Chapter 7.4, Problem 80E

To determine

whether the two system of linear equation yield same solution and if so find the solution of system of linear equations.

Answer to Problem 80E

Two system of linear equations yield the same solution and the solution of the two system of linear equation is $x=-3,y=6,z=-4$

Explanation of Solution

Given information: Two system of linear equations

(a)

  $x+3y-z=19$ Equation 1

  $y+6z=-18$ Equation 2

  $z=-4$ Equation 3

(b)

  $x-y+3z=-21$ Equation 1

  $y-2z=14$ Equation 2

  $z=-4$ Equation 3

Concept used:Gaussian Elimination is the process of using valid row operations on a matrix until it is in reduced row echelon form

Calculation:

The solution of the first system is obtained as:

(a)

  $x+3y-z=19$ Equation 1

  $y+6z=-18$ Equation 2

  $z=-4$ Equation 3

From equation 3, the value of $z$ is known. To solve for y , substitute $z=-4$ into equation 2 to obtain,

  $\begin{array}{l}y+6\left(-4\right)=-18\\ y=-18+24\\ y=6\end{array}$

Finally, substitute $y=6$ and $z=-4$ into equation 1 to obtain

  $\begin{array}{l}x+3\left(6\right)-\left(-4\right)=19\\ x+18+4=19\\ x=19-22\\ x=-3\end{array}$

The solution of the first system is $x=-3,y=6,z=-4$

Now, check solution of second system,

(b) $x-y+3z=-21$ Equation 1

  $y-2z=14$ Equation 2

  $z=-4$ Equation 3

From equation 3the value of $z$ is known. To solve for y , substitute $z=-4$ into equation 2 to obtain

  $\begin{array}{l}y-2\left(-4\right)=14\\ y=14-8\\ y=6\end{array}$

Finally, substitute $y=6$ and $z=-4$ into equation 1 to obtain

  $\begin{array}{l}x-\left(6\right)+3\left(-4\right)=-21\\ x=-21+6+12\\ x=-3\end{array}$

The solution of a is $x=-3,y=6,z=-4$

Therefore, two system of linear equations yield the same solution