Chapter 7.3, Problem 56E

Solution

To state: The solution for the system of equation using any method:

  $\begin{array}{c}1.25x+z=-2\\ y-5.5z=-2.75\\ 3x-1.5y=-6\end{array}$

The required solution is $\left(-2,0,0.5\right)$ .

Given information:

The given system is:

  $\begin{array}{c}1.25x+z=-2\\ y-5.5z=-2.75\\ 3x-1.5y=-6\end{array}$

Explanation:

Consider the given system:

  $\begin{array}{c}1.25x+z=-2\cdots \left(1\right)\\ y-5.5z=-2.75\cdots \left(2\right)\\ 3x-1.5y=-6\cdots \left(3\right)\end{array}$

Multiply the first equation by 5.5 and add it to the second equation to eliminate variable z ,

  $\left[1.25x+z=-2\cdots \left(1\right)\right]\times 5.5=6.875x+5.5z=-11$

  $\begin{array}{r}6.875x+5.5z=-11\\ \underset{¯}{y-5.5z=-2.75}\\ 6.875x+y=-13.75\end{array}$

Multiply the resultant equation to 1.5:

  $\begin{array}{l}\left[6.875x+y=-13.75\right]\times 1.5\\ =10.3125x+1.5y=-26.625\cdots \left(4\right)\end{array}$

Add equations (3) and (4),

  $\begin{array}{l}\underset{¯}{\begin{array}{c}10.3125x+1.5y=-20.625\\ 3x-1.5y=-6\end{array}}\\ 13.3125x=-26.625\\ x=-2\end{array}$

Substitute the value of x as -2 in equation (1),

  $\begin{array}{c}1.25\left(-2\right)+z=-2\\ -2.5+z=-2\\ z=0.5\end{array}$

Now substitute the value of x as -2 in equation (3),

  $\begin{array}{c}3\left(-2\right)-1.5y=-6\\ -6-1.5y=-6\\ -1.5y=0\\ y=0\end{array}$

Therefore, the resultant solution for the system of equation is $\left(-2,0,0.5\right)$ .