Chapter 7, Problem 102RE

To determine

To find a linear system that satisfies the given triplet as solution.

Answer to Problem 102RE

One possible linear system of equation that satisfies $\left(\frac{-1}{2},-2,\frac{-3}{4}\right)$ as solution is:

  $\begin{array}{l}2x+y+4z=-6\hfill \\ 2x-y+4z=-2\hfill \\ 2x-y-4z=4\hfill \end{array}$

Explanation of Solution

Given:

The ordered triplet is:

  $\left(-\frac{1}{2},-2,-\frac{3}{4}\right)$

The general form a linear equation with three variables is $ax+by+cz=d$

In the above general form, arbitrarily select a, b and c and find d. There can be many choices.

Let us first put, $a=2,b=-1,c=4$ now we get $2x+y+4z=d$

Put the values $\left(\frac{-1}{2},-2,\frac{-3}{4}\right)$ for $\left(x,y,z\right)$ which gives value $das-6$

Now put, $a=2,b=-1,c=4$ we get $2x-y+4z=d$

Substituting the values $\left(\frac{-1}{2},-2,\frac{-3}{4}\right)$ for $\left(x,y,z\right)$ , which gives $d=-2$

Now put, $a=2,b=-1,c=-4$ we get $2x-y+4z=d$

Substituting the values $\left(\frac{-1}{2},-2,\frac{-3}{4}\right)$ for $\left(x,y,z\right)$ , which gives $d=4$

So one possible linear system of equation that satisfies $\left(\frac{-1}{2},-2,\frac{-3}{4}\right)$ as solution is:

  $\begin{array}{l}2x+y+4z=-6\hfill \\ 2x-y+4z=-2\hfill \\ 2x-y-4z=4\hfill \end{array}$