{9}$

Thus $x=3,y=-\frac{8}{3},z=\frac{1}{9}$ is the solution for the given system of equations.