Q. 165 - Intermediate Algebra

Question

To determine whether the ordered triple is a solution to the system.

$$\begin{gathered} x+3 y-z=15 \\ y=\frac{2}{3}x-2 \\ x-3y+z=-2 \\ \left(a\right) (-6,5,\frac{1}{2} ) \\ \left(b\right) (5,\frac{4}{3},-3) \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The ordered triple $\left(-6,5,\frac{1}{2}\right)$ is not a solution of the system of linear equations.

Part (b) The ordered triple $\left(5,\frac{4}{3},-3\right)$ is not a solution of the system of linear equations. 

Part (a) Step 1. Given information

We have been given system of linear equations:

$$\begin{gathered} x+3 y-z=15 \\ y=\frac{2}{3}x-2 \\ x-3y+z=-2 \\ \left(a\right) (-6,5,\frac{1}{2} ) \end{gathered}$$

Part (a) Step 2. Testing

Consider the system of linear equations:

First substitute 

$$\begin{gathered} x=-6, \\ y=5, \\ z=\frac{1}{2} \end{gathered}$$ into the first equation

$$\begin{gathered} x+3 y-z=15 \\ -6+3\left(5\right)-\frac{1}{2}=15 \\ -6+15-\frac{1}{2}=15 \\ 9-\frac{1}{2}=15 \\ \frac{18-1}{2}=15 \\ \frac{17}{2}=15 \end{gathered}$$

This is not true.

The ordered triple $\left(-6,5,\frac{1}{2}\right)$ is not a solution of the system of linear equations.

part (b) Step 1. Given information

We have been given system of linear equations:

$$\begin{gathered} x+3 y-z=15 \\ y=\frac{2}{3}x-2 \\ x-3y+z=-2 \end{gathered}$$

 And coordinate$\left(5,\frac{4}{3},-3\right)$

Part (b) Step 2. Testing

Lets substitute 

$$\begin{gathered} x=5, \\ y=\frac{4}{3}, \\ z=-3 \end{gathered}$$ into the equation one.

$$\begin{gathered} x+3 y-z =15 \\ 5+3\frac{4}{3}-(-3) \&=15 \\ 5+4+3 \&=15 \\ 12 =15 \end{gathered}$$

This is not true.

The ordered triple  is not a solution of the system of linear equations.