Chapter 7.3, Problem 9E

a.

To determine

The solution of the system of equations.

Answer to Problem 9E

No.

Explanation of Solution

Given information:

The linear equations are:

  $\{\begin{array}{l}4x+y-z=0\\ -8x-6y+z=\frac{-7}{4}\\ 3x-y=\frac{-9}{4}\end{array}$

  $(a)\left(\frac{1}{2},\frac{-3}{4},\frac{-7}{4}\right)$

Calculation:

The linear equations are:

  $\{\begin{array}{l}4x+y-z=0\\ -8x-6y+z=\frac{-7}{4}\\ 3x-y=\frac{-9}{4}\end{array}$

To find whether each ordered triple is a solution of the system of equations or not, plug the ordered triple $\left(\frac{1}{2},\frac{-3}{4},\frac{-7}{4}\right)$ into the first equation in the system of equations.

  $\begin{array}{l}4x+y-z=0\\ 4\left(\frac{1}{2}\right)+\left(\frac{-3}{4}\right)-\left(\frac{-7}{4}\right)=0\\ 2-\frac{3}{4}+\frac{7}{4}=0\\ 3\ne 0\end{array}$

Both sides are not equivalent. It mean the ordered triple are not satisfies the system of equation. Therefore, the ordered triple $\left(\frac{1}{2},\frac{-3}{4},\frac{-7}{4}\right)$ is not a solution of the system of equations.

b.

To determine

The solution of the system of equations.

Answer to Problem 9E

No.

Explanation of Solution

Given information:

The linear equations are:

  $\{\begin{array}{l}4x+y-z=0\\ -8x-6y+z=\frac{-7}{4}\\ 3x-y=\frac{-9}{4}\end{array}$

  $(b)\left(\frac{-3}{2},\frac{5}{4},\frac{-5}{4}\right)$

Calculation:

The linear equations are:

  $\{\begin{array}{l}4x+y-z=0\\ -8x-6y+z=\frac{-7}{4}\\ 3x-y=\frac{-9}{4}\end{array}$

To find whether each ordered triple is a solution of the system of equations or not, plug the ordered triple $\left(\frac{-3}{2},\frac{5}{4},\frac{-5}{4}\right)$ into the first equation in the system of equations.

  $\begin{array}{l}4x+y-z=0\\ 4\left(\frac{-3}{2}\right)+\left(\frac{5}{4}\right)-\left(\frac{-5}{4}\right)=0\\ \frac{-12}{2}+\frac{5}{4}+\frac{5}{4}=0\\ \frac{-24}{4}+\frac{10}{4}=0\\ \frac{-14}{4}\ne 0\\ \frac{-7}{2}\ne 0\end{array}$

Both sides are not equivalent. It mean the ordered triple are not satisfies the system of equation. Therefore, the ordered triple $\left(\frac{-3}{2},\frac{5}{4},\frac{-5}{4}\right)$ is not a solution of the system of equations.

c.

To determine

The solution of the system of equations.

Answer to Problem 9E

Yes.

Explanation of Solution

Given information:

The linear equations are:

  $\{\begin{array}{l}4x+y-z=0\\ -8x-6y+z=\frac{-7}{4}\\ 3x-y=\frac{-9}{4}\end{array}$

  $(c)(\frac{-1}{2},\frac{3}{4},\frac{-5}{4})$

Calculation:

The linear equations are:

  $\{\begin{array}{l}4x+y-z=0\\ -8x-6y+z=\frac{-7}{4}\\ 3x-y=\frac{-9}{4}\end{array}$

To find whether each ordered triple is a solution of the system of equations or not, plug the ordered triple $(\frac{-1}{2},\frac{3}{4},\frac{-5}{4})$ into the first equation in the system of equations.

  $4\left(\frac{-1}{2}\right)+\left(\frac{3}{4}\right)-\left(\frac{-5}{4}\right)=0\Rightarrow \left(-\frac{8}{4}\right)+\left(\frac{3}{4}\right)+\left(\frac{5}{4}\right)=0$

Substitute all the three possible values of $x,y,$ and $z$ into the second and third system of equations to check for equality.

  $\begin{array}{l}8\left(\frac{-1}{2}\right)-6\left(\frac{3}{4}\right)+\left(\frac{-5}{4}\right)=\frac{-7}{4}\Rightarrow \left(\frac{16}{4}\right)-\left(\frac{18}{4}\right)-\left(\frac{5}{4}\right)=\frac{-7}{4}\\ 3\left(\frac{-1}{2}\right)-\left(\frac{3}{4}\right)=\frac{-9}{4}\Rightarrow -\frac{6}{4}-\frac{3}{4}=\frac{-9}{4}\end{array}$

The ordered triple fits all the three equations so, all the three system of equations are equivalent.

Thus, it is a solution to the system of equations.

d.

To determine

The solution of the system of equations.

Answer to Problem 9E

No.

Explanation of Solution

Given information:

The linear equations are:

  $\{\begin{array}{l}4x+y-z=0\\ -8x-6y+z=\frac{-7}{4}\\ 3x-y=\frac{-9}{4}\end{array}$

  $(d)(1,-2,2)$

Calculation:

The linear equations are:

  $\{\begin{array}{l}4x+y-z=0\\ -8x-6y+z=\frac{-7}{4}\\ 3x-y=\frac{-9}{4}\end{array}$

To find whether each ordered triple is a solution of the system of equations or not, plug the ordered triple $\left(\frac{-1}{2},\frac{1}{6},-\frac{3}{4}\right)$ into the first equation in the system of equations.

  $\begin{array}{l}4x+y-z=0\\ 4\left(\frac{-1}{2}\right)+\left(\frac{1}{6}\right)-\left(-\frac{3}{4}\right)=0\\ \frac{-24}{12}+\frac{2}{12}+\frac{9}{12}\ne 0\\ \frac{-13}{12}\ne 0\end{array}$

Both sides are not equivalent. It mean the ordered triple are not satisfies the system of equation. Therefore, the ordered triple $\left(\frac{-1}{2},\frac{1}{6},-\frac{3}{4}\right)$ is not a solution of the system of equations.