Chapter 11.1, Problem 13AYU

To determine

The solution of the given system of equations.

Answer to Problem 13AYU

Solution:

  $(x,y)=(1,-1,2)$ is the solution.

Explanation of Solution

Given:

  $3x+3y+2z=4$

  $x-y-z=0$

  $2y-3z=-8$

  $x=1,y=-1,z=2$

Calculation:

The system of equations is,

  $3x+3y+2z=4$

  $x-y-z=0$

  $2y-3z=-8$

From the third equation we get,

  $2y=3z-8$

  $y=\frac{3}{2}z-4$

Substituting this value of $y$ in the second equation we get.

  $x-\left(\frac{3}{2}z-4\right)-z=0$

  $x-\frac{3}{2}z+4-z=0$

  $x-\frac{5}{2}z+4=0$

  $x=\frac{5}{2}z-4$

Substituting these values of $x$ and $y$ in the first equation we get,

  $3\left(\frac{5}{2}z-4\right)+3\left(\frac{3}{2}z-4\right)+2z=4$

  $\frac{15}{2}z-12+\frac{9}{2}z-12+2z=4$

  $\frac{15}{2}z+\frac{9}{2}z+2z=4+12+12$

  $14z=28$

  $z=\frac{28}{14}$

  $z=2$

Substituting $z=2$ in the expressions for $x$ and $y$ we get,

  $x=\frac{5}{2}(2)-4=5-4=1$

  $y=\frac{3}{2}(2)-4=3-4=-1$

Thus the solution is $x=1,y=-1,z=2$

Therefore the given point (1,-1,2) satisfies the given system of equations and hence is the solution.